Free math worksheets

- An interior cell represents students who belong to both its row category and its column category. - A row total is found at the far right of that row.

1. The Walk row and Late column meet at the cell containing \(3\). 2. The total at the end of the Walk row is \(12\).

a) \(3\) students. b) \(12\) students.

- Work within the Grade \(7\) row. - Subtract the known interior count from the row total.

1. The two Grade \(7\) interior counts must add to the Grade \(7\) row total of \(30\). 2. The missing count is \(30-18=12\).

\(12\).

- A two-way table places the categories of one variable along rows and the categories of the other along columns. - Check whether every interior position in each layout represents one grade category paired with one travel category. - Think about what changes, and what stays the same, when a table is transposed.

1. Each layout places all categories of one variable on one dimension and all categories of the other variable on the other dimension, so both layouts are valid. 2. In Layout A, the count belongs in the Grade 8 row and Bus column. 3. In Layout B, the same count belongs in the Bus row and Grade 8 column. 4. Switching rows and columns changes only the table's orientation; each student still contributes the same grade-and-travel pair.

a) Both Layout A and Layout B are valid. b) In Layout A, the count is in the Grade 8 row and Bus column. In Layout B, it is in the Bus row and Grade 8 column. c) Switching rows and columns changes the orientation, not which two categories are paired for each student.

- Identify which variable each displayed heading belongs to. - In a two-way table, one axis represents one categorical variable and the other axis represents the other variable. - Do not mix a grade category and a travel category on the same axis.

1. A valid two-way table puts all categories of one variable on one axis and all categories of the other variable on the other axis. 2. Maya's rows mix Walk, which is a travel-method category, with Grade \(7\), which is a grade-level category. 3. Her columns likewise mix Bus with Grade \(8\). 4. One valid organization is Grade \(7\) and Grade \(8\) as the rows and the travel-method categories as the columns. The reverse orientation is also valid.

The layout is invalid because each axis mixes categories from the two different variables. Put all grade categories on one axis and all travel-method categories on the other; for example, Grade \(7\) and Grade \(8\) as rows and the travel-method categories as columns.

- Add across each row and down each column. - Read an interior cell by combining its row and column labels. - “Exactly one” uses the two off-diagonal interior cells.

1. The physics row total is \(120+130=250\), and the not-physics row total is \(180+70=250\). 2. The mathematics column total is \(120+180=300\), and the not-mathematics column total is \(130+70=200\). 3. The grand total is \(250+250=500\). 4. The count \(120\) represents students interested in both subjects. The count \(70\) represents students interested in neither subject. 5. Exactly one subject includes \(130+180=310\) students.

a) <table><tr><th></th><th>Interested in mathematics</th><th>Not interested in mathematics</th><th>Total</th></tr><tr><th>Interested in physics</th><td>\(120\)</td><td>\(130\)</td><td>\(250\)</td></tr><tr><th>Not interested in physics</th><td>\(180\)</td><td>\(70\)</td><td>\(250\)</td></tr><tr><th>Total</th><td>\(300\)</td><td>\(200\)</td><td>\(500\)</td></tr></table> b) \(120\) students are interested in both subjects; \(70\) students are interested in neither subject. c) \(310\) students.

- Treat each hyphenated response as one paired observation. - Tally one travel row at a time, then check that all observations are counted. - Use row and column sums to verify the grand total.

1. Tally each paired response into one row and one column. 2. The row counts are Bike: \(3\) early and \(2\) late; Walk: \(3\) early and \(1\) late; Bus: \(2\) early and \(1\) late. 3. Add row and column totals to complete the table.

<table><tr><th></th><th>Early</th><th>Late</th><th>Total</th></tr><tr><th>Bike</th><td>\(3\)</td><td>\(2\)</td><td>\(5\)</td></tr><tr><th>Walk</th><td>\(3\)</td><td>\(1\)</td><td>\(4\)</td></tr><tr><th>Bus</th><td>\(2\)</td><td>\(1\)</td><td>\(3\)</td></tr><tr><th>Total</th><td>\(8\)</td><td>\(4\)</td><td>\(12\)</td></tr></table>

- Choose one variable for rows and the other for columns. - Enter the category counts before calculating totals. - Check that the row-total sum equals the column-total sum.

1. Use grade levels as rows and the three lunch choices as columns. 2. Enter the six stated counts. 3. Row totals are \(30\) and \(30\); column totals are \(24\), \(21\), and \(15\), for \(60\) students.

<table><tr><th></th><th>Cafeteria</th><th>Packed</th><th>Skipped</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(14\)</td><td>\(9\)</td><td>\(7\)</td><td>\(30\)</td></tr><tr><th>Grade \(8\)</th><td>\(10\)</td><td>\(12\)</td><td>\(8\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(24\)</td><td>\(21\)</td><td>\(15\)</td><td>\(60\)</td></tr></table>

- Each path from the root identifies one grade-and-travel combination. - Put the terminal count from each path into the matching interior cell of the table. - Add the completed interior cells to obtain the margins.

1. Follow the Grade 7 branch: \(18\) students ride the bus and \(14\) walk, for a row total of \(32\). 2. Follow the Grade 8 branch: \(10\) students ride the bus and \(18\) walk, for a row total of \(28\). 3. The bus column total is \(18+10=28\), and the walk column total is \(14+18=32\). 4. The grand total is \(32+28=60\), agreeing with the tree.

