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mel-nik [20]
2 years ago
8

Phoebe surveyed teens and adults to find the average number of hours of sleep they get during the week. Her results are shown be

low. Which statement regarding her survey results is true?
Hours of Sleep for Teens
A box-and-whisker plot. The number line goes from 1 to 12. The whiskers range from 2.5 to 12, and the box ranges from 5 to 8.5. A line divides the box at 7.

Hours of Sleep for Adults
A box-and-whisker plot. The number line goes from 1 to 12. The whiskers range from 3 to 11, and the box ranges from 5 to 7.5. A line divides the box at 7.
Adults get more hours of sleep on average than teens.
Adults have less variability in the hours of sleep.
Teens have less variability in the hours of sleep.
Teens get less sleep on average than adults.
Mathematics
2 answers:
Thepotemich [5.8K]2 years ago
8 0

Answer:

its the second option

Step-by-step explanation:

i did the test and got 100

Keith_Richards [23]2 years ago
8 0

Answer:

B: Adults have less variability in the hours of sleep.

Step-by-step explanation:

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Cristine's middle school has a total of 900 students and 45 teachers. The local high school has 110 teachers and a student-teach
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Write and solve an equation of ratios:

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Cross multiplying, we get (900)(110) / 45, or

                                                                 
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A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?
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Answer:

(a) Selection = 6724520

(b) At\ most\ 12 = 6553976

(c) At\ most\ 8 = 6066720

(d) At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

Step-by-step explanation:

Given

Colors = 8

Balloons = 28 --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

n => 28 + 8-1= 35

r = 28

Selection = ^{35}^C_{28

Selection = \frac{35!}{(35 - 28)!28!}

Selection = \frac{35!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29}{7!}

Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}

Selection = \frac{33891580800}{5040}

Selection = 6724520

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

n => 15 + 8 -1 = 22

r = 15

Selection of at least 13 =

At\ least\ 13 = ^{22}C_{15}

At\ least\ 13 = \frac{22!}{(22-15)!15!}

At\ least\ 13 = \frac{22!}{7!15!}

At\ least\ 13 = 170544

Ways of selecting at most 12  =

At\ most\ 12 = Total - At\ least\ 13 --- Complement rule

At\ most\ 12 = 6724520- 170544

At\ most\ 12 = 6553976

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

n => 19+ 8 -1 = 26

r = 19

Selection of at least 9 =

At\ least\ 9 = ^{26}C_{19}

At\ least\ 9 = \frac{26!}{(26-19)!19!}

At\ least\ 9 = \frac{26!}{7!19!}

At\ least\ 9 = 657800

Ways of selecting at most 8  =

At\ most\ 8 = Total - At\ least\ 9 --- Complement rule

At\ most\ 8 = 6724520- 657800

At\ most\ 8 = 6066720

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

n =6+6-1 = 11

r = 6

Selection =

^{11}C_6 = \frac{11!}{(11-6)!6!}

^{11}C_6 = \frac{11!}{5!6!}

^{11}C_6 = 462

Using inclusion/exclusion rule of two sets:

Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)

Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462

Only\ 12\ red\ and\ only\ 8\ blue\ = 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  6724520 - 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

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A. There are not 15 successes and 15 failures. A confidence interval can be computed by adding 2 successes and 2 failures.

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