Answers:
1.)
C - Final exams: midterm = Q3 92 - Q1 80 = 12 and finals = Q3 85 - Q1 78 = 7, so finals exam has the smaller IQR.
2.)
B - exam median is much higher than the class median: refer to the attached image of class and exam box and whisker graph
3.)
A - <span>
IQR is a better measure of spread for movies than it is for basketball games: because the data for movies are quite wide or apart from each other than the data in the basketball games
4.)
A - </span><span>
mean for April is higher than October's mean: mean in April is 67 and mean in October is 60
5.)
A - </span><span>
Neither data set has suspected outliers: meaning there's no data that is further apart from the group or set
6.)
B - </span>
There is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical
<span>
7.)
C - </span>
college spread is best described by the IQR. The high school spread is best described by the standard deviation: this is because of the wide range of data in the college than the data in the high school.
Answer:
![\displaystyle 2^{\frac{6}{5}}=2\sqrt[5]{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%5E%7B%5Cfrac%7B6%7D%7B5%7D%7D%3D2%5Csqrt%5B5%5D%7B2%7D)
Step-by-step explanation:
<u>Fractional Exponents</u>
An expression like

can be expressed as a radical of the form:
![\sqrt[m]{a^n}](https://tex.z-dn.net/?f=%5Csqrt%5Bm%5D%7Ba%5En%7D)
We have the expression:

Its equivalent radical form is:
![\displaystyle 2^{\frac{6}{5}}=\sqrt[5]{2^6}](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%5E%7B%5Cfrac%7B6%7D%7B5%7D%7D%3D%5Csqrt%5B5%5D%7B2%5E6%7D)
Since the exponent is greater than the index of the radical, we can take 2 out of it by following the procedure:
![\sqrt[5]{2^6}=\sqrt[5]{2^5\cdot 2}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E6%7D%3D%5Csqrt%5B5%5D%7B2%5E5%5Ccdot%202%7D)
Taking out
from the radical:
![\sqrt[5]{2^6}=2\sqrt[5]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E6%7D%3D2%5Csqrt%5B5%5D%7B2%7D)
Thus:
![\displaystyle 2^{\frac{6}{5}}=2\sqrt[5]{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%5E%7B%5Cfrac%7B6%7D%7B5%7D%7D%3D2%5Csqrt%5B5%5D%7B2%7D)
9514 1404 393
Answer:
32 yd²
Step-by-step explanation:
Use the area formula with the given dimensions.
A = 1/2(b1 +b2)h . . . . . base lengths b1, b2; height h
A = 1/2(2 +6)8 = 1/2(8)(8) = 32 . . . . square yards
The area of the trapezoid is 32 square yards.
Add up all the numbers that have xs on them and you should be good
Answer:
x=11/2
Step-by-step explanation: