Free Math Worksheets

- Picture the temperatures on a thermometer. - Which temperature is farthest below zero? - Where does zero belong relative to the negative temperatures?

1. Among the negative temperatures, the value farther left on a number line is colder: \(-12<-7<-2\). 2. Zero is greater than every negative number, and \(3\) is greater than zero. 3. Therefore, \(-12<-7<-2<0<3\).

\(-12\,^\circ\text{F}<-7\,^\circ\text{F}<-2\,^\circ\text{F}<0\,^\circ\text{F}<3\,^\circ\text{F}\)

- Locate zero and determine which direction contains lesser numbers. - Compare how far each number is from zero. - The number farther to the left is less.

1. On a number line, the number farther to the left is less. 2. Because \(-450\) lies farther left than \(-405\), \(-450<-405\).

\(-450\) is less because it lies farther to the left on the number line.

- Convert each fraction to a decimal. - Compare the decimal values digit by digit. - A rounded decimal can be used as long as it clearly separates the values.

1. Write each fraction as a decimal: \(\frac{2}{3}\approx0.667\), \(\frac{3}{5}=0.6\), \(\frac{4}{7}\approx0.571\), and \(\frac{5}{8}=0.625\). 2. Since \(0.667\) is the greatest decimal, \(\frac{2}{3}\) is the greatest fraction.

a) \(\frac{2}{3}\)

- Rewrite both boundary fractions with denominator \(12\). - Identify the integer numerators between the two boundary numerators. - Simplify every resulting fraction.

1. Rewrite the bounds with denominator \(12\): \(\frac{1}{2}=\frac{6}{12}\) and \(\frac{5}{6}=\frac{10}{12}\). 2. The numerators strictly between \(6\) and \(10\) are \(7\), \(8\), and \(9\). 3. The fractions are \(\frac{7}{12}\), \(\frac{8}{12}\), and \(\frac{9}{12}\). 4. Simplify: \(\frac{7}{12}\), \(\frac{2}{3}\), and \(\frac{3}{4}\).

\(\frac{7}{12},\frac{2}{3},\frac{3}{4}\)

- Rewrite the fractions with a common denominator. - For negative numbers, the value farther left on the number line is smaller. - Compare the equivalent numerators carefully.

1. Use a common denominator of \(16\): \(-\frac{3}{4} = -\frac{12}{16}\), \(-\frac{5}{8} = -\frac{10}{16}\), and \(-\frac{11}{16}\) is unchanged. 2. Compare the numerators: \(-12 < -11 < -10\). 3. Therefore, \(-\frac{3}{4} < -\frac{11}{16} < -\frac{5}{8}\).

\(-\frac{3}{4} < -\frac{11}{16} < -\frac{5}{8}\)

- Locate each number between two consecutive integers. - Convert a fraction to a decimal when useful. - Compare the distances to the two neighboring integers.

1. For a), \(12\frac{4}{9}\) lies between \(12\) and \(13\). Since \(\frac{4}{9}<\frac{1}{2}\), it is closer to \(12\). 2. For b), \(-8.7\) is \(0.3\) from \(-9\) and \(0.7\) from \(-8\), so the nearest integer is \(-9\). 3. For c), \(-\frac{21}{5}=-4.2\), which is \(0.2\) from \(-4\), so the nearest integer is \(-4\). 4. For d), \(-\frac{15}{4}=-3.75\), which is \(0.25\) from \(-4\), so the nearest integer is \(-4\).

a) \(12\) b) \(-9\) c) \(-4\) d) \(-4\)

- Convert each pair to the same form. - Of two negative numbers, the one closer to zero is greater. - A repeating decimal continues beyond the shown digits.

1. For a), \(-\frac{3}{5} = -0.6\). Since \(-0.6 < -0.5\), the correct symbol is \(<\). 2. For b), \(1\frac{1}{8} = 1 + 0.125 = 1.125\), so the values are equal. 3. For c), \(-\frac{1}{3} = -0.3333\ldots\). Since \(-0.3300\ldots > -0.3333\ldots\), \(-0.33 > -\frac{1}{3}\).

a) \(-\frac{3}{5} < -0.5\) b) \(1\frac{1}{8} = 1.125\) c) \(-0.33 > -\frac{1}{3}\)

- Convert the mixed number to a decimal. - For two negative values, the one closer to zero is greater. - Think about which point lies farther right on the number line.

1. Convert the mixed number: \(-4\frac{2}{3} = -4.6666\ldots\). 2. Compare the absolute values: \(4.6 < 4.6666\ldots\). 3. Of two negative numbers, the one with the smaller absolute value lies farther right and is greater. 4. Therefore, \(-4.6 > -4\frac{2}{3}\).

\(-4.6\) is greater than \(-4\frac{2}{3}\).

- Add trailing zeros to align decimal places. - For negative numbers, the value farther left is smaller. - Convert fractions to decimals when useful. - A negative number with greater absolute value is smaller.

1. For a), compare \(0.708\) and \(0.710\). Since \(708 < 710\) thousandths, \(0.708 < 0.71\). 2. For b), \(-0.520\) lies to the left of \(-0.502\), so \(-0.52 < -0.502\). 3. For c), \(\frac{3}{8} = 3 \div 8 = 0.375\), so the values are equal. 4. For d), \(-\frac{2}{3} = -0.6666\ldots\), which is less than \(-0.6000\ldots\). Therefore, \(-\frac{2}{3} < -0.6\).

a) \(0.708 < 0.71\) b) \(-0.52 < -0.502\) c) \(\frac{3}{8} = 0.375\) d) \(-\frac{2}{3} < -0.6\)

- Write all three values in each set in the same form, such as percents. - Divide the numerator by the denominator to write a fraction as a decimal. - Multiply a decimal by \(100\) to write it as a percent.

1. For a), \(\frac{3}{8} = 0.375 = 37.5\%\), \(0.37 = 37\%\), and \(38\%\) is already a percent. Since \(37\% < 37.5\% < 38\%\), the order is \(0.37 < \frac{3}{8} < 38\%\). 2. For b), \(0.6 = 60\%\), \(\frac{5}{8} = 0.625 = 62.5\%\), and \(61\%\) is already a percent. Since \(60\% < 61\% < 62.5\%\), the order is \(0.6 < 61\% < \frac{5}{8}\).

a) \(0.37 < \frac{3}{8} < 38\%\) b) \(0.6 < 61\% < \frac{5}{8}\)

- Would it help to write all the numbers in the same form? - Which form—fractions, decimals, or percents—is easiest for you to compare?

1. Write each value as a percent: \(0.3 = 30\%\) \(\frac{1}{4} = \frac{25}{100} = 25\%\) \(28\%\) is already a percent. \(\frac{7}{20} = \frac{35}{100} = 35\%\) 2. Compare the percents: \(25\% < 28\% < 30\% < 35\%\). 3. Replace the percents with the original values: \(\frac{1}{4} < 28\% < 0.3 < \frac{7}{20}\).

\(\frac{1}{4} < 28\% < 0.3 < \frac{7}{20}\)

- How many hundredths is \(0.12\)? - Write each fraction with denominator \(100\), or divide to find its decimal value. - Are any two values equal?

1. Convert each value to a percent: \(0.12 = 12\%\) \(\frac{1}{8} = 0.125 = 12.5\%\) \(13\%\) is already a percent. \(\frac{3}{25} = \frac{12}{100} = 12\%\) 2. Compare the percents: \(12\% = 12\% < 12.5\% < 13\%\). 3. The original values are ordered as \(0.12 = \frac{3}{25} < \frac{1}{8} < 13\%\).

Percents: \(12\%\); \(12.5\%\); \(13\%\); \(12\%\) Order: \(0.12 = \frac{3}{25} < \frac{1}{8} < 13\%\)

- Divide the numerator by the denominator to convert a fraction to a decimal. - Multiply a decimal by \(100\) to convert it to a percent. - Use the hundredths digit of the percent to round to the nearest tenth.

1. For a), \(\frac{1}{22} = 0.04545\ldots = 4.545\ldots\% \approx 4.5\%\). This is less than \(5\%\). 2. For b), \(0.049 = 4.9\%\). This is less than \(5\%\). 3. For c), \(\frac{3}{55} = 0.05454\ldots = 5.454\ldots\% \approx 5.5\%\). This is greater than \(5\%\). 4. For d), \(0.051 = 5.1\%\). This is greater than \(5\%\).

a) \(\frac{1}{22} \approx 4.5\%\) b) \(0.049 = 4.9\%\) c) \(\frac{3}{55} \approx 5.5\%\) d) \(0.051 = 5.1\%\) The values greater than \(5\%\) are c) and d).

- Convert both sides of each comparison to the same form. - Translate phrases such as “one out of every twenty” into fractions. - Use decimal or percent form to compare.

