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Irina18 [472]
4 years ago
13

Joshua wants to burn at least 400 calories per day, but no more than 600. He does this by walking and playing basketball. Assumi

ng he burns 4 calories per minute walking, w, and 5 calories per minute spent playing basketball, b, the situation can be modeled using these inequalities: 4w + 5b ≥ 400 4w + 5b ≤ 600 Which are possible solutions for the number of minutes Joshua can participate in each activity? Check all that apply.
40 minutes walking,
 40 minutes basketball 60 minutes walking,
 20 minutes basketball 20 minutes walking,
 60 minutes basketball 50 minutes walking,
50 minutes basketball 60 minutes walking,
80 minutes basketball 70 minutes walking,
60 minutes basketball
Mathematics
2 answers:
Artyom0805 [142]4 years ago
7 0

Answer:

The answers are D and F on edge.

Step-by-step explanation:

pentagon [3]4 years ago
3 0
50 minutes walking 50 minutes basketball and 70 minutes walking 60 minutes basketball
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W + 13 = 71. What's the number to replace the variable w
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<span>W + 13 = 71
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answer
w = 58
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Use the t-distribution to find a confidence interval for a difference in means μ1-μ2 given the relevant sample results. Give the
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Answer:

(a) The best estimate of \mu_{1}-\mu_{2} is 13.2.

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(c) The 95% confidence interval for the difference between two means is (7.90, 18.50).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means using a <em>t</em>-interval is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}

(a)

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean \bar x is a point-estimate of the population mean μ.

Similarly the point estimate of the difference between two means is:

\bar x_{1}-\bar x_{2}

Compute the point estimate of \mu_{1}-\mu_{2} as follows:

E(\mu_{1}-\mu_{2})=\bar x_{1}-\bar x_{2}\\=79.0-65.8\\=13.2

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(b)

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S_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}=\frac{(35-1)10.5^{2}+(20-1)7.2^{2}}{35+20-2}=89.311

Compute the critical value of <em>t</em> as follows:

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (35+20-2)}=t_{0.025, 53}=2.00

*Use a <em>t</em>-table.

Compute the margin of error as follows:

MOE=t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}

          =2.00\times\sqrt{89.311\times [\frac{1}{35}+\frac{1}{20}]} \\=5.298\\\approx5.30

Thus, the margin of error is 5.30.

(c)

Compute the 95% confidence interval for the difference between two means as follows:

CI=(\bar x_{1}-\bar x_{2})\pm MOE

      =13.2\pm 5.298\\=(7.902, 18.498)\\\approx (7.90, 18.50)

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