<table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(18\)</td><td>\(14\)</td><td>\(32\)</td></tr><tr><th>Grade \(8\)</th><td>\(10\)</td><td>\(18\)</td><td>\(28\)</td></tr><tr><th>Total</th><td>\(28\)</td><td>\(32\)</td><td>\(60\)</td></tr></table>

- Check each row and column independently. - Only one total is wrong, so the four interior counts should remain unchanged.

1. The row totals are correct: \(16+44=60\) and \(9+31=40\). 2. The returned-column total is correct: \(16+9=25\). 3. The kept-column total should be \(44+31=75\), not \(74\).

The incorrect entry is the kept-column total. It should be \(75\).

- Decode both letters before tallying. - Mark each code in exactly one cell. - Check that the four cell counts total the number of codes.

1. Tally codes by their first and second letters. 2. North has \(6\) hybrid and \(3\) in-person students; South has \(4\) hybrid and \(3\) in-person students. 3. The column totals are \(10\) hybrid and \(6\) in-person, for \(16\) students.

<table><tr><th></th><th>Hybrid</th><th>In person</th><th>Total</th></tr><tr><th>North</th><td>\(6\)</td><td>\(3\)</td><td>\(9\)</td></tr><tr><th>South</th><td>\(4\)</td><td>\(3\)</td><td>\(7\)</td></tr><tr><th>Total</th><td>\(10\)</td><td>\(6\)</td><td>\(16\)</td></tr></table>

- Compute the grand total in two independent ways. - A valid table cannot count a different number of observations by rows and columns.

1. The row totals sum to \(28+22=50\). 2. The column totals sum to \(31+20=51\). 3. A two-way table must have the same grand total from rows and columns, so these margins are inconsistent.

No. The row totals give \(50\), while the column totals give \(51\).

- Place the stated equal counts before subtracting from the row totals. - Complete each grade row separately. - Add the two columns and verify the grand total.

1. Place \(18\) in both stated equal cells: Grade \(7\) e-books and Grade \(8\) printed books. 2. Grade \(7\) printed books: \(34-18=16\). 3. Grade \(8\) e-books: \(26-18=8\). 4. The column totals are \(18+8=26\) for e-books and \(16+18=34\) for printed books.

<table><tr><th></th><th>E-books</th><th>Printed books</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(18\)</td><td>\(16\)</td><td>\(34\)</td></tr><tr><th>Grade \(8\)</th><td>\(8\)</td><td>\(18\)</td><td>\(26\)</td></tr><tr><th>Total</th><td>\(26\)</td><td>\(34\)</td><td>\(60\)</td></tr></table>

- Add each new response only to its matching grade-and-club cell. - Recalculate both row totals after updating the four interior cells. - Add each column and verify the new grand total.

1. Update Grade \(7\): robotics becomes \(12+3=15\), and art becomes \(8+1=9\). 2. Update Grade \(8\): robotics remains \(9\), and art becomes \(11+4=15\). 3. Each grade row totals \(24\). 4. Each club column totals \(24\), for \(48\) responses altogether.

<table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(15\)</td><td>\(9\)</td><td>\(24\)</td></tr><tr><th>Grade \(8\)</th><td>\(9\)</td><td>\(15\)</td><td>\(24\)</td></tr><tr><th>Total</th><td>\(24\)</td><td>\(24\)</td><td>\(48\)</td></tr></table>

- Add the three known interior counts before using the grand total. - Subtract that sum from \(50\) to find the fourth interior cell. - Complete the row and column margins only after all four interior counts are known.

1. The three known interior cells total \(19+8+14=41\). 2. The missing South noncomposting count is \(50-41=9\). 3. The row totals are \(19+8=27\) for North and \(14+9=23\) for South. 4. The column totals are \(19+14=33\) who compost and \(8+9=17\) who do not, and \(33+17=50\).

<table><tr><th></th><th>Composts</th><th>Does not compost</th><th>Total</th></tr><tr><th>North</th><td>\(19\)</td><td>\(8\)</td><td>\(27\)</td></tr><tr><th>South</th><td>\(14\)</td><td>\(9\)</td><td>\(23\)</td></tr><tr><th>Total</th><td>\(33\)</td><td>\(17\)</td><td>\(50\)</td></tr></table>

- Convert each complete group of five tally marks before counting any extra marks. - Place each count at the intersection of its practice group and equipment choice. - Add rows and columns to verify the grand total.

1. Convert the four tally groups to counts: \(8\), \(5\), \(6\), and \(9\). 2. The Red row total is \(8+5=13\), and the Blue row total is \(6+9=15\). 3. The cones column total is \(8+6=14\), and the hurdles column total is \(5+9=14\). 4. The grand total is \(13+15=28\).

<table><tr><th></th><th>Cones</th><th>Hurdles</th><th>Total</th></tr><tr><th>Red</th><td>\(8\)</td><td>\(5\)</td><td>\(13\)</td></tr><tr><th>Blue</th><td>\(6\)</td><td>\(9\)</td><td>\(15\)</td></tr><tr><th>Total</th><td>\(14\)</td><td>\(14\)</td><td>\(28\)</td></tr></table>

- Complete each book-format column from its total. - Add across each grade row only after both columns are filled. - Verify the grand total from the two column totals.

1. Grade \(8\) e-book preferences: \(45-19=26\). 2. Grade \(7\) printed-book preferences: \(35-12=23\). 3. The Grade \(7\) total is \(19+23=42\), and the Grade \(8\) total is \(26+12=38\). 4. The grand total is \(45+35=80\).

<table><tr><th></th><th>E-books</th><th>Printed books</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(19\)</td><td>\(23\)</td><td>\(42\)</td></tr><tr><th>Grade \(8\)</th><td>\(26\)</td><td>\(12\)</td><td>\(38\)</td></tr><tr><th>Total</th><td>\(45\)</td><td>\(35\)</td><td>\(80\)</td></tr></table>

- A two-way table organizes two categorical variables at the same time. - Ask whether each interior position would identify one category from each variable. - Separate marginal totals from joint counts that describe paired categories.

1. Priya's display lists the marginal counts for pet type and homework location separately. 2. It does not show how the categories pair for the same students, so it does not cross the two variables. 3. One correct organization is pet type as the rows and homework location as the columns; the reverse orientation would also be valid. 4. To fill the interior cells, the paired responses are needed, such as how many dog owners usually work at home, how many dog owners usually work at the library, and so on.

The display is not a two-way table because it does not cross pet type with homework location. A correct table can use Dog, Cat, and No pet as rows and Home and Library as columns, or the reverse. The joint counts for each paired category combination are needed to fill the interior cells.

- In a standard two-way count table, one person should contribute to one category of each variable. - Check what happens to a student who belongs to both proposed sport groups. - Also check whether every surveyed student belongs to at least one proposed row category.

1. The proposed row categories overlap because a student who plays both soccer and basketball would belong in both rows. 2. They are also not exhaustive because a student who plays neither sport would belong in no row. 3. Therefore, the two proposed categories do not form a valid one-variable row partition for a standard two-way table in which each student contributes to exactly one interior cell. 4. One valid row variable is sports-participation group with the categories Soccer only, Basketball only, Both, and Neither.

No. “Plays soccer” and “Plays basketball” overlap for students who play both and omit students who play neither. One valid row-category set is Soccer only, Basketball only, Both, and Neither.

- Match each interior count to the same pair of category labels in both tables. - A transpose changes where a cell appears, but not which two categories that cell represents. - Check the Bus row and Walk row separately rather than comparing row totals alone.

1. Transposing a two-way table switches the row and column headings but keeps each paired category count attached to the same two categories. 2. The North-Bus count \(14\) and South-Bus count \(8\) are correctly placed in Table B. 3. Table A has North-Walk \(9\), so Table B should place \(9\) in the Walk row and North column. 4. Table A has South-Walk \(11\), so Table B should place \(11\) in the Walk row and South column. 5. Table B has those two Walk counts reversed, so it does not represent exactly the same paired data.

No. In Table B, the Walk-North entry should be \(9\), and the Walk-South entry should be \(11\). Those two counts are reversed.

- Row relative frequencies divide by the total of the row being examined. - Ask which groups should each supply their own denominator for the planned comparison. - Separate whether an orientation is valid from whether it is convenient for a particular conditional comparison.

1. Row relative frequencies use each row total as the denominator. 2. To compare device preferences within each grade, each grade should therefore form one row. 3. Grade level should be the row variable, and preferred device type should be the column variable. 4. The opposite orientation would still be a valid two-way count table because it would contain the same paired counts, but row relative frequencies in that orientation would compare grades within each device type instead of devices within each grade.

Put grade level in the rows and device type in the columns. The opposite orientation is still a valid two-way count table, but its row relative frequencies would answer a different conditional question.

- Begin with the count for participants who are in both categories. - Subtract the “both” count from each category total to find the two “only” cells. - Use the grand total after the other three interior cells are known.

1. Place \(400\) in the cell for participants who exercise regularly and eat a balanced diet. 2. The number who exercise regularly but do not eat a balanced diet is \(600-400=200\). 3. The number who eat a balanced diet but do not exercise regularly is \(550-400=150\). 4. The number who do neither is \(1000-(400+200+150)=250\). 5. The number who do exactly one activity is \(200+150=350\).

a) <table><tr><th></th><th>Eats a balanced diet</th><th>Does not eat a balanced diet</th><th>Total</th></tr><tr><th>Exercises regularly</th><td>\(400\)</td><td>\(200\)</td><td>\(600\)</td></tr><tr><th>Does not exercise regularly</th><td>\(150\)</td><td>\(250\)</td><td>\(400\)</td></tr><tr><th>Total</th><td>\(550\)</td><td>\(450\)</td><td>\(1000\)</td></tr></table> b) \(250\) participants. c) \(350\) participants.

- Use the row and column totals to complete one missing entry at a time. - An interior cell combines its row category and column category. - “At least one” can be found by subtracting the “neither” count from the total.

1. The cat-owner row total is \(200-110=90\), and the no-dog column total is \(200-85=115\). 2. The number who own a cat but not a dog is \(90-30=60\). 3. The number who own a dog but not a cat is \(85-30=55\). 4. The count \(30\) represents owners of both a dog and a cat. The count \(55\) in the lower-right cell represents owners of neither pet. 5. The number who own at least one pet is \(200-55=145\).

a) <table><tr><th></th><th>Owns a dog</th><th>Does not own a dog</th><th>Total</th></tr><tr><th>Owns a cat</th><td>\(30\)</td><td>\(60\)</td><td>\(90\)</td></tr><tr><th>Does not own a cat</th><td>\(55\)</td><td>\(55\)</td><td>\(110\)</td></tr><tr><th>Total</th><td>\(85\)</td><td>\(115\)</td><td>\(200\)</td></tr></table> b) \(30\) respondents own both pets; \(55\) respondents own neither pet. c) \(145\) respondents.

- Use the “neither” count with the row and column totals to find an adjacent cell. - Continue using row and column totals until all four interior cells are known. - For part c), divide the relevant cell by all \(25\) students.