1. For part a, \(\frac{1}{20} = 0.05 = 5\%\), so \(5\% > 4\%\). 2. For part b, \(\frac{15}{1000} = 0.015 = 1.5\%\), so the quantities are equal. 3. For part c, \(\frac{3}{4} = 0.75 = 75\%\), so \(75\% > 70\%\). 4. For part d, \(\frac{1}{40} = 0.025\), so the quantities are equal.

a) \(>\) b) \(=\) c) \(>\) d) \(=\)

- Can you write every value in hundredths? - How does the decimal point move when you write a decimal as a percent? - Put all values in the same form before comparing them.

1. Convert each value to a percent: \(0.5 = 50\%\), \(\frac{1}{4} = 25\%\), \(0.2 = 20\%\), and \(\frac{3}{10} = 30\%\). 2. Compare the percent values: \(15\% < 20\% < 25\% < 30\% < 50\%\). 3. The original values are ordered as \(15\% < 0.2 < \frac{1}{4} < \frac{3}{10} < 0.5\).

\(15\% < 0.2 < \frac{1}{4} < \frac{3}{10} < 0.5\)

- How can you simplify or scale each fraction to get denominator \(100\)? - What does “percent” mean? - Once the denominator is \(100\), the numerator gives the percent.

1. Convert each fraction: \(\frac{3}{4} = \frac{75}{100} = 75\%\), \(\frac{13}{50} = \frac{26}{100} = 26\%\), \(\frac{7}{25} = \frac{28}{100} = 28\%\), and \(\frac{120}{1000} = \frac{12}{100} = 12\%\). 2. Compare the percents: \(12\% < 26\% < 28\% < 75\%\).

a) \(75\%\) b) \(26\%\) c) \(28\%\) d) \(12\%\) Order: \(12\% < 26\% < 28\% < 75\%\)

- What percent is \(0.4\)? - What can you multiply \(25\) by to get \(100\)? - Once both values are percents, compare them.

1. Convert \(0.4\) to a percent: \(0.4 = \frac{40}{100} = 40\%\). 2. Write \(\frac{9}{25}\) with denominator \(100\): \(\frac{9}{25} = \frac{36}{100} = 36\%\). 3. Since \(40\% > 36\%\), \(0.4\) is greater than \(\frac{9}{25}\).

\(0.4\) is greater because \(40\% > 36\%\).

- Convert the fractions to decimals. - Add trailing zeros mentally so each decimal has the same number of places. - Compare tenths, then hundredths, then thousandths.

1. Convert the fractions to decimals: \(\frac{2}{5}=0.4\) and \(\frac{11}{25}=\frac{44}{100}=0.44\). 2. Compare the decimals: \(0.4<0.405<0.44<0.45\). 3. Write the original values in that order: \(\frac{2}{5}<0.405<\frac{11}{25}<0.45\).

\(\frac{2}{5}<0.405<\frac{11}{25}<0.45\)

- Convert fractions to decimals. - On a number line, the value farther right is greater. - For negative numbers, greater absolute value means a smaller number.

1. For a), \(-\frac{3}{5} = -0.6\). Since \(-0.6 < -0.58\), the correct symbol is \(<\). 2. For b), \(-2.07\) is closer to zero than \(-2.7\), so \(-2.07 > -2.7\). 3. For c), \(-\frac{1}{8} = -0.125\), so the values are equal. 4. For d), \(-\frac{1}{3} = -0.3333\ldots\). Since \(-0.33\) is closer to zero, \(-0.33 > -\frac{1}{3}\).

a) \(-\frac{3}{5} < -0.58\) b) \(-2.07 > -2.7\) c) \(-\frac{1}{8} = -0.125\) d) \(-0.33 > -\frac{1}{3}\)

- Add trailing zeros to align decimal places. - Think about the positions on a number line. - A negative number with greater absolute value is smaller.

1. Write the values with the same number of decimal places: \(-5.400\), \(-5.040\), \(-5.440\), \(-5.404\), and \(-5.000\). 2. For negative numbers, the value with the greater absolute value is smaller. 3. Therefore, \(-5.44 < -5.404 < -5.4 < -5.04 < -5\).

\(-5.44 < -5.404 < -5.4 < -5.04 < -5\)

- Choose one form—fraction, decimal, or percent—and rewrite all four values in that form. - To write a fraction as a decimal, divide the numerator by the denominator. - To write a decimal as a percent, multiply by \(100\) and include the percent sign.

1. Convert the values to decimals or percents: \(0.35 = 35\%\) \(\frac{3}{8} = 3 \div 8 = 0.375 = 37.5\%\) \(36\% = 0.36\) \(\frac{1}{3} = 1 \div 3 = 0.\overline{3} \approx 33.3\%\) 2. Compare the equivalent percents: \(33.\overline{3}\% < 35\% < 36\% < 37.5\%\). 3. Therefore, the original values are ordered as \(\frac{1}{3} < 0.35 < 36\% < \frac{3}{8}\).

\(\frac{1}{3} < 0.35 < 36\% < \frac{3}{8}\)

- “To the left” means “less than.” - Select every number less than \(-20\). - Order only the selected numbers.

1. Numbers to the left of \(-20\) are less than \(-20\). The qualifying numbers are \(-22,-35,-21\). 2. From least to greatest, they are \(-35<-22<-21\).

\(-35<-22<-21\)

- Picture the numbers on a number line. - Which number lies farthest to the left? - For negative numbers, greater distance below zero means a lesser value.

1. For negative integers, the number with the greater absolute value is less. 2. Comparing their positions on a number line gives \(-111<-110<-101<-100<-11<-10<-1\).

\(-111<-110<-101<-100<-11<-10<-1\)

- Picture each pair on a number line. - For negative numbers, how does distance from zero affect order? - How does any negative number compare with zero?

1. \(-45>-54\) because \(-45\) lies to the right of \(-54\). 2. \(-12<0\) because every negative number is less than zero. 3. \(-100<-99\) because \(-100\) lies farther left. 4. \(7>-70\) because every positive number is greater than every negative number. 5. \(-1=-1\) because the values are identical. 6. \(-18>-19\) because \(-18\) lies closer to zero.

a) \(-45>-54\) b) \(-12<0\) c) \(-100<-99\) d) \(7>-70\) e) \(-1=-1\) f) \(-18>-19\)

- “Strictly between” does not include the endpoints. - Move through the integers in order on a number line. - Distinguish \(<\) from \(\leq\).

1. Part a) requires \(-5<x<1\), so the integers are \(-4,-3,-2,-1,0\). 2. Part b) requires \(-12<x<-7\). The numbers from the list that satisfy both inequalities are \(-11,-10,-9,-8\).

a) \(-4,-3,-2,-1,0\) b) \(-11,-10,-9,-8\)

- Picture the years on a timeline. - BCE years occur before CE years. - Use the signed-integer model given in the problem to compare the years.

1. Represent the years as \(-384,10,1048,1643,1918\). 2. Order the integers: \(-384<10<1048<1643<1918\). 3. Match the ordered years to the names: Aristotle, Hero of Alexandria, Omar Khayyam, Isaac Newton, Katherine Johnson.

Aristotle, Hero of Alexandria, Omar Khayyam, Isaac Newton, Katherine Johnson

- Picture the integers on a number line. - Negative numbers are less than zero and positive numbers. - Among negative numbers, the value farther from zero is less.

1. Order the negative numbers: \(-25<-18<-3\). 2. Zero is greater than all negative numbers and less than all positive numbers. 3. Order the positive numbers: \(4<7<12\). 4. Combining the groups gives \(-25<-18<-3<0<4<7<12\).

\(-25<-18<-3<0<4<7<12\)

- Compare place values and absolute values carefully. - Among negative integers, a greater absolute value means a lesser number. - Place zero between the negative and positive values.

1. Order the negative integers: \(-210<-120<-105\). 2. Place zero between the negative and positive integers. 3. Order the positive integers: \(105<201\). 4. The completed chain is \(-210<-120<-105<0<105<201\).

\(-210<-120<-105<0<105<201\)

- First write the integer represented by each description. - Which value lies farthest left on a number line? - Among negative integers, a greater absolute value means a lesser number.

1. The integer before \(-10\) is \(-11\), the integer after \(-10\) is \(-9\), and the integer before \(-12\) is \(-13\). 2. Order the resulting values: \(-13<-11<-10<-9<0\).

\(-13<-11<-10<-9<0\)

- Negative numbers lie to the left of zero, and positive numbers lie to the right. - Among negative numbers, the one farther from zero is less. - The pointed side of \(<\) faces the lesser value.

1. Both numbers in part a) are negative, and \(-25\) lies farther left than \(-17\). Therefore, \(-25<-17<0\). 2. In part b), the negative number lies below zero and the positive number lies above zero. Therefore, \(-4<0<14\). 3. In part c), \(-1\) is below zero and \(1\) is above zero. Therefore, \(-1<0<1\).

a) \(-25<-17<0\) b) \(-4<0<14\) c) \(-1<0<1\)

- The integer immediately before a number is one unit to its left. - The integer immediately after a number is one unit to its right. - Compare each pair by imagining its position on a number line.

