-6m+13=0 (put all like terms together)
-6m=-13 (subtracted 13 from both sides to leave m alone)
m=2.17 (divided both sides by -6)
So Jeremy had x amount of money to begin with
he then spent $17 and was left with $75 to spend for camp so
x-17=75
add 17 to both sides
x=92
congrats, you're doing algebra
Answer:
![Mean = 68.9](https://tex.z-dn.net/?f=Mean%20%3D%2068.9)
--- Variance
Step-by-step explanation:
Given
![\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcccccc%7D%7BClass%7D%20%26%20%7B51-58%7D%20%26%20%7B59-66%7D%20%26%20%7B67-74%7D%20%26%20%7B75-82%7D%20%26%20%7B83-90%7D%20%5C%20%5C%5C%20%7BFrequency%7D%20%26%20%7B6%7D%20%26%20%7B3%7D%20%26%20%7B11%7D%20%26%20%7B13%7D%20%26%20%7B4%7D%20%5C%20%5Cend%7Barray%7D)
Solving (a): Calculate the mean.
The given data is a grouped data. So, first we calculate the class midpoint (x)
For 51 - 58.
![x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2851%2B58%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28109%29%20%3D%2054.5)
For 59 - 66
![x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2859%2B66%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28125%29%20%3D%2062.5)
For 67 - 74
![x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2867%2B74%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28141%29%20%3D%2070.5)
For 75 - 82
![x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2875%2B82%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28157%29%20%3D%2078.5)
For 83 - 90
![x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2883%2B90%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28173%29%20%3D%2086.5)
So, the table becomes:
![\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcccccc%7D%7Bx%7D%20%26%20%7B54.5%7D%20%26%20%7B62.5%7D%20%26%20%7B70.5%7D%20%26%20%7B78.5%7D%20%26%20%7B86.5%7D%20%5C%20%5C%5C%20%7BFrequency%7D%20%26%20%7B6%7D%20%26%20%7B3%7D%20%26%20%7B11%7D%20%26%20%7B13%7D%20%26%20%7B4%7D%20%5C%20%5Cend%7Barray%7D)
The mean is then calculated as:
![Mean = \frac{\sum fx}{\sum f}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cfrac%7B%5Csum%20fx%7D%7B%5Csum%20f%7D)
![Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cfrac%7B54.5%2A4%2B62.5%2A3%2B70.5%2A11%2B78.5%2A13%2B86.5%2A4%7D%7B6%2B3%2B11%2B13%2B4%7D)
![Mean = \frac{2547.5}{37}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cfrac%7B2547.5%7D%7B37%7D)
-- approximated
Solving (b): The sample variance:
This is calculated as:
![s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}](https://tex.z-dn.net/?f=s%5E2%20%3D%5Cfrac%7B%5Csum%20%28x%20-%20%5Coverline%20x%29%5E2%7D%7B%5Csum%20f%20-%201%7D)
So, we have:
![s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}](https://tex.z-dn.net/?f=s%5E2%20%3D%5Cfrac%7B%2854.5-68.9%29%5E2%2B%2862.5-68.9%29%5E2%2B%2870.5-68.9%29%5E2%2B%2878.5-68.9%29%5E2%2B%2886.5-68.9%29%5E2%7D%7B37%20-%201%7D)
![s^2 =\frac{652.8}{36}](https://tex.z-dn.net/?f=s%5E2%20%3D%5Cfrac%7B652.8%7D%7B36%7D)
-- approximated
If you’re simplifying it’s 100x^2+200x+5
The answer is C because:
It is keeping it identity through the multiplication.