1. The number who do not play an instrument is \(25-12=13\), and the number who do not play a sport regularly is \(25-15=10\). 2. Since \(4\) students do neither, the number who play a sport but no instrument is \(13-4=9\). 3. The number who do both is \(15-9=6\). 4. The number who play an instrument but no sport is \(12-6=6\). 5. The required percentage is \(\frac{9}{25}=0.36=36\%\).

a) <table><tr><th></th><th>Plays a sport</th><th>Does not play a sport</th><th>Total</th></tr><tr><th>Plays an instrument</th><td>\(6\)</td><td>\(6\)</td><td>\(12\)</td></tr><tr><th>Does not play an instrument</th><td>\(9\)</td><td>\(4\)</td><td>\(13\)</td></tr><tr><th>Total</th><td>\(15\)</td><td>\(10\)</td><td>\(25\)</td></tr></table> b) \(6\) students. c) \(36\%\).

- Convert the percentage within the Android group to a count first. - Use the Android total to find the other Android cell. - Complete the remaining row and column using their totals.

1. The number of Android phones sold without a screen protector is \(0.70\cdot120=84\). 2. Therefore, the number of Android phones sold with a screen protector is \(120-84=36\). 3. The number of non-Android phones is \(200-120=80\), and the number sold without a screen protector is \(200-90=110\). 4. The number of non-Android phones sold with a screen protector is \(90-36=54\). 5. The remaining cell is \(80-54=26\).

a) \(36\) Android phones. b) <table><tr><th></th><th>With screen protector</th><th>Without screen protector</th><th>Total</th></tr><tr><th>Android</th><td>\(36\)</td><td>\(84\)</td><td>\(120\)</td></tr><tr><th>Non-Android</th><td>\(54\)</td><td>\(26\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(90\)</td><td>\(110\)</td><td>\(200\)</td></tr></table> c) \(54\) non-Android phones.

- Find the total with a material defect before completing its row or column. - Add the two shape-defect cells for part b). - “At most one” excludes only the cell for both defects.

1. Since \(920\) components do not have a material defect, \(1000-920=80\) have a material defect. 2. Of those \(80\), \(20\) also have a shape defect, so \(80-20=60\) have a material defect but no shape defect. 3. The number with a shape defect is \(20+35=55\). 4. The number with neither defect is \(920-35=885\). 5. Having at most one defect means every component except the \(20\) with both defects, so \(1000-20=980\).

a) <table><tr><th></th><th>Material defect</th><th>No material defect</th><th>Total</th></tr><tr><th>Shape defect</th><td>\(20\)</td><td>\(35\)</td><td>\(55\)</td></tr><tr><th>No shape defect</th><td>\(60\)</td><td>\(885\)</td><td>\(945\)</td></tr><tr><th>Total</th><td>\(80\)</td><td>\(920\)</td><td>\(1000\)</td></tr></table> b) \(55\) components. c) \(980\) components; subtract the \(20\) components with both defects from all \(1000\) components.

- Treat each pie chart as a within-row percentage distribution for one instrument-status group. - Use the stated group total with the percentage shown for each slice to recover the corresponding count. - After finding the four interior counts, add rows and columns to check the grand total.

1. In panel a), \(60\%\) of the \(40\) students belong to a club, so \(0.60\cdot40=24\). The remaining \(16\) do not belong to a club. 2. In panel b), \(30\%\) of the \(50\) students belong to a club, so \(0.30\cdot50=15\). The remaining \(35\) do not belong to a club. 3. The club-membership column total is \(24+15=39\), the nonmembership total is \(16+35=51\), and the grand total is \(90\).

<table><tr><th></th><th>Belongs to a club</th><th>Does not belong to a club</th><th>Total</th></tr><tr><th>Plays an instrument</th><td>\(24\)</td><td>\(16\)</td><td>\(40\)</td></tr><tr><th>Does not play an instrument</th><td>\(15\)</td><td>\(35\)</td><td>\(50\)</td></tr><tr><th>Total</th><td>\(39\)</td><td>\(51\)</td><td>\(90\)</td></tr></table>

- Split the “exactly one” count into chess only and art only. - Put the both and neither counts in the two diagonal cells they describe. - Add rows and columns, then verify that the four interior cells total the survey size.

1. Chess only is \(13\), so art only is \(22-13=9\). 2. Place \(9\) in the both cell and \(7\) in the neither cell. 3. The chess row total is \(9+13=22\), and the not-chess row total is \(9+7=16\). 4. The column totals are \(18\) and \(20\), and \(9+13+9+7=38\), matching the survey size.

<table><tr><th></th><th>In art club</th><th>Not in art club</th><th>Total</th></tr><tr><th>In chess club</th><td>\(9\)</td><td>\(13\)</td><td>\(22\)</td></tr><tr><th>Not in chess club</th><td>\(9\)</td><td>\(7\)</td><td>\(16\)</td></tr><tr><th>Total</th><td>\(18\)</td><td>\(20\)</td><td>\(38\)</td></tr></table>

- Convert the within-group percentage into a count using that row total. - Complete each row before adding column totals.

1. First-time finishers: \(0.60\cdot 40=24\); first-time nonfinishers: \(40-24=16\). 2. Returning nonfinishers: \(40-18=22\). 3. Column totals are \(42\) finishers and \(38\) nonfinishers.

<table><tr><th></th><th>Finished</th><th>Did not finish</th><th>Total</th></tr><tr><th>First-time</th><td>\(24\)</td><td>\(16\)</td><td>\(40\)</td></tr><tr><th>Returning</th><td>\(18\)</td><td>\(22\)</td><td>\(40\)</td></tr><tr><th>Total</th><td>\(42\)</td><td>\(38\)</td><td>\(80\)</td></tr></table>

- The new “Not car” category combines two original rows. - Add corresponding cells before calculating totals.