1. One unit before and after \(-1\) are \(-2\) and \(0\). One unit before and after \(-50\) are \(-51\) and \(-49\). One unit before and after \(-199\) are \(-200\) and \(-198\). 2. Comparing positions on a number line gives \(-45>-54\), \(-12<2\), \(0>-7\), and \(-101<-100\).

a) For \(-1\): before \(-2\), after \(0\); for \(-50\): before \(-51\), after \(-49\); for \(-199\): before \(-200\), after \(-198\) b) \(-45>-54\); \(-12<2\); \(0>-7\); \(-101<-100\)

- Convert each fraction to a decimal. - Remember that a repeating decimal continues beyond the digits shown. - Add trailing zeros when comparing decimal place values.

1. \(\frac{3}{4}=0.75\), so \(\frac{3}{4}>0.7\). 2. \(\frac{1}{8}=0.125\), so \(0.12<\frac{1}{8}\). 3. \(2\frac{1}{2}=2.5=2.50\). 4. \(\frac{1}{3}=0.333\ldots\), so \(0.33<\frac{1}{3}\). 5. \(\frac{9}{20}=\frac{45}{100}=0.45\), so the values are equal.

a) \(\frac{3}{4}>0.7\) b) \(0.12<\frac{1}{8}\) c) \(2\frac{1}{2}=2.50\) d) \(0.33<\frac{1}{3}\) e) \(\frac{9}{20}=0.45\)

- Convert the mixed numbers to decimals. - Add trailing zeros to align decimal places. - For negative numbers, greater absolute value means a smaller value. - The number closest to zero is greatest.

1. Convert the mixed numbers: \(-4\frac{1}{4} = -4.25\) and \(-4\frac{3}{10} = -4.3\). 2. Compare \(-4.20\), \(-4.25\), \(-4.08\), \(-4.30\), and \(-4.19\). 3. The least number has the greatest absolute value, so it is \(-4.3 = -4\frac{3}{10}\). 4. The greatest number is closest to zero, so it is \(-4.08\).

a) \(-4\frac{3}{10}\) b) \(-4.08\)

- Convert every value to either a decimal or a percent. - Compare the values after they are written in the same form. - Check whether any two representations have the same value.

1. Write all values as decimals: \(\frac{3}{8}=0.375\), \(35\%=0.35\), \(0.4=0.4\), and \(\frac{2}{5}=0.4\). 2. Compare the decimals: \(0.35<0.375<0.4=0.4\). 3. Therefore, \(35\%<\frac{3}{8}<0.4=\frac{2}{5}\).

\(35\%<\frac{3}{8}<0.4=\frac{2}{5}\)

- Write both quotients as fractions. - Simplify each fraction before comparing. - Compare the whole-number parts first.

1. Write \(A\) as a fraction: \(A=\frac{250}{12}=\frac{125}{6}=20\frac{5}{6}\). 2. Evaluate \(B\): \(B=\frac{315}{15}=21\). 3. Since \(21>20\frac{5}{6}\), \(B>A\).

\(B\) is greater because \(21>20\frac{5}{6}\).

- Write all four numbers in the same form. - Use a common denominator or convert all values to decimals. - Compare the equivalent values before writing the original numbers in order.

1. Convert the mixed numbers to improper fractions: \(3\frac{1}{4}=\frac{13}{4}\) and \(3\frac{1}{2}=\frac{7}{2}\). 2. Use the common denominator \(60\): \(\frac{13}{4}=\frac{195}{60}\), \(\frac{10}{3}=\frac{200}{60}\), \(\frac{17}{5}=\frac{204}{60}\), and \(\frac{7}{2}=\frac{210}{60}\). 3. Since \(195<200<204<210\), the order is \(3\frac{1}{4}<\frac{10}{3}<\frac{17}{5}<3\frac{1}{2}\).

\(3\frac{1}{4}<\frac{10}{3}<\frac{17}{5}<3\frac{1}{2}\)

- Convert each mixed number to an improper fraction, or convert the improper fraction to a mixed number. - Simplify or rewrite fractions so the pair has a common denominator. - Compare only after both values are in the same form.

1. For a), \(3\frac{4}{7}=\frac{3\times7+4}{7}=\frac{25}{7}\). Since \(\frac{25}{7}>\frac{24}{7}\), the symbol is \(>\). 2. For b), \(\frac{45}{12}=\frac{15}{4}\), and \(3\frac{3}{4}=\frac{15}{4}\). The symbol is \(=\). 3. For c), \(6\frac{2}{5}=\frac{32}{5}=\frac{64}{10}\). Since \(\frac{64}{10}>\frac{62}{10}\), the symbol is \(>\).

a) \(>\) b) \(=\) c) \(>\)

- Find the next whole number greater than each fraction. - Subtract the fractional part from \(1\) to find each distance. - The smaller distance identifies the closer number. - Use a common denominator to compare the distances.

1. Convert \(\frac{25}{4}\): \(25\div4=6\) remainder \(1\), so \(\frac{25}{4}=6\frac{1}{4}\). It needs \(1-\frac{1}{4}=\frac{3}{4}\) to reach \(7\). 2. Convert \(\frac{31}{5}\): \(31\div5=6\) remainder \(1\), so \(\frac{31}{5}=6\frac{1}{5}\). It needs \(1-\frac{1}{5}=\frac{4}{5}\) to reach \(7\). 3. Use a common denominator: \(\frac{3}{4}=\frac{15}{20}\) and \(\frac{4}{5}=\frac{16}{20}\). Since \(\frac{15}{20}<\frac{16}{20}\), \(\frac{25}{4}\) is closer to \(7\).

\(\frac{25}{4}=6\frac{1}{4}\) is closer to \(7\) because its distance, \(\frac{3}{4}\), is less than \(\frac{4}{5}\), the distance from \(\frac{31}{5}=6\frac{1}{5}\) to \(7\).

- Compare how far fractions near \(1\) are from \(1\). - Try rewriting a pair with equal numerators. - For improper fractions, compare their whole-number and fractional parts.

1. For a), \(\frac{7}{8}=1-\frac{1}{8}\) and \(\frac{8}{9}=1-\frac{1}{9}\). Since \(\frac{1}{9}<\frac{1}{8}\), \(\frac{8}{9}\) is closer to \(1\), so \(\frac{8}{9}>\frac{7}{8}\). 2. For b), rewrite with numerator \(20\): \(\frac{4}{9}=\frac{20}{45}\) and \(\frac{5}{11}=\frac{20}{44}\). With equal numerators, the smaller denominator gives the greater fraction, so \(\frac{5}{11}>\frac{4}{9}\). 3. For c), \(\frac{12}{5}=2\frac{2}{5}\) and \(\frac{15}{7}=2\frac{1}{7}\). Since \(\frac{2}{5}>\frac{1}{7}\), \(\frac{12}{5}>\frac{15}{7}\).

a) \(\frac{8}{9}>\frac{7}{8}\) b) \(\frac{5}{11}>\frac{4}{9}\) c) \(\frac{12}{5}>\frac{15}{7}\)

- With equal numerators, compare the denominators. - Convert the mixed number to an improper fraction. - For fractions close to \(1\), compare the missing unit fractions.

1. For a), the fractions have equal numerators. The fraction with the larger denominator is smaller, so \(\frac{7}{15}<\frac{7}{13}\). 2. For b), \(2\frac{3}{5}=\frac{2\times5+3}{5}=\frac{13}{5}\), so the values are equal. 3. For c), \(\frac{8}{9}=1-\frac{1}{9}\) and \(\frac{9}{10}=1-\frac{1}{10}\). Since \(\frac{1}{10}<\frac{1}{9}\), \(\frac{9}{10}\) is closer to \(1\), so \(\frac{8}{9}<\frac{9}{10}\).

a) \(\frac{7}{15}<\frac{7}{13}\) b) \(2\frac{3}{5}=\frac{13}{5}\) c) \(\frac{8}{9}<\frac{9}{10}\)

- Represent the unknown fraction as \(\frac{x}{8}\). - Multiply the two boundary values by \(8\). - Identify the natural-number numerator in the resulting interval.

1. Let the fraction be \(\frac{x}{8}\). Then \(\frac{2}{5}<\frac{x}{8}<\frac{3}{5}\). 2. Multiply every part by \(8\): \(3.2<x<4.8\). 3. The only natural-number value in this interval is \(x=4\). 4. Therefore, the fraction is \(\frac{4}{8}=\frac{1}{2}\).

\(\frac{4}{8}\), which simplifies to \(\frac{1}{2}\)

- Convert an improper fraction to a mixed number when that makes the whole-number parts easy to compare. - For fractions near \(1\), compare the amounts missing from \(1\). - Write tenths as decimals when comparing with a decimal.

