a random sample of 21 students from the U and at A&T were surveyed about their music listening habits. Their responses (in hours, each square represents 0.5 hour) were recorded and displayed using the following histograms. The left represents U. The right represents SHSAT. Which data set(U or SHSAT students) has a larger standard deviation? Explain.

 U

Step-by-step explanation:

Looking at the graphs, U is skewed right and SHSAT is more of a normal distribution. Generally, this means the standard deviation would be greater for the data set that is skewed.

Let's find the standard deviation for each data set to confirm our theory.

Step 1) Find the mean number of hours

Step 2) Subtract each data value from the mean

Step 3) Find the mean of Step 2

U

hours x frequency: 1(3) + 2(5) + 3(4) + 4(3) + 5(3) + 6(2) + 9(1) = 72

Step 1: mean = (hours x frequency)/frequency = 72/21 = 3.4

Step 2: 2.4(3) + 1.4(5) + 0.4(4) + 0.6(3) + 1.6(3) + 2.6(2) + 5.6(1) = 33.2

Step 3: standard deviation = (sum from step 2)/ frequency = 33.2/21 = 1.58

SHSAT

hours x frequency: 1(1) + 2(3) + 3(4) + 4(5) + 5(3) + 6(3) + 7(2) = 86

Step 1: mean = (hours x frequency)/frequency = 86/21 = 4.1

Step 2: 3.1(1) + 2.1(3) + 1.1(4) + 0.1(5) +0.95(3) + 1.9(3) + 2.9(2) = 28.5

Step 3: standard deviation = (sum from step 2)/ frequency = 28.5/21 = 1.36

The calculations confirm that U has the greater standard deviation

no

step-by-step explanation:

2-7=-5

7-2=5

when you combing like terms for ech expression, 5 and -5. these numbers are not equal to each other.

the ratio of balloons to tables is 5: 2 and the constant variation is 5.

the equation is y=5x