1. Keep the Car row as \(8\) and \(5\). 2. Combine Bus and Walk: rain \(11+7=18\), no rain \(6+3=9\). 3. The row totals are \(13\) and \(27\); column totals are \(26\) and \(14\).

<table><tr><th></th><th>Rain</th><th>No rain</th><th>Total</th></tr><tr><th>Car</th><td>\(8\)</td><td>\(5\)</td><td>\(13\)</td></tr><tr><th>Not car</th><td>\(18\)</td><td>\(9\)</td><td>\(27\)</td></tr><tr><th>Total</th><td>\(26\)</td><td>\(14\)</td><td>\(40\)</td></tr></table>

- Translate “twice as many” into a \(2:1\) ratio for the adult row. - Use the overall morning total after completing the adult row. - Complete the youth row, then verify the evening and grand totals.

1. Split the \(45\) adults in a \(2:1\) ratio: \(30\) attend in the morning and \(15\) attend in the evening. 2. Youth morning attendance is \(46-30=16\). 3. Youth evening attendance is \(30-16=14\). 4. The evening total is \(15+14=29\), and the grand total is \(45+30=75\).

<table><tr><th></th><th>Morning</th><th>Evening</th><th>Total</th></tr><tr><th>Adults</th><td>\(30\)</td><td>\(15\)</td><td>\(45\)</td></tr><tr><th>Youths</th><td>\(16\)</td><td>\(14\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(46\)</td><td>\(29\)</td><td>\(75\)</td></tr></table>

- Complete each class row from its total of \(20\). - After the interior cells are complete, add down each option column. - Check that the row and column totals agree on the grand total.

1. Class A project students: \(20-12=8\). 2. Class B lab students: \(20-11=9\). 3. Class C lab students: \(20-5=15\). 4. The lab total is \(12+9+15=36\), the project total is \(8+11+5=24\), and the grand total is \(60\).

<table><tr><th></th><th>Lab</th><th>Project</th><th>Total</th></tr><tr><th>Class A</th><td>\(12\)</td><td>\(8\)</td><td>\(20\)</td></tr><tr><th>Class B</th><td>\(9\)</td><td>\(11\)</td><td>\(20\)</td></tr><tr><th>Class C</th><td>\(15\)</td><td>\(5\)</td><td>\(20\)</td></tr><tr><th>Total</th><td>\(36\)</td><td>\(24\)</td><td>\(60\)</td></tr></table>

- First determine how many responses are usable. - Complete the Grade \(7\) row, then use the grand total to find the Grade \(8\) row total. - Complete the second row and add the columns.

1. The table includes \(70-6=64\) usable responses. 2. Grade \(7\) “No” responses: \(34-22=12\). 3. Grade \(8\) has \(64-34=30\) usable responses. 4. Grade \(8\) “No” responses: \(30-18=12\). 5. The column totals are \(22+18=40\) “Yes” responses and \(12+12=24\) “No” responses.

<table><tr><th></th><th>Yes</th><th>No</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(22\)</td><td>\(12\)</td><td>\(34\)</td></tr><tr><th>Grade \(8\)</th><td>\(18\)</td><td>\(12\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(40\)</td><td>\(24\)</td><td>\(64\)</td></tr></table>

- Apply each percentage only to its own group total. - Subtract each pool-user count from its row total to find the nonusers. - Add the two rows by columns and verify the grand total.

1. Weekday pool users: \(0.70\cdot40=28\). 2. Weekday nonusers: \(40-28=12\). 3. Weekend pool users: \(0.36\cdot50=18\). 4. Weekend nonusers: \(50-18=32\). 5. The column totals are \(28+18=46\) pool users and \(12+32=44\) nonusers, for \(90\) visitors.

<table><tr><th></th><th>Uses the pool</th><th>Does not use the pool</th><th>Total</th></tr><tr><th>Weekday</th><td>\(28\)</td><td>\(12\)</td><td>\(40\)</td></tr><tr><th>Weekend</th><td>\(18\)</td><td>\(32\)</td><td>\(50\)</td></tr><tr><th>Total</th><td>\(46\)</td><td>\(44\)</td><td>\(90\)</td></tr></table>

- Convert the fraction of the morning row into a count. - Complete the morning row, then use the overall pass total. - Complete the evening row and verify both columns.

1. Morning passes: \(\frac{2}{3}\cdot42=28\). 2. Morning failures: \(42-28=14\). 3. Evening passes: \(43-28=15\). 4. Evening failures: \(30-15=15\). 5. The failure total is \(14+15=29\), and the grand total is \(72\).

<table><tr><th></th><th>Pass</th><th>Fail</th><th>Total</th></tr><tr><th>Morning</th><td>\(28\)</td><td>\(14\)</td><td>\(42\)</td></tr><tr><th>Evening</th><td>\(15\)</td><td>\(15\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(43\)</td><td>\(29\)</td><td>\(72\)</td></tr></table>

- Keep each paired count attached to the same two category labels. - Switching orientation transposes the interior cells but does not change the data.

1. With location as rows, place Urban \((13, 7)\) and Rural \((5, 15)\). 2. With transportation as rows, transpose the interior counts: Bus \((13, 5)\) and Train \((7, 15)\). 3. Both tables represent the same \(40\) observations.

