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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:

You might be interested in
Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. The results in the screen disp
e-lub [12.9K]

Answer:

D. is the correct answer, hope this helps!


4 0
2 years ago
Read 2 more answers
Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of
nikitadnepr [17]

Answer:

a) Reject H₀

b) [0.31; 3.35]

Step-by-step explanation:

Hello!

a) The objective of this example is to compare if the population means of the production rate of the assembly lines A, B and C. To do so the data of the production of each line were recorded and an ANOVA was run using it.

The study variable is:

Y: Production rate of an assembly line.

Assuming that the study variable has a normal distribution for each population, the observations are independent and the population variances are equal, you can apply a parametric ANOVA with the hypothesis:

H₀ μ₁= μ₂= μ₃

H₁: At least one of the population means is different from the others

Where:

Population 1: line A

Population 2: line B

Population 3: line C

α: 0.01

This test is always one-tailed to the right. The statistic is the Snedecor's F, constructed as the MSTr divided by the MSEr if the value of the statistic is big, this means that there is a greater variance due to the treatments than to the error, this means that the population means are different. If the value of F is small, it means that the differences between populations are not significant ( may differ due to error and not treatment).

The critical region is:

F_{k-1;n-k; 1-\alpha } = F_{2;15; 0.99} = 6.36

If F ≥ 3.36, the decision is to reject the null hypothesis.

Looking at the given data:

F= \frac{MSTr}{MSEr}= 11.32653

With this value the decision is to reject the null hypothesis.

Using the p-value method:

p-value: 0.001005

α: 0.01

The p-value is less than the significance level, the decision is to reject the null hypothesis.

At a level of 5%, there is significant evidence to say that at least one of the population means of the production ratio of the assembly lines A, B and C is different than the others.

b) In this item, you have to stop paying attention to the production ratio of the assembly line A to compare the population means of the production ratio of lines B and C.

(I'll use the same subscripts to be congruent with part a.)

The parameter to estimate is μ₂ - μ₃

The populations are the same as before, so you can still assume that the study variables have a normal distribution and their population variances are unknown but equal. The statistic to use under these conditions, since the sample sizes are 6 for both assembly lines, is a pooled-t for two independent variables with unknown but equal population variances.

t=  (X[bar]₂ - X[bar]₃) - ( μ₂ - μ₃) ~t_{n_2+n_3-2}

Sa√(1/n₂+1/n₃)

The formula for the interval is:

(X[bar]₂ - X[bar]₃) ± t_{n_2+n_3-2; 1 - \alpha /2}* Sa\sqrt{*\frac{1}{n_2} + \frac{1}{n_3} }

Sa^{2} = \frac{(n_2-1)*S_2^2+ (n_3-1)*S_3^2}{n_2+n_3-2}

Sa^{2} = \frac{(5*0.67)+ (5*0.7)}{6+6-2}

Sa^{2} = 0.685

Sa= 0.827 ≅ 0.83

t_{n_2+n_3-2;1-\alpha /2}= t_{10;0.995} = 3.169

X[bar]₂ = 43.33

X[bar]₃ = 41.5

(43.33-41.5) ± 3.169 * *0.83\sqrt{*\frac{1}{6} + \frac{1}{6} }

1.83 ± 3.169 * 0.479

[0.31; 3.35]

With a confidence level of 99% you'd expect that the difference of the population means of the production rate of the assemly lines B and C.

I hope it helps!

8 0
2 years ago
Match the following guess solutions ypyp for the method of undetermined coefficients with the second-order nonhomogeneous linear
Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

1 ) Given that

(d^2y/dx^2) + 4y = x - x^2 + 20\\\\ (d^2y/dx^2) + 4y =  - x^2 + x + 20

For a non homogeneous part - x^2 + x + 20 , we assume the particular solution is

y_p(x) = Ax^2 + Bx + C

2 ) Given that

d^2y/dx^2 + 6dy/dx + 8y = e^{2x}

For a non homogeneous part   e^{2x} , we assume the particular solution is

y_p(x) = Ae^{2x}

3 ) Given that

y′′ + 4y′ + 20y = −3sin(2x)

For a non homogeneous part −3sin(2x) , we assume the particular solution is

y_p(x) =  Acos(2x)+Bsin(2x)

4 ) Given that

y′′ − 2y′ − 15y = 3xcos(2x)

For a non homogeneous part  3xcos(2x)  , we assume the particular solution is

y_p(x) = (Ax+B)cos2x+(Cx+D)sin2x

4 0
2 years ago
Belinda is thinking about buying a car for $38,000. The table below shows the projected value of two different cars for three ye
Veseljchak [2.6K]

Answer:

108.01

Step-by-step explanation:

3 0
2 years ago
The graph shows the solution to the initial value problem y'(t)=mt, y(t0)= -4
Charra [1.4K]
y'(t)=mt
\\
\\ y=\int {mt} \, dt=m \int{t} \,dt=m \frac{t^2}{2} +C \\ \\y= \frac{mt^2}{2}+C

Now find equation of the graph. It passes through the point (2,3), and intersects y-axis at -2.

y=at^2+b
\\
\\3=a\times 2^2-2
\\
\\3=4a-2
\\
\\5=4a
\\
\\a= \frac{5}{4} 
\\
\\y=\frac{5}{4} t^2-2 \\ \\y= \frac{m}{2} t^2+C
\\ \\  \frac{m}{2} = \frac{5}{4} 
\\
\\m= \frac{5}{2} 
\\
\\C=-2
\\
\\y(t_0)=-4
\\
\\-4= \frac{5}{4} t_0^2-2
\\
\\ -2= \frac{5}{4} t_0^2
\\
\\ t_0^2=- \frac{8}{5} 
\\
\\t_0=\pm \sqrt{- \frac{8}{5} }



3 0
2 years ago
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