Answer:

Step-by-step explanation:
we know that
A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form
or
In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin
In this problem we have

Substitute the values and solve for k


therefore
the equation of the line is equal to

15.2
16 + 12 + 8 + 22 + 25 + 8 = 91 / 6 = 15.16 = 15.2
Answer:

Step-by-step explanation:
![\sf{ [ (4 \frac{1}{6} + 2 \frac{1}{3} ) \div 4 \frac{1}{3}] - 1\frac{1}{2} }](https://tex.z-dn.net/?f=%20%5Csf%7B%20%5B%20%284%20%5Cfrac%7B1%7D%7B6%7D%20%20%2B%202%20%5Cfrac%7B1%7D%7B3%7D%20%29%20%20%5Cdiv%204%20%5Cfrac%7B1%7D%7B3%7D%5D%20-%20%201%5Cfrac%7B1%7D%7B2%7D%20%7D)
Convert the mixed numbers into improper fraction
![\longrightarrow{ \sf{ [ ( \frac{25}{6} + \frac{7}{3} ) \div \frac{13}{3}] - \frac{3}{2}}}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%28%20%5Cfrac%7B25%7D%7B6%7D%20%20%2B%20%20%5Cfrac%7B7%7D%7B3%7D%20%29%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%7D%20)
Add the fractions : 25 / 6 and 7 / 3
While performing addition or subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fraction.
To do so, first take the L.C.M of 6 and 3 which results to 6
![\longrightarrow\sf{ [( \frac{25 + 7 \times 2}{6} ) \div \frac{13}{3} ] - \frac{3}{2}}](https://tex.z-dn.net/?f=%20%20%5Clongrightarrow%5Csf%7B%20%5B%28%20%5Cfrac%7B25%20%2B%207%20%5Ctimes%202%7D%7B6%7D%20%29%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%20)
![\longrightarrow{ \sf{ [( \frac{25 + 14}{6} ) \div \frac{13}{3} ] - \frac{3}{2} }}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%28%20%5Cfrac%7B25%20%2B%2014%7D%7B6%7D%20%29%20%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%20%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%20%7D%7D)
![\longrightarrow{ \sf{ [ \frac{39}{6} \div \frac{13}{3}] - \frac{3}{2} }}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%5Cfrac%7B39%7D%7B6%7D%20%20%5Cdiv%20%20%5Cfrac%7B13%7D%7B3%7D%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%20%7D%7D)
Multiply the dividend by the reciprocal of the divisor.
Reciprocal of any number or fraction can be obtained by interchanging the position of numerator and denominator
![\longrightarrow{ \sf{ [ \frac{39}{6} \times \frac{3}{13} ] - \frac{3}{2}}}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%20%5Cfrac%7B39%7D%7B6%7D%20%20%5Ctimes%20%20%5Cfrac%7B3%7D%7B13%7D%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%7D%20)
To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term
![\longrightarrow{ \sf{ [ \frac{39 \times 3}{6 \times 13} ] - \frac{3}{2}}}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%20%5Cfrac%7B39%20%5Ctimes%203%7D%7B6%20%5Ctimes%2013%7D%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%7D%7D%20)
![\longrightarrow{ \sf{ [ \frac{117}{78} ] - \frac{3}{2} }}](https://tex.z-dn.net/?f=%20%5Clongrightarrow%7B%20%5Csf%7B%20%5B%20%20%5Cfrac%7B117%7D%7B78%7D%20%20%5D%20-%20%20%5Cfrac%7B3%7D%7B2%7D%20%7D%7D)

While performing the addition or subtraction of like fractions , you just have to add or subtract the numerator respectively in which the denominator is retained same

Subtract 3 from 3

Divide 0 by 2

Hope I helped!
Best regards!
Answer:
Answer is : $411.16
Step-by-step explanation:
How much she earns per hour which is $10.82 times by the total hours she have worked 38h then you get the answer which is $411.16
$10.82 X 38 hours = $411.16 total
Both methods assume a normal distribution of the data, but the z-tests are most useful when the standard deviation is known.
OR u can use , z-tests are used when we have large sample sizes (n > 30), whereas t-tests are most helpful with a smaller sample size (n < 30). Both methods assume a normal distribution of the data, but the z-tests are most useful when the standard deviation is known.