Location as rows: <table><tr><th></th><th>Bus</th><th>Train</th><th>Total</th></tr><tr><th>Urban</th><td>\(13\)</td><td>\(7\)</td><td>\(20\)</td></tr><tr><th>Rural</th><td>\(5\)</td><td>\(15\)</td><td>\(20\)</td></tr><tr><th>Total</th><td>\(18\)</td><td>\(22\)</td><td>\(40\)</td></tr></table> Transportation as rows: <table><tr><th></th><th>Urban</th><th>Rural</th><th>Total</th></tr><tr><th>Bus</th><td>\(13\)</td><td>\(5\)</td><td>\(18\)</td></tr><tr><th>Train</th><td>\(7\)</td><td>\(15\)</td><td>\(22\)</td></tr><tr><th>Total</th><td>\(20\)</td><td>\(20\)</td><td>\(40\)</td></tr></table>

- Convert each row ratio into its total number of equal parts. - Use the corresponding row total to find one part. - Complete both rows before adding the columns.

1. For local visitors, the \(3:2\) ratio has \(5\) equal parts, so each part is \(40\div5=8\). This gives \(24\) indoor and \(16\) outdoor visitors. 2. For out-of-town visitors, the \(1:2\) ratio has \(3\) equal parts, so each part is \(45\div3=15\). This gives \(15\) indoor and \(30\) outdoor visitors. 3. The indoor total is \(24+15=39\). 4. The outdoor total is \(16+30=46\), and the grand total is \(85\).

<table><tr><th></th><th>Indoor</th><th>Outdoor</th><th>Total</th></tr><tr><th>Local</th><td>\(24\)</td><td>\(16\)</td><td>\(40\)</td></tr><tr><th>Out-of-town</th><td>\(15\)</td><td>\(30\)</td><td>\(45\)</td></tr><tr><th>Total</th><td>\(39\)</td><td>\(46\)</td><td>\(85\)</td></tr></table>

- Apply the fraction to the total number of “Yes” responses. - Complete the “Yes” column before subtracting within each age-group row. - Add the “No” column and verify the grand total.

1. Teen “Yes” responses: \(\frac{1}{3}\cdot36=12\). 2. Adult “Yes” responses: \(36-12=24\). 3. Teen “No” responses: \(30-12=18\). 4. Adult “No” responses: \(42-24=18\). The “No” total is \(36\), and the grand total is \(72\).

<table><tr><th></th><th>Yes</th><th>No</th><th>Total</th></tr><tr><th>Teens</th><td>\(12\)</td><td>\(18\)</td><td>\(30\)</td></tr><tr><th>Adults</th><td>\(24\)</td><td>\(18\)</td><td>\(42\)</td></tr><tr><th>Total</th><td>\(36\)</td><td>\(36\)</td><td>\(72\)</td></tr></table>

- Each club row must total \(20\). - Find the Research count separately for each club. - Add each session column and verify the grand total.

1. Club A Research: \(20-8-5=7\). 2. Club B Research: \(20-6-9=5\). 3. Club C Research: \(20-10-4=6\). 4. The Coding total is \(8+6+10=24\), and the Design total is \(5+9+4=18\). 5. The Research total is \(7+5+6=18\), and the grand total is \(60\).

<table><tr><th></th><th>Coding</th><th>Design</th><th>Research</th><th>Total</th></tr><tr><th>Club A</th><td>\(8\)</td><td>\(5\)</td><td>\(7\)</td><td>\(20\)</td></tr><tr><th>Club B</th><td>\(6\)</td><td>\(9\)</td><td>\(5\)</td><td>\(20\)</td></tr><tr><th>Club C</th><td>\(10\)</td><td>\(4\)</td><td>\(6\)</td><td>\(20\)</td></tr><tr><th>Total</th><td>\(24\)</td><td>\(18\)</td><td>\(18\)</td><td>\(60\)</td></tr></table>

- Use the total of \(100\) to convert each overall percentage directly to a count. - Find the fourth interior count from the grand total. - Add the grade rows and travel columns to complete the margins.

1. Because the survey total is \(100\), the three given percentages correspond to counts of \(30\), \(18\), and \(12\). 2. The missing Grade \(8\) walking count is \(100-(30+18+12)=40\). 3. The grade totals are \(30+18=48\) for Grade \(7\) and \(12+40=52\) for Grade \(8\). 4. The travel totals are \(30+12=42\) bus riders and \(18+40=58\) walkers.

<table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(30\)</td><td>\(18\)</td><td>\(48\)</td></tr><tr><th>Grade \(8\)</th><td>\(12\)</td><td>\(40\)</td><td>\(52\)</td></tr><tr><th>Total</th><td>\(42\)</td><td>\(58\)</td><td>\(100\)</td></tr></table>

- A corrected classification moves one response rather than adding or deleting it. - Adjust the two affected Grade \(7\) cells in opposite directions. - Recalculate the two choice columns while keeping both grade totals fixed.

1. Reclassifying one response moves it between cells; it does not change the number of Grade \(7\) responses. 2. Grade \(7\) cafeteria decreases from \(20\) to \(19\), and Grade \(7\) packed lunch increases from \(10\) to \(11\). 3. The Grade \(8\) counts remain \(14\) and \(16\). 4. The corrected column totals are \(19+14=33\) cafeteria choices and \(11+16=27\) packed lunches, for \(60\) responses.

<table><tr><th></th><th>Cafeteria</th><th>Packed lunch</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(19\)</td><td>\(11\)</td><td>\(30\)</td></tr><tr><th>Grade \(8\)</th><td>\(14\)</td><td>\(16\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(33\)</td><td>\(27\)</td><td>\(60\)</td></tr></table>

- Keep the Agree column unchanged. - Combine Maybe and Disagree separately within each attendance group. - Recalculate the margins after recoding.

