unit 1 test review\n1. $\frac{3}{5}+\frac{1}{10}$\n2. $\frac{1}{9}div\frac{1}{2}$\n3. $\frac{4}{6}-\frac{1}{2…

Explanation:

Step1: Find a common - denominator for $\frac{3}{5}+\frac{1}{10}$

The common denominator of 5 and 10 is 10. Rewrite $\frac{3}{5}$ as $\frac{6}{10}$. Then $\frac{3}{5}+\frac{1}{10}=\frac{6}{10}+\frac{1}{10}=\frac{6 + 1}{10}=\frac{7}{10}$.

Step2: Solve $\frac{1}{9}\div\frac{1}{2}$

Dividing by a fraction is the same as multiplying by its reciprocal. So $\frac{1}{9}\div\frac{1}{2}=\frac{1}{9}\times\frac{2}{1}= \frac{2}{9}$.

Step3: Simplify $\frac{4}{6}\times\frac{1}{2}$

First, simplify $\frac{4}{6}$ to $\frac{2}{3}$. Then $\frac{2}{3}\times\frac{1}{2}=\frac{2\times1}{3\times2}=\frac{1}{3}$.

Step4: Multiply $4.3932\times1.33791$

$4.3932\times1.33791 = 5.87977\approx5.88$ (rounded to the hundredths place).

Step5: Multiply $4.3932\times1.33791$ and round to the tenths place

$4.3932\times1.33791 = 5.87977\approx5.9$ (rounded to the tenths place).

Step6: Analyze the box - and - whisker plot

The minimum is 68, the maximum is 194, the median is 142, the range is $194 - 68=126$, and a possible value for the lower quarter could be 110.

Step7: Determine the relationship

As the weight (in lbs) increases, the height (in inches) generally increases. So there is a direct relationship.

Step8: (No clear division problem given for the 100 - square grid, assuming a general division concept)

If we consider a division problem like $a\div b$, we can represent it on the 100 - square grid by shading appropriate number of squares. But without a specific division problem, we can't give a definite answer.

Answer:

1. $\frac{7}{10}$
2. $\frac{2}{9}$
3. $\frac{1}{3}$
4. $5.88$
5. $5.9$
6. Minimum: 68, Maximum: 194, Median: 142, Range: 126, Lower - quarter (possible value): 110
7. Direct
8. No specific answer without a given division problem for the grid.