1. For a), \(\frac{11}{4}=2\frac{3}{4}\). Since \(\frac{3}{5}<\frac{3}{4}\), \(2\frac{3}{5}<\frac{11}{4}\). 2. For b), \(\frac{9}{10}=1-\frac{1}{10}\) and \(\frac{10}{11}=1-\frac{1}{11}\). Since \(\frac{1}{11}<\frac{1}{10}\), \(\frac{10}{11}\) is closer to \(1\), so \(\frac{9}{10}<\frac{10}{11}\). 3. For c), \(\frac{4}{10}=0.40\). Since \(0.45>0.40\), \(0.45>\frac{4}{10}\).

a) \(2\frac{3}{5}<\frac{11}{4}\) b) \(\frac{9}{10}<\frac{10}{11}\) c) \(0.45>\frac{4}{10}\)

- Evaluate the absolute values first. - Convert the fraction to a decimal or the decimals to fractions. - On a number line, a value farther left is smaller.

1. Evaluate the absolute values: \(b = \frac{3}{4} = 0.75\) and \(d = 0.6\). 2. Write all values as decimals: \(a = -1.2\), \(b = 0.75\), \(c = -0.5\), and \(d = 0.6\). 3. Compare the negative values: \(-1.2 < -0.5\). 4. Compare the positive values: \(0.6 < 0.75\). 5. Therefore, \(a < c < d < b\).

\(-1.2 < -\frac{1}{2} < \lvert0.6\rvert < \left\lvert-\frac{3}{4}\right\rvert\)

- Simplify \(-\frac{8}{2}\) first. - Identify which values have no fractional part. - Place the values from left to right on a number line. - Remember that \(-4 < -2.5\).

1. Simplify the fraction: \(-\frac{8}{2} = -4\). 2. The integers in the list are \(4\), \(0\), \(-4\), and \(-1\). Excluding the positive whole number \(4\) leaves \(0\), \(-4\), and \(-1\). 3. For ordering, use the decimal values \(-2.5\), \(4\), \(0\), \(-4\), \(0.5\), \(-1\), and \(1.2\). 4. The order is \(-4 < -2.5 < -1 < 0 < 0.5 < 1.2 < 4\).

a) \(0\), \(-1\), and \(-\frac{8}{2}\) b) \(-\frac{8}{2} < -2.5 < -1 < 0 < \frac{1}{2} < 1.2 < 4\)

- For two negative numbers, the one closer to zero is greater. - Compare the absolute values. - Rewrite \(\frac{4}{9}\) and \(\frac{5}{11}\) with a common denominator.

1. The absolute values are \(\lvert a\rvert = 3\frac{4}{9}\) and \(\lvert b\rvert = 3\frac{5}{11}\). 2. Compare the fractional parts using denominator \(99\): \(\frac{4}{9} = \frac{44}{99}\) and \(\frac{5}{11} = \frac{45}{99}\). 3. Since \(\frac{44}{99} < \frac{45}{99}\), \(\lvert a\rvert < \lvert b\rvert\). 4. Of two negative numbers, the one with the smaller absolute value is closer to zero and lies farther right. Therefore, \(a\) lies farther right.

\(a = -3\frac{4}{9}\) lies farther right because \(\left|-3\frac{4}{9}\right| < \left|-3\frac{5}{11}\right|\).

- Find the number halfway between \(-2\) and \(-3\). - Use that midpoint to determine which endpoint is closer. - Convert each fraction to a decimal or compare its exact distance from each endpoint.

1. The point halfway between \(-2\) and \(-3\) is \(-2.5\). Numbers to the right of \(-2.5\) are closer to \(-2\), and numbers to the left are closer to \(-3\). 2. Convert or estimate the fractions: \(-\frac{11}{4}=-2.75\), \(-\frac{7}{3}\approx-2.33\), \(-\frac{13}{5}=-2.6\), and \(-\frac{9}{4}=-2.25\). 3. The values greater than \(-2.5\) are \(-\frac{7}{3}\) and \(-\frac{9}{4}\), so these are closer to \(-2\). 4. Their distances confirm the result: \(\left|-\frac{7}{3}-(-2)\right|=\frac{1}{3}<\frac{2}{3}\), and \(\left|-\frac{9}{4}-(-2)\right|=\frac{1}{4}<\frac{3}{4}\).

\(-\frac{7}{3}\) and \(-\frac{9}{4}\)

- Values are easier to compare when they are written in the same form. - You could write every value as a decimal or as a percent. - Pay close attention to place value in the decimals.

1. Convert each value to a decimal: \(0.45 = 0.45\) \(4.5\% = 0.045\) \(\frac{2}{5} = \frac{4}{10} = 0.4\) \(\frac{1}{20} = \frac{5}{100} = 0.05\) \(44\% = 0.44\) 2. Compare the decimals: \(0.045 < 0.05 < 0.4 < 0.44 < 0.45\). 3. Replace the decimals with the original values: \(4.5\% < \frac{1}{20} < \frac{2}{5} < 44\% < 0.45\).

\(4.5\% < \frac{1}{20} < \frac{2}{5} < 44\% < 0.45\)

- Convert all values to decimals or all to fractions. - Order negative and positive values separately first. - A value farther left on the number line is smaller. - Trailing zeros can help align decimal places.

1. Convert the fractions to decimals: \(\frac{1}{2} = 0.5\), \(-\frac{5}{8} = -0.625\), and \(-\frac{1}{10} = -0.1\). 2. Order the negative values: \(-0.625 < -0.6 < -0.1\). 3. Order the positive values: \(0.45 < 0.5\). 4. Combining the groups gives \(-\frac{5}{8} < -0.6 < -\frac{1}{10} < 0.45 < \frac{1}{2}\).

\(-\frac{5}{8} < -0.6 < -\frac{1}{10} < 0.45 < \frac{1}{2}\)

- Convert the fraction to a decimal. - The greater number lies farther right on the number line. - Absolute value is distance from zero. - For negative numbers, being greater and having greater absolute value are not the same.

1. Convert \(a\): \(-\frac{7}{4} = -1.75\). 2. Since \(-1.75 > -1.8\), \(a\) lies farther to the right. 3. The absolute values are \(\lvert a\rvert = 1.75\) and \(\lvert b\rvert = 1.8\). 4. Since \(1.8 > 1.75\), \(b\) has the greater absolute value.

1. \(a = -\frac{7}{4}\) lies farther right. 2. \(b = -1.8\) has the greater absolute value.

- Find each number’s distance from \(1\). - Convert values to the same form when useful. - Compare the two nonnegative distances in each part.

1. The distance from a number \(x\) to \(1\) is \(|x-1|\). 2. For a), \(\left|\frac{3}{7}-1\right|=\frac{4}{7}\approx0.571\), while \(|1.6-1|=0.6\). Since \(\frac{4}{7}<0.6\), \(\frac{3}{7}\) is closer to \(1\). 3. For b), \(\left|-\frac{1}{4}-1\right|=\frac{5}{4}=1.25\), while \(|2.1-1|=1.1\). Since \(1.1<1.25\), \(2.1\) is closer to \(1\).

a) \(\frac{3}{7}\) b) \(2.1\)

- Convert the percent and fraction to comparable forms. - Compute the absolute difference between each value and \(-2\). - The smallest absolute difference gives the closest number.

1. Find each distance from \(-2\): \(A:\ \left|-\frac{7}{3}-(-2)\right|=\frac{1}{3}\approx0.333\) \(B:\ |-1.65-(-2)|=0.35\) \(C:\ |-2.2-(-2)|=0.2\) \(D:\ -175\%=-1.75\), so \(|-1.75-(-2)|=0.25\). 2. The smallest distance is \(0.2\), so \(C=-2.2\) is closest to \(-2\).

\(C=-2.2\)

- Convert the fractions to decimals. - For negative numbers, the value closer to zero is greater. - Compare absolute values in the reverse order. - Use enough decimal places to distinguish close values.

1. Convert the fractions: \(-\frac{3}{7} \approx -0.4286\) and \(-\frac{4}{9} = -0.4444\ldots\). 2. Compare the absolute values: \(0.4 < 0.4286 < 0.44 < 0.4444\ldots < 0.45\). 3. For negative numbers, the value with the smaller absolute value is greater. 4. Therefore, \(-0.4 > -\frac{3}{7} > -0.44 > -\frac{4}{9} > -0.45\).

\(-0.4 > -\frac{3}{7} > -0.44 > -\frac{4}{9} > -0.45\)

- All the values lie between \(-1\) and \(-0.9\). - Convert the fractions to enough decimal places to compare them. - For negative values, the one closer to \(-1\) is smaller. - Compare absolute values in reverse order.

1. Convert the fractions to decimals: \(-\frac{9}{10} = -0.9\), \(-\frac{11}{12} \approx -0.9167\), and \(-\frac{12}{13} \approx -0.9231\). 2. Compare the values. For negative numbers, the value with the greater absolute value is smaller. 3. The order is \(-0.95 < -0.9231\ldots < -0.9166\ldots < -0.915 < -0.9\). 4. Using the original forms, \(-0.95 < -\frac{12}{13} < -\frac{11}{12} < -0.915 < -\frac{9}{10}\).