1. Keep the Agree counts of \(12\) and \(8\). 2. Combine Maybe and Disagree: Online has \(5+3=8\), and In person has \(4+6=10\) responses in “Does not agree.” 3. The row totals are \(20\) and \(18\); the column totals are \(20\) and \(18\), for \(38\) students.

<table><tr><th></th><th>Agree</th><th>Does not agree</th><th>Total</th></tr><tr><th>Online</th><td>\(12\)</td><td>\(8\)</td><td>\(20\)</td></tr><tr><th>In person</th><td>\(8\)</td><td>\(10\)</td><td>\(18\)</td></tr><tr><th>Total</th><td>\(20\)</td><td>\(18\)</td><td>\(38\)</td></tr></table>

- Find the total number of “Yes” responses from the grand total and the “No” total. - Split that column equally between the two grades. - Complete each grade row and verify the “No” column.

1. The grand total is \(27+33=60\), so the “Yes” total is \(60-34=26\). 2. Equal “Yes” counts split \(26\) into \(13\) for each grade. 3. Grade \(7\) “No” responses: \(27-13=14\). 4. Grade \(8\) “No” responses: \(33-13=20\), and \(14+20=34\) checks the stated total.

<table><tr><th></th><th>Yes</th><th>No</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(13\)</td><td>\(14\)</td><td>\(27\)</td></tr><tr><th>Grade \(8\)</th><td>\(13\)</td><td>\(20\)</td><td>\(33\)</td></tr><tr><th>Total</th><td>\(26\)</td><td>\(34\)</td><td>\(60\)</td></tr></table>

- Use the bus column total to find the missing bus count. - Complete each neighborhood row from its row total. - Add the train column after the interior counts are complete.

1. Neighborhood \(3\) bus responses: \(40-16-9=15\). 2. Neighborhood \(1\) train responses: \(25-16=9\). 3. Neighborhood \(2\) train responses: \(20-9=11\). 4. Neighborhood \(3\) train responses: \(30-15=15\). 5. The train total is \(9+11+15=35\), and the grand total is \(75\).

<table><tr><th></th><th>Bus</th><th>Train</th><th>Total</th></tr><tr><th>Neighborhood \(1\)</th><td>\(16\)</td><td>\(9\)</td><td>\(25\)</td></tr><tr><th>Neighborhood \(2\)</th><td>\(9\)</td><td>\(11\)</td><td>\(20\)</td></tr><tr><th>Neighborhood \(3\)</th><td>\(15\)</td><td>\(15\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(40\)</td><td>\(35\)</td><td>\(75\)</td></tr></table>

- Treat “Not recorded” as an explicit category rather than discarding it. - Use the equal split before adding row totals.

1. Split the \(4\) not-recorded conditions equally: \(2\) indoor and \(2\) outdoor. 2. Indoor total is \(18+7+2=27\); outdoor total is \(12+9+2=23\). 3. Condition totals are \(30\) working, \(16\) needs repair, and \(4\) not recorded.

<table><tr><th></th><th>Working</th><th>Needs repair</th><th>Not recorded</th><th>Total</th></tr><tr><th>Indoor</th><td>\(18\)</td><td>\(7\)</td><td>\(2\)</td><td>\(27\)</td></tr><tr><th>Outdoor</th><td>\(12\)</td><td>\(9\)</td><td>\(2\)</td><td>\(23\)</td></tr><tr><th>Total</th><td>\(30\)</td><td>\(16\)</td><td>\(4\)</td><td>\(50\)</td></tr></table>

- Check student IDs before tallying campus and travel categories. - Remove only the repeated record, not both copies of that student. - Tally the remaining records and verify the grand total decreased by one.

1. ID \(104\) appears twice with the same campus and travel method, so count it once. 2. North has \(3\) bus riders and \(1\) walker. 3. South has \(1\) bus rider and \(3\) walkers. 4. Each campus row and each travel column totals \(4\), for \(8\) students.

<table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>North</th><td>\(3\)</td><td>\(1\)</td><td>\(4\)</td></tr><tr><th>South</th><td>\(1\)</td><td>\(3\)</td><td>\(4\)</td></tr><tr><th>Total</th><td>\(4\)</td><td>\(4\)</td><td>\(8\)</td></tr></table>

- Think about what limits the overlap: it cannot be larger than either defect group. - For the smallest overlap, try to place as many components as possible in the two single-defect cells while keeping the grand total fixed. - Check each proposed extreme by completing all four interior cells and making sure no count is negative.

1. Let \(b\) be the number with both defects. Then material-only is \(28-b\), surface-only is \(31-b\), and neither is \(50-[b+(28-b)+(31-b)]=b-9\). 2. Every cell count must be nonnegative. From \(b-9\ge 0\), the overlap must be at least \(9\). This minimum gives material-only \(19\), surface-only \(22\), and neither \(0\). 3. The overlap cannot exceed either defect total, so it cannot exceed \(28\). This maximum gives material-only \(0\), surface-only \(3\), and neither \(19\). 4. Both endpoint tables have the required row total \(28\), column total \(31\), and grand total \(50\).

a) Smallest possible overlap: \(9\). <table><tr><th></th><th>Surface defect</th><th>No surface defect</th><th>Total</th></tr><tr><th>Material defect</th><td>\(9\)</td><td>\(19\)</td><td>\(28\)</td></tr><tr><th>No material defect</th><td>\(22\)</td><td>\(0\)</td><td>\(22\)</td></tr><tr><th>Total</th><td>\(31\)</td><td>\(19\)</td><td>\(50\)</td></tr></table> b) Largest possible overlap: \(28\). <table><tr><th></th><th>Surface defect</th><th>No surface defect</th><th>Total</th></tr><tr><th>Material defect</th><td>\(28\)</td><td>\(0\)</td><td>\(28\)</td></tr><tr><th>No material defect</th><td>\(3\)</td><td>\(19\)</td><td>\(22\)</td></tr><tr><th>Total</th><td>\(31\)</td><td>\(19\)</td><td>\(50\)</td></tr></table>

- Represent the two East Branch cells with expressions whose difference is \(8\). - Use the East row total to determine those two counts. - Then use the online total and West row total to finish the table.

