4. x^10
5. x^3
I hope this helped! Mark me Brainliest! :) -Raven❤️
Answer:
Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795
Step-by-step explanation:
Given
See attachment for class
Solving (a): Fill the midpoint of each class.
Midpoint (M) is calculated as:
Where
Lower class interval
Upper class interval
So, we have:
Class 63-65:
Class 66 - 68:
When the computation is completed, the frequency distribution will be:
Solving (b): Mean and standard deviation using 1-VarStats
Using 1-VarStats, the solution is:
<em>See attachment for result of 1-VarStats</em>
Answer:
Step-by-step explanation:
-7x + 12 + (-8x + 48) = 180
-15x + 60 = 180
-15x = 120
x = -8
m<BDC= -7(-8)+ 12 = 56 + 12 = 68
m<CBA = -8(-8)+48 = 64+48 = 112
The sample % of these two populations would be 100/size (of student body at each school) x 100 so this would compare the two student bodies preferences for the particular type of candy bar. However, the actual % of the whole student body at each school would be a factor also. If the high school only had 200 students then this would be 50% representative but if the middle school had say 500 students this would only be 20% representative so this would have to be taken into account too. It might be more representative to have the same % of the student bodies respectively for the sample.