\(-0.95 < -\frac{12}{13} < -\frac{11}{12} < -0.915 < -\frac{9}{10}\)

- Add trailing zeros to make place values easier to compare. - Remember the order of negative numbers on a number line. - Convert \(\frac{3}{4}\) to a decimal before choosing values.

1. For a), examples include \(0.41\) and \(0.45\), since \(0.4 < 0.41 < 0.45 < 0.5\). 2. For b), examples include \(-1.19\) and \(-1.15\), since \(-1.2 < -1.19 < -1.15 < -1.1\). 3. For c), convert \(\frac{3}{4} = 0.75\). Examples include \(0.76\) and \(0.78\).

Answers will vary. One possible set is: a) \(0.41\) and \(0.42\) b) \(-1.18\) and \(-1.12\) c) \(0.77\) and \(0.79\)

- Convert one fifth to a decimal. - Write the interval as a double inequality. - Use a number line to compare negative values.

1. Convert the fraction: \(-\frac{1}{5} = -0.2\). 2. A number \(x\) is in the interval when \(-0.2 < x < -0.1\). 3. Since \(-0.2 < -0.15 < -0.1\), \(-0.15\) is in the interval. 4. Since \(-0.25 < -0.2\), \(-0.25\) is outside the interval. 5. Since \(-0.05 > -0.1\), \(-0.05\) is outside the interval.

Only \(-0.15\) lies in the interval because \(-0.2 < -0.15 < -0.1\).

- What mathematical quantity measures how close two numbers are? - It may help to write every number as a decimal. - Picture the numbers on a number line.

1. Write each value as a decimal: \(A = -2.25\), \(B = -1.8\), \(C = -2.1\), \(D = -2.2\), and \(E = -1.95\). 2. Find each distance from \(-2\): \(|-2.25 - (-2)| = 0.25\), \(|-1.8 - (-2)| = 0.2\), \(|-2.1 - (-2)| = 0.1\), \(|-2.2 - (-2)| = 0.2\), and \(|-1.95 - (-2)| = 0.05\). 3. The smallest distance is \(0.05\), so \(E = -1.95\) is closest to \(-2\).

The number \(E = -1.95\) is closest to \(-2\).

- On the number line, the value farther right is greater. - The midpoint is the average of the two values. - Convert each fraction to a decimal.

1. For a), \(-1.4\) lies to the right of \(-1.5\), so \(x > y\). 2. For b), calculate the midpoint: \(z = \frac{-1.4 + (-1.5)}{2} = \frac{-2.9}{2} = -1.45\). 3. For c), \(-\frac{7}{5} = -1.4\) and \(-\frac{3}{2} = -1.5\). 4. Thus, \(-\frac{7}{5} > -\frac{3}{2}\), and \(-\frac{7}{5}\) equals \(x\).

a) \(-1.4 > -1.5\) b) \(z = -1.45\) c) \(-\frac{7}{5} > -\frac{3}{2}\); \(-\frac{7}{5}\) equals \(x\).

- A number to the left on a number line is smaller. - Convert the fraction and mixed number to decimals. - For negative numbers, the value with greater absolute value is smaller.

1. Convert the fraction and mixed number: \(B=-\frac{13}{4}=-3.25\) and \(C=-3\frac{1}{10}=-3.1\). 2. Numbers to the left of \(-3.1\) are less than \(-3.1\). These are \(A=-3.15\), \(B=-3.25\), and \(D=-3.2\). Point \(C\) is exactly at \(-3.1\), so it is not included. 3. From least to greatest, \(-3.25<-3.2<-3.15\). 4. In the original forms, \(-\frac{13}{4}<-3.2<-3.15\).

\(-\frac{13}{4}<-3.2<-3.15\)

- Write both values in each pair as fractions, decimals, or percents. - For fractions such as \(\frac{5}{6}\) and \(\frac{1}{11}\), divide the numerator by the denominator. - Compare the decimal place values carefully.

1. For a), \(0.15 = 15\%\) and \(\frac{4}{25} = \frac{16}{100} = 16\%\). Since \(15\% < 16\%\), \(0.15 < \frac{4}{25}\). 2. For b), \(\frac{5}{6} = 0.8333\ldots = 83.333\ldots\%\). Since \(83.333\ldots\% > 83\%\), \(\frac{5}{6} > 83\%\). 3. For c), \(0.09 = 9\%\) and \(\frac{1}{11} = 0.0909\ldots = 9.09\ldots\%\). Since \(9\% < 9.09\ldots\%\), \(0.09 < \frac{1}{11}\).

a) \(0.15 < \frac{4}{25}\) b) \(\frac{5}{6} > 83\%\) c) \(0.09 < \frac{1}{11}\)

- Find how many hundredths each value represents. - Divide \(2\) by \(9\) to write \(\frac{2}{9}\) as a decimal. - Then order the results as they would appear on a number line.

1. Convert each value to a percent: \(\frac{2}{9} = 0.222\ldots \approx 22.2\%\) \(0.22 = 22\%\) \(22.5\%\) is already a percent. \(\frac{1}{4} = 0.25 = 25\%\) 2. Compare the percents: \(22\% < 22.2\% < 22.5\% < 25\%\). 3. Replace the percents with the original values: \(0.22 < \frac{2}{9} < 22.5\% < \frac{1}{4}\).

\(0.22 < \frac{2}{9} < 22.5\% < \frac{1}{4}\)

- Would it be easier to compare all the values as decimals or all as percents? - How can you quickly write \(\frac{3}{4}\) and \(\frac{13}{20}\) as percents? - Think about how many hundredths each value represents.

1. Write each value as a percent: \(0.7 = 70\%\) \(\frac{3}{4} = \frac{75}{100} = 75\%\) \(72\%\) is already a percent. \(\frac{13}{20} = \frac{65}{100} = 65\%\) \(0.68 = 68\%\) 2. Compare the percents: \(65\% < 68\% < 70\% < 72\% < 75\%\). 3. Replace the percents with the original values: \(\frac{13}{20} < 0.68 < 0.7 < 72\% < \frac{3}{4}\).

\(\frac{13}{20} < 0.68 < 0.7 < 72\% < \frac{3}{4}\)

- Write all three portions in the same form, such as percents. - How do you write \(0.51\) as a percent? - What can you multiply \(25\) by to get \(100\)?

1. Convert Jordan's portion: \(\frac{13}{25} = \frac{52}{100} = 52\%\). 2. Convert Leo's portion: \(0.51 = 51\%\). 3. Compare the percents: \(51\% < 52\% < 54\%\). 4. The order is Leo, Jordan, Sarah. Sarah saved the greatest portion.

Leo \((51\%)\), Jordan \((52\%)\), Sarah \((54\%)\). Sarah saved the greatest portion.

- Can you simplify a fraction before writing it with denominator \(100\)? - How can you convert a fraction whose denominator does not divide \(100\) evenly? - After converting, compare the percent values.

1. Sparkling apple drink: \(\frac{120}{200} = \frac{60}{100} = 60\%\). 2. Tropical juice: \(\frac{45}{60} = \frac{3}{4} = \frac{75}{100} = 75\%\). 3. Berry blend: \(\frac{18}{25} = \frac{72}{100} = 72\%\). 4. Lemonade: \(\frac{3}{40} = 0.075 = 7.5\%\). 5. From greatest to least, the order is tropical juice, berry blend, sparkling apple drink, lemonade.

Fruit content: tropical juice \(75\%\), berry blend \(72\%\), sparkling apple drink \(60\%\), lemonade \(7.5\%\). Ranking: 1. tropical juice, 2. berry blend, 3. sparkling apple drink, 4. lemonade.

- What does it mean for a number to lie strictly between two values? - Write both endpoints and every test value in the same form. - Check whether a value equals an endpoint or lies inside the interval.

1. Write the endpoints as percents: the lower endpoint is \(25\%\), and \(\frac{2}{5} = \frac{40}{100} = 40\%\). 2. Convert the test values: \(0.3 = 30\%\) \(\frac{7}{20} = \frac{35}{100} = 35\%\) \(42\%\) is already a percent. \(0.24 = 24\%\) \(\frac{1}{4} = \frac{25}{100} = 25\%\) 3. Only \(30\%\) and \(35\%\) are greater than \(25\%\) and less than \(40\%\). The value \(25\%\) is an endpoint, so it is not strictly between the endpoints.

\(0.3\), which is \(30\%\), and \(\frac{7}{20}\), which is \(35\%\), lie strictly between \(25\%\) and \(\frac{2}{5} = 40\%\).

- Which form makes all four values easiest to compare? - Subtract the least percent from the greatest percent.

