If they scored 5 times, they got 1 touchdown and 4 field goals.
Equation:
7 + 3 + 3 + 3 + 3 = 19.
Jason's scores : 80 90 95 85 70 and
Jill's score : 70 75 90 100 95.
Mean of Jason's scores = 
Mean of Jill's scores = 
Now, in order to find the mean absolute deviation, need to find the difference of each score from means.
<u>Mean absolute deviation for Jason's scores.</u>
|84-80| = 4
|84-90| = 6
|84-95| = 9
|84-85| = 1
|84-70|= 14

<u>Mean absolute deviation for Jill's scores</u>
|86-70| = 16
|86-75| = 11
|86-90| = 4
|86-100| = 14
|86-95|= 9

Jill got average quiz score 86 and Jason got 84.
Therefore, Jill got better quiz average.
Also, the mean absolute deviation for Jason scores is less that is 6.8 than 10.8.
Therefore, Jason got more consistent grades.
Answer:
3/5
Step-by-step explanation:
cosine is adjacent/hypotenuse
Part A
The expression "ma" is the same as "m*a" (m times a). To isolate 'a', we divide both sides by m. Division is the inverse operation of multiplication. Think of it as undoing multiplication
F = ma
F = m*a
F/m = m*a/m
F/m = a
a = F/m
Answer: a = F/m
Note: the slash "/" without quotes means "divide"
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Part B
Use the result from part A. We will plug F = -24 and m = 10 into that equation
a = F/m
a = -24/10
a = -2.4
Answer: -2.4
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Part C
We will use the equation F = m*a
F = m*a
24 = m*12 ... plug in F = 24 and a = 12
24/12 = m*12/12
2 = m
m = 2
Answer: 2