1. Let East in-person registrations be \(x\). Then East online registrations are \(x+8\). 2. Since the East total is \(34\), \(x+(x+8)=34\), so \(2x=26\), \(x=13\), and East online registrations are \(21\). 3. West online registrations are \(35-21=14\). 4. West in-person registrations are \(26-14=12\). The in-person column total is \(13+12=25\), and the grand total is \(60\).

<table><tr><th></th><th>Online</th><th>In person</th><th>Total</th></tr><tr><th>East Branch</th><td>\(21\)</td><td>\(13\)</td><td>\(34\)</td></tr><tr><th>West Branch</th><td>\(14\)</td><td>\(12\)</td><td>\(26\)</td></tr><tr><th>Total</th><td>\(35\)</td><td>\(25\)</td><td>\(60\)</td></tr></table>

- Write an expression for each workshop column total. - Use the statement that the two column totals are equal. - Substitute the solved value before calculating row and grand totals.

1. The Coding column total is \(x+26\), and the Design column total is \(18+2x\). 2. Set the equal column totals equal: \(x+26=18+2x\). 3. Solving gives \(x=8\), so the four interior counts are \(8\), \(18\), \(26\), and \(16\). 4. The row totals are \(26\) and \(42\); each column total is \(34\), and the grand total is \(68\).

\(x=8\). <table><tr><th></th><th>Coding</th><th>Design</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(8\)</td><td>\(18\)</td><td>\(26\)</td></tr><tr><th>Grade \(8\)</th><td>\(26\)</td><td>\(16\)</td><td>\(42\)</td></tr><tr><th>Total</th><td>\(34\)</td><td>\(34\)</td><td>\(68\)</td></tr></table>

- Choose one grade-and-travel cell and use the margins to force the other three. - Try a different starting count while keeping all entries nonnegative. - If both completions satisfy every margin, the interior table is not unique.

1. The row and column totals constrain the table, but one interior count can still vary. 2. One valid completion has Grade \(7\) counts \((15, 15)\) and Grade \(8\) counts \((10, 10)\). 3. Another valid completion has Grade \(7\) counts \((20, 10)\) and Grade \(8\) counts \((5, 15)\). Both have the required margins.

No. One completion is <table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(15\)</td><td>\(15\)</td><td>\(30\)</td></tr><tr><th>Grade \(8\)</th><td>\(10\)</td><td>\(10\)</td><td>\(20\)</td></tr><tr><th>Total</th><td>\(25\)</td><td>\(25\)</td><td>\(50\)</td></tr></table> Another is <table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(20\)</td><td>\(10\)</td><td>\(30\)</td></tr><tr><th>Grade \(8\)</th><td>\(5\)</td><td>\(15\)</td><td>\(20\)</td></tr><tr><th>Total</th><td>\(25\)</td><td>\(25\)</td><td>\(50\)</td></tr></table>

- Determine the two age-group totals first. - Convert the overall response ratio into “Yes” and “No” column totals. - Use the within-teen fraction, then subtract from the row and column margins to complete the adult row.

1. The teen total is \(0.40\cdot90=36\), so the adult total is \(90-36=54\). 2. The \(5:4\) response ratio has \(9\) equal parts, so each part represents \(90\div9=10\) responses. 3. The overall response totals are \(5\cdot10=50\) “Yes” and \(4\cdot10=40\) “No.” 4. Teen “Yes” responses: \(\frac{2}{3}\cdot36=24\). 5. Teen “No” responses: \(36-24=12\). 6. Adult “Yes” responses: \(50-24=26\), and adult “No” responses: \(40-12=28\). These total \(54\) adults.

<table><tr><th></th><th>Yes</th><th>No</th><th>Total</th></tr><tr><th>Teens</th><td>\(24\)</td><td>\(12\)</td><td>\(36\)</td></tr><tr><th>Adults</th><td>\(26\)</td><td>\(28\)</td><td>\(54\)</td></tr><tr><th>Total</th><td>\(50\)</td><td>\(40\)</td><td>\(90\)</td></tr></table>

- Apply the two corrections in the order stated. - Moving a student between campuses changes row totals; changing travel method within one campus changes column totals. - Verify that the grand total remains constant.

1. Move one walker from North to South: North becomes \((10, 13)\), and South becomes \((16, 9)\). 2. Reclassify one South response from bus to walking: South becomes \((15, 10)\). 3. The final campus totals are \(23\) and \(25\); the travel totals are \(25\) bus riders and \(23\) walkers. 4. The grand total remains \(48\) because neither correction adds or removes a student.

<table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>North</th><td>\(10\)</td><td>\(13\)</td><td>\(23\)</td></tr><tr><th>South</th><td>\(15\)</td><td>\(10\)</td><td>\(25\)</td></tr><tr><th>Total</th><td>\(25\)</td><td>\(23\)</td><td>\(48\)</td></tr></table>