1. Write each portion as a percent: \(\frac{3}{4} = \frac{75}{100} = 75\%\) \(0.72 = 72\%\) \(78\%\) is already a percent. \(\frac{19}{25} = \frac{76}{100} = 76\%\) 2. Compare the values: \(72\% < 75\% < 76\% < 78\%\). 3. The greatest portion is \(78\%\), and the least portion is \(0.72\), or \(72\%\). 4. The difference is \(78 - 72 = 6\) percentage points.

The greatest portion is \(78\%\). The least portion is \(0.72\). The difference is \(6\) percentage points.

- Convert all four quantities to decimals before comparing them. - Divide the numerator by the denominator for each fraction. - Align decimal places carefully when ordering the values.

1. Convert each concentration to a decimal. Sample A: \(0.28\% = 0.0028\) Sample B: \(0.0022\) Sample C: \(\frac{1}{450} = 0.002\overline{2}\) Sample D: \(\frac{5}{2000} = 0.0025\) 2. Compare the decimals: \(0.0022 < 0.002\overline{2} < 0.0025 < 0.0028\). 3. Therefore, the order is Sample B, Sample C, Sample D, Sample A. Sample A has the greatest concentration.

From lowest to highest: Sample B, Sample C, Sample D, Sample A. Sample A has the greatest concentration.

- Convert every portion to the same form. - Write the percent as a decimal. - Divide the numerator by the denominator to convert the fraction.

1. The gold portion is \(\frac{3}{800} = 0.00375\). 2. The silver portion is \(0.35\% = 0.0035\). 3. The platinum portion is already written as \(0.004\). 4. Since \(0.004 > 0.00375 > 0.0035\), platinum has the greatest portion. The jeweler’s claim is correct.

The jeweler’s claim is correct. Platinum has the greatest portion because \(0.004 > 0.00375 > 0.0035\).

- Write both values in each pair in the same form. - Calculate enough decimal places to round the percent correctly. - Percent means “per hundred.”

1. For a), \(\frac{5}{7} = 0.714285\ldots = 71.4285\ldots\% \approx 71.4\%\). Also, \(0.72 = 72.0\%\). Since \(71.4\% < 72.0\%\), \(\frac{5}{7} < 0.72\). 2. For b), \(\frac{1}{9} = 0.111\ldots = 11.111\ldots\% \approx 11.1\%\). Also, \(0.11 = 11.0\%\). Since \(11.1\% > 11.0\%\), \(\frac{1}{9} > 0.11\).

a) \(\frac{5}{7} < 0.72\), because \(71.4\% < 72.0\%\) b) \(\frac{1}{9} > 0.11\), because \(11.1\% > 11.0\%\)

- Write all the values in the same form before comparing them. - Convert the fractions and decimal to percents. - Compare the tenths digits of the percent values carefully.

1. Convert each value to a percent: \(0.62 = 62.0\%\), \(\frac{5}{8} = 0.625 = 62.5\%\), \(62.8\%\) is already a percent, and \(\frac{11}{18} = 0.6111\ldots = 61.111\ldots\% \approx 61.1\%\). 2. Compare the percents: \(61.1\% < 62.0\% < 62.5\% < 62.8\%\). 3. Replace the percents with the original values: \(\frac{11}{18} < 0.62 < \frac{5}{8} < 62.8\%\).

\(\frac{11}{18} < 0.62 < \frac{5}{8} < 62.8\%\)

- Convert every quantity to the same form, such as percent. - Write verbal portions as fractions first. - Simplify fractions before converting them.

1. Quantity A is \(\frac{3}{500} = 0.006 = 0.6\%\). 2. Quantity B is \(\frac{1}{4} = 0.25 = 25\%\). 3. Quantity C is \(\frac{12}{80} = \frac{3}{20} = 0.15 = 15\%\). 4. Quantity D is \(0.007 = 0.7\%\). 5. Therefore, \(0.6\% < 0.7\% < 15\% < 25\%\), so the order is A, D, C, B.

A \((0.6\%)\) < D \((0.7\%)\) < C \((15\%)\) < B \((25\%)\)

- Divide the numerator by the denominator to write a fraction as a decimal. - What does “percent” mean? - For a repeating decimal, compare enough digits to decide which value is greater.

1. For a), \(\frac{3}{8} = 0.375 = 37.5\%\). Since \(37.5\% > 37\%\), \(\frac{3}{8} > 37\%\). 2. For b), \(0.06 = 6\%\) and \(\frac{6}{100} = 6\%\). Therefore, \(0.06 = \frac{6}{100}\). 3. For c), \(\frac{2}{3} = 0.666\ldots = 66.666\ldots\%\). Since \(66.666\ldots\% > 66\%\), \(\frac{2}{3} > 66\%\). 4. For d), \(0.125 = 12.5\%\) and \(\frac{1}{8} = 0.125 = 12.5\%\). Therefore, \(0.125 = \frac{1}{8}\).

a) \(\frac{3}{8} > 37\%\) b) \(0.06 = \frac{6}{100}\) c) \(\frac{2}{3} > 66\%\) d) \(0.125 = \frac{1}{8}\)

- Write both proportions in the same form before comparing them. - Divide the numerator by the denominator, then convert the decimal to a percent.

1. Convert the fraction to a decimal: \(13\div18=0.7\overline{2}\). 2. Convert the decimal to a percent: \(0.7\overline{2}\times100\%\approx72.2\%\). 3. Since \(72.2\%>72\%\), Class 6A had the greater proportion.

Class 6A had the greater proportion because \(\frac{13}{18}\approx72.2\%\), which is greater than \(72\%\).

- Convert the fractions to decimals. - For negative numbers, the value with greater absolute value is smaller. - A number line can help organize the values.

1. Convert the fractions: \(B = -\frac{4}{5} = -0.8\) and \(D = -\frac{3}{4} = -0.75\). 2. Order the negative values: \(-0.8 < -0.75 < -0.7\). 3. Order the positive values: \(0.75 < 0.8\). 4. Therefore, \(B < D < E < A < C\).

\(B < D < E < A < C\)

- Convert each fraction to a decimal with enough places to compare. - Write \(0.62\) as \(0.620\) when looking for a number with three decimal places. - Check that your number is strictly between the two endpoints.

1. Convert the fractions to decimals: \(\frac{4}{7}\approx0.571\) and \(\frac{5}{8}=0.625\). 2. Therefore, \(\frac{4}{7}<0.62<\frac{5}{8}<0.63\). 3. Since \(0.62=0.620\), any of \(0.621\), \(0.622\), \(0.623\), or \(0.624\) lies between \(0.620\) and \(0.625\).

a) \(\frac{4}{7}<0.62<\frac{5}{8}<0.63\) b) One possible answer is \(0.621\).

- Convert the mixed numbers to decimals. - Write all decimals with the same number of decimal places by adding trailing zeros. - The problem asks for descending order.

1. Convert the mixed numbers to decimals: Trail A is \(7.5\,\text{mi}\), and Trail C is \(7.6\,\text{mi}\). 2. Compare the four values: \(7.6>7.505>7.5>7.45\). 3. Therefore, the order from longest to shortest is Trail C, Trail D, Trail A, Trail B.

Trail C, Trail D, Trail A, Trail B

- First compare the corresponding positive magnitudes. - Remember that among negative numbers, the number with the greater magnitude is smaller. - Add trailing zeros when helpful so the decimal places line up.

1. For part a), \(-4.1\square>-4.15\) means the positive value \(4.1\square\) must be less than \(4.15\). The possible hundredths digits are \(0,1,2,3,4\). 2. For part b), write \(-0.4\) as \(-0.40\). For \(-0.\square7<-0.40\), the positive value \(0.\square7\) must be greater than \(0.40\). This happens for \(\square=4,5,6,7,8,9\). 3. For part c), write \(-8.2\) as \(-8.20\). The inequality requires \(8.20>8.\square1\), which is true only when \(\square=0\) or \(\square=1\).

a) \(\square\in\{0,1,2,3,4\}\) b) \(\square\in\{4,5,6,7,8,9\}\) c) \(\square\in\{0,1\}\)

- Convert all values to decimals with the same number of places. - Order them as they would appear from left to right on a number line. - The average of two numbers lies halfway between them.

1. Convert the fractions: \(-\frac{7}{4} = -1.75\) and \(-\frac{3}{2} = -1.5\). 2. Compare the values: \(-1.75 < -1.705 < -1.6 < -1.5\). 3. Therefore, \(-\frac{7}{4} < -1.705 < -1.6 < -\frac{3}{2}\). 4. The two greatest values are \(-1.6\) and \(-1.5\). Their midpoint is \(\frac{-1.6 + (-1.5)}{2} = -1.55\), which lies between them.

a) \(-\frac{7}{4} < -1.705 < -1.6 < -\frac{3}{2}\) b) Answers will vary. One possible answer is \(-1.55\).

- Think about the order of negative numbers on a number line. - What happens to inequality symbols when every number is multiplied by \(-1\)? - Check the result by substituting the digit into all three numbers.

