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gogolik [260]
3 years ago
7

Deon does a weekly exercise program consisting of cardiovascular work and weight training. Each week, he exercises for at least

12 hours. He spends at most 10 hours doing cardiovascular work. He spends at most 6 hours on weight training. Let x denote the time (in hours) that Deon spends doing cardiovascular work. Let y denote the time (in hours) that he spends on weight training. Shade the region corresponding to all values of x and y that satisfy these requirements

Mathematics
1 answer:
LenKa [72]3 years ago
7 0

Answer:

  see attached

Step-by-step explanation:

The line at the lower left of the shaded region is x+y=12. The shaded area is above that line because is represents a minimum amount of exercise.

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After a sleepover, 3/4 of a large pepperoni remained. The next day, he ate 1/8 of what was left over. What fraction of a large p
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3 years ago
How do I do this Q11
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3 years ago
What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean
Elena-2011 [213]

Answer:

0.658 is the probability that a sample 90 test takers will provide a sample mean test score within 10 points of the population mean of 502.

Step-by-step explanation:

The following information is missing:

The standard deviation of population is 100.

We are given the following information in the question:

Population mean, μ = 502

Standard Deviation, σ = 100

Sample size, n  = 90

Standard error =

\dfrac{\sigma}{\sqrt{n}} = \dfrac{100}{\sqrt{90}} =  10.54

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(test score within 10 points)

P(492 \leq x \leq 512) \\\\= P(\displaystyle\frac{492 - 502}{10.54} \leq z \leq \displaystyle\frac{512-502}{10.54}) \\\\= P(-0.9487 \leq z \leq 0.9487)\\= P(z \leq 0.9487) - P(z < -0.9487)\\= 0.829 -0.171 = 0.658 = 65.8\%

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0.658 is the probability that a sample 90 test takers will provide a sample mean test score within 10 points of the population mean of 502.

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