1. Reverse all three signs by multiplying each expression by \(-1\): \(k.6>3.4>k.2\). 2. The inequality \(k.6>3.4\) requires \(k\ge 3\). When \(k=3\), \(3.6>3.4\). 3. The inequality \(3.4>k.2\) requires \(k\le 3\). When \(k=3\), \(3.4>3.2\). 4. The only digit satisfying both conditions is \(k=3\). Substitution gives \(-3.6<-3.4<-3.2\), which is true.

The only possible digit is \(k=3\).

- Find the opposite of each negative number. - Convert the fraction to a decimal. - Order the negative and positive values separately. - Reversing every sign reverses the order.

1. Convert \(b\): \(b = -\frac{5}{4} = -1.25\). 2. The opposites are \(-a = 1.2\), \(-b = 1.25\), and \(-c = 1.15\). 3. Order the negative values: \(-1.25 < -1.2 < -1.15\), so \(b < a < c\). 4. Order the positive values: \(1.15 < 1.2 < 1.25\), so \(-c < -a < -b\). 5. Combining them gives \(b < a < c < -c < -a < -b\).

\(b < a < c < -c < -a < -b\)

- Divide the numerator by the denominator. What pattern appears in the decimal? - Convert the decimal to a percent. - Compare the percent values digit by digit.

1. Write the fraction as a decimal: \(1 \div 6 = 0.1666\ldots = 0.1\overline{6}\). 2. Convert the decimal to a percent: \(0.1\overline{6} = 16.\overline{6}\%\). 3. Since \(16.\overline{6}\%\) is not equal to \(16\%\), Tim's statement is incorrect. 4. Because \(16.\overline{6} > 16\), it follows that \(\frac{1}{6} > 16\%\).

Tim's statement is incorrect. \(\frac{1}{6} = 16.\overline{6}\% \approx 16.67\%\), which is greater than \(16\%\). Therefore, \(\frac{1}{6} > 16\%\).

- Choose either fractions or decimals as a common form for each comparison. - When comparing fractions, use a common denominator. - Calculate enough decimal places to make the comparison clear.

1. For a), \(\frac{2}{5} = 0.4\). Since \(0.4 > 0.38\), \(\frac{2}{5} > 0.38\). 2. For b), \(\frac{3}{25} = \frac{12}{100} = 0.12\). Therefore, \(0.12 = \frac{3}{25}\). 3. For c), \(15\% = \frac{15}{100} = \frac{9}{60}\), while \(\frac{1}{6} = \frac{10}{60}\). Therefore, \(15\% < \frac{1}{6}\). 4. For d), \(\frac{7}{8} = 0.875\). Since \(0.875 > 0.87\), \(\frac{7}{8} > 0.87\).

a) \(>\) b) \(=\) c) \(<\) d) \(>\)

- Rewrite all four values in one common form, such as decimals. - Identify the greatest and least values before subtracting. - A difference is found by subtraction.

1. Rewrite each value as a decimal: \(A = \frac{4}{5} = 0.80\) \(B = 0.75\) \(C = 82\% = 0.82\) \(D = \frac{17}{20} = \frac{85}{100} = 0.85\) 2. Since \(0.75 < 0.80 < 0.82 < 0.85\), the order is \(B < A < C < D\). 3. The difference between the greatest and least values is \(0.85 - 0.75 = 0.10\). 4. As a fraction, \(0.10 = \frac{10}{100} = \frac{1}{10}\).

a) \(B < A < C < D\) b) Decimal: \(0.10\); fraction: \(\frac{1}{10}\)

- Colder temperatures are farther left on a number line. - List the integers strictly between \(-14\) and \(-11\). - Opposites have equal absolute values and different signs.

1. Ordering the temperatures gives \(-14<-11<-3<2\), so the cities from coldest to warmest are B, A, D, C. 2. An integer strictly between \(-14\) and \(-11\) can be \(-13\) or \(-12\). Either is a possible temperature for City E. 3. The opposite of \(-14\) is \(14\).

a) City B, City A, City D, City C b) One possible answer is \(-13\,^\circ\text{F}\) or \(-12\,^\circ\text{F}\). c) \(14\)

- Identify the number multiplying the variable in each expression. - Rewrite fractions as decimals to compare the coefficients. - Place negative numbers carefully when ordering from least to greatest.

1. The coefficients are \(A: 0.4\), \(B: \frac{1}{5}=0.2\), \(C: -2\), \(D: \frac{3}{4}=0.75\), and \(E: -1\). 2. Compare the coefficients: \(-2<-1<0.2<0.4<0.75\). 3. Match each coefficient to its expression. The order is \(C,E,B,A,D\).

Coefficients: \(A: 0.4\); \(B: 0.2\); \(C: -2\); \(D: 0.75\); \(E: -1\) Order from least to greatest: \(C,E,B,A,D\)

- Convert each fraction to a decimal. - Write all decimals with the same number of places when comparing. - Find the distance from \(0.9\) by subtraction.

1. Convert the fractions to decimals: \(\frac{7}{8}=0.875\), \(\frac{4}{5}=0.8\), and \(\frac{17}{20}=0.85\). 2. Order the values: \(0.8<0.85<0.87<0.875<0.88\). 3. Therefore, \(\frac{4}{5}<\frac{17}{20}<0.87<\frac{7}{8}<0.88\). 4. The closest value to \(0.9\) is \(0.88\), and the difference is \(0.9-0.88=0.02\).

a) \(\frac{4}{5}<\frac{17}{20}<0.87<\frac{7}{8}<0.88\) b) \(0.88\), with a difference of \(0.02\)

- Compare each measured altitude with the scheduled altitude. - Picture the deviations on a number line. - Which value is farthest to the left?

1. Check 1 is \(10{,}200-10{,}000=+200\,\text{ft}\) from the target. 2. Check 2 is \(9750-10{,}000=-250\,\text{ft}\) from the target. 3. Check 3 is \(9900-10{,}000=-100\,\text{ft}\) from the target. 4. From least to greatest, \(-250<-100<+200\).

a) Check 1: \(+200\,\text{ft}\); Check 2: \(-250\,\text{ft}\); Check 3: \(-100\,\text{ft}\) b) \(-250<-100<+200\)

- Use common denominators for fraction pairs. - Add trailing zeros to compare decimals. - Convert between fractions and decimals when useful. - For negative numbers, smaller absolute value means a greater number.

1. Use denominator \(36\): \(-\frac{5}{9} = -\frac{20}{36}\) and \(-\frac{7}{12} = -\frac{21}{36}\). Since \(-20 > -21\), \(-\frac{5}{9} > -\frac{7}{12}\). 2. Write \(-3.11\) as \(-3.110\). Since \(-3.101\) is closer to zero, \(-3.101 > -3.11\). 3. Convert \(-\frac{19}{20} = -0.95\). Therefore, \(-\frac{19}{20} < -0.9\). 4. Convert to fractions with denominator \(24\): \(-1\frac{2}{3} = -\frac{5}{3} = -\frac{40}{24}\), and \(-\frac{13}{8} = -\frac{39}{24}\). Thus, \(-1\frac{2}{3} < -\frac{13}{8}\).

1) \(-\frac{5}{9} > -\frac{7}{12}\) 2) \(-3.101 > -3.11\) 3) \(-\frac{19}{20} < -0.9\) 4) \(-1\frac{2}{3} < -\frac{13}{8}\)

- Separate the values into positive numbers, zero, and negative numbers. - Convert the fraction and mixed number to decimals. - For negative values, the one closer to zero is greater.

1. Convert the fraction and mixed number: \(-\frac{13}{4} = -3.25\) and \(3\frac{1}{5} = 3.2\). 2. Order the positive values: \(3.25 > 3.2\). 3. Zero is less than every positive value and greater than every negative value. 4. Order the negative values: \(-0.03 > -3.05 > -3.1 > -3.25\). 5. Therefore, \(3.25 > 3\frac{1}{5} > 0 > -0.03 > -3.05 > -3.1 > -\frac{13}{4}\).

\(3.25 > 3\frac{1}{5} > 0 > -0.03 > -3.05 > -3.1 > -\frac{13}{4}\)

- Convert the fractions to decimals. - Think of the values as temperatures on a thermometer. - Use the original forms in the final inequality. - Add trailing zeros if that helps align decimal places.

1. Convert the fractions: \(-\frac{1}{2} = -0.5\) and \(-\frac{3}{5} = -0.6\). 2. Compare the decimal values \(-0.5\), \(-0.45\), \(-0.6\), \(-0.55\), and \(-0.4\). 3. From least to greatest, \(-0.6 < -0.55 < -0.5 < -0.45 < -0.4\). 4. Using the original forms, \(-\frac{3}{5} < -0.55 < -\frac{1}{2} < -0.45 < -0.4\).

\(-\frac{3}{5} < -0.55 < -\frac{1}{2} < -0.45 < -0.4\)

- Convert the fractions to decimals. - Use a number line to decide which values lie farther left. - For part b), check both strict inequalities. - Find the least listed value before choosing an even smaller number.

1. Convert the fractions: \(-\frac{1}{2} = -0.5\) and \(-\frac{9}{4} = -2.25\). 2. The numbers less than \(-1\) are \(-4.2\), \(-5\), and \(-\frac{9}{4}\). 3. The values strictly between \(-3\) and \(-0.5\) are \(-0.75\) and \(-\frac{9}{4}\). The endpoint \(-\frac{1}{2} = -0.5\) is not included. 4. The least listed value is \(-5\). Any rational number less than \(-5\), such as \(-6\), answers part c).

a) \(-4.2\), \(-5\), and \(-\frac{9}{4}\) b) \(-0.75\) and \(-\frac{9}{4}\) c) Answers will vary. One example is \(-6\).

- Deeper positions are farther below zero. - Values farther left on the number line are smaller. - For part b), choose values strictly inside the interval.

1. Deeper positions correspond to smaller numbers. Two examples less than \(-12.5\) are \(-13\) and \(-15\). 2. A valid position must satisfy \(-12.5 < y < -10\). One decimal example is \(-11.5\), and one fraction example is \(-\frac{21}{2} = -10.5\). 3. Since \(-13\) lies to the left of \(-12.5\), \(-13 < -12.5\).

a) Answers will vary. One possible pair is \(-13\) and \(-15\). b) Answers will vary. One decimal is \(-11.5\), and one fraction is \(-\frac{21}{2}\). c) \(-13\) is less than \(-12.5\).

- Read the marked points in fourths. - For part b), determine how many \(\frac{1}{250}\)-length intervals fit in a span of \(2\). - Do not count the two endpoints.

1. The number line is divided into fourths. Point \(A\) is three fourths past \(10\), so \(A=10\frac{3}{4}\). Point \(B\) is two fourths past \(11\), so \(B=11\frac{2}{4}=11\frac{1}{2}\). 2. The interval from \(10\) to \(12\) has length \(2\). Dividing each whole into \(250\) equal parts creates \(2\times250=500\) equal intervals. 3. The interior tick marks are one fewer than the number of intervals, so there are \(500-1=499\) fractions with denominator \(250\) strictly between the endpoints.

a) \(A=10\frac{3}{4}\), \(B=11\frac{1}{2}\) b) \(499\) fractions

- Find a number divisible by all four denominators. - Rewrite each fraction with that common denominator. - Match each new numerator to its original fraction before ordering.

1. The least common multiple of \(18\), \(24\), \(36\), and \(12\) is \(72\). 2. Rewrite each fraction: \(\frac{13}{18}=\frac{52}{72}\), \(\frac{19}{24}=\frac{57}{72}\), \(\frac{25}{36}=\frac{50}{72}\), and \(\frac{11}{12}=\frac{66}{72}\). 3. Since \(50<52<57<66\), the order is \(\frac{25}{36}<\frac{13}{18}<\frac{19}{24}<\frac{11}{12}\).

\(\frac{25}{36}<\frac{13}{18}<\frac{19}{24}<\frac{11}{12}\)

- Rewrite the condition as an interval for \(k\) in terms of \(n\). - Test possible denominators from smallest to largest. - Look for the first interval that contains a natural number.

1. The condition is \(\frac{1}{5}<\frac{k}{n}<\frac{1}{4}\), so \(\frac{n}{5}<k<\frac{n}{4}\). 2. If \(n\le4\), then every positive fraction \(\frac{k}{n}\) is at least \(\frac{1}{4}\), so no value works. For \(n=5,6,7,8\), there is no natural number strictly between \(\frac{n}{5}\) and \(\frac{n}{4}\). 3. For \(n=9\), the inequality is \(1.8<k<2.25\), so \(k=2\). 4. The first possible fraction is \(\frac{2}{9}\).

a) \(n=9\) b) \(\frac{2}{9}\)

- First identify which fractions are below \(1\) and which are above \(1\). - For fractions just below \(1\), compare the amount missing from \(1\). - A smaller missing amount gives a fraction closer to \(1\).

1. \(B=\frac{124}{123}>1\), while \(A\) and \(C\) are both less than \(1\). Therefore, \(B\) is greatest. 2. Write the other fractions by their distances from \(1\): \(A=1-\frac{1}{124}\) and \(C=1-\frac{1}{125}\). 3. Since \(\frac{1}{124}>\frac{1}{125}\), \(A\) is farther below \(1\) than \(C\). Thus, \(A<C<B\).

\(A<C<B\), or \(\frac{123}{124}<\frac{124}{125}<\frac{124}{123}\).

- Start with the pair that has equal denominators. - Then compare the pair that has equal numerators. - For the last comparison, determine how far each fraction is above \(\frac{1}{2}\).

1. Since \(\frac{7}{15}\) and \(\frac{8}{15}\) have the same denominator, \(\frac{7}{15}<\frac{8}{15}\). 2. Since \(\frac{7}{15}\) and \(\frac{7}{13}\) have the same numerator, the fraction with the larger denominator is smaller. Thus, \(\frac{7}{15}<\frac{7}{13}\). 3. Compare the remaining pair with \(\frac{1}{2}\): \(\frac{8}{15}=\frac{1}{2}+\frac{1}{30}\) and \(\frac{7}{13}=\frac{1}{2}+\frac{1}{26}\). Since \(\frac{1}{26}>\frac{1}{30}\), \(\frac{7}{13}>\frac{8}{15}\). 4. Therefore, \(\frac{7}{15}<\frac{8}{15}<\frac{7}{13}\).

\(\frac{7}{15}<\frac{8}{15}<\frac{7}{13}\)

- Ask whether fractions have immediate neighbors on a number line. - Rewrite the endpoints with a much larger common denominator. - Consider whether the denominator can keep increasing.

1. The student is not correct. 2. Between any two different rational numbers, there are infinitely many other rational numbers. 3. For example, \(\frac{1}{100}=\frac{10}{1000}\) and \(\frac{2}{100}=\frac{20}{1000}\), so fractions such as \(\frac{11}{1000}\), \(\frac{12}{1000}\), and many others lie between them. Rewriting with still larger denominators produces more fractions, and this process can continue without end.

The student is not correct. Infinitely many fractions lie between any two different fractions, no matter how close the endpoints are.

- Rewrite both boundary fractions with denominator \(20\). - List the possible integer numerators strictly between the two boundary numerators. - Check which numerators share no factor of \(2\) or \(5\) with \(20\).

1. Rewrite the boundaries with denominator \(20\): \(\frac{2}{5}=\frac{8}{20}\) and \(\frac{4}{5}=\frac{16}{20}\). 2. The fractions in the interval are \(\frac{9}{20}\), \(\frac{10}{20}\), \(\frac{11}{20}\), \(\frac{12}{20}\), \(\frac{13}{20}\), \(\frac{14}{20}\), and \(\frac{15}{20}\). 3. A fraction with denominator \(20\) is in simplest form when its numerator has no common factor with \(20\) other than \(1\). 4. The numerators \(9\), \(11\), and \(13\) are relatively prime to \(20\). Therefore, the fractions in simplest form are \(\frac{9}{20}\), \(\frac{11}{20}\), and \(\frac{13}{20}\).

\(\frac{9}{20}\), \(\frac{11}{20}\), and \(\frac{13}{20}\)

- Follow the proposed numerator-and-denominator rule exactly. - Use a common denominator to check whether the result lies between the endpoints. - Find the arithmetic mean by adding the two endpoints and dividing by \(2\).

1. Add the numerators and denominators: \(\frac{1+1}{4+2}=\frac{2}{6}=\frac{1}{3}\). 2. Using denominator \(12\), \(\frac{1}{4}=\frac{3}{12}\), \(\frac{1}{3}=\frac{4}{12}\), and \(\frac{1}{2}=\frac{6}{12}\). Thus, \(\frac{1}{4}<\frac{1}{3}<\frac{1}{2}\). 3. The arithmetic mean is \(\left(\frac{1}{4}+\frac{1}{2}\right)\div2=\frac{3}{4}\div2=\frac{3}{8}\). 4. Since \(\frac{1}{3}=\frac{8}{24}\) and \(\frac{3}{8}=\frac{9}{24}\), the new fraction is not exactly halfway between the original fractions.

a) \(\frac{1}{3}\) b) Yes, because \(\frac{1}{4}<\frac{1}{3}<\frac{1}{2}\). c) No. The exact midpoint is \(\frac{3}{8}\).

- First list every group of three that includes zero. - For each group, identify the least, middle, and greatest values. - The number of negative values determines whether zero is last or in the middle.

1. The three-number selections that include zero are \(\{-10,-2,0\}\), \(\{-10,5,0\}\), and \(\{-2,5,0\}\). 2. Ordering each selection gives \(-10<-2<0\), \(-10<0<5\), and \(-2<0<5\).

\(-10<-2<0\) \(-10<0<5\) \(-2<0<5\)