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Anna [14]
2 years ago
15

A rectangle has a width of 59 centimeters and a perimeter of 238 centimeters. What is the rectangle's length?

Mathematics
1 answer:
Mademuasel [1]2 years ago
5 0

Answer:

60 cm

Step-by-step explanation:

2x the width and 2x the length gives you the perimeter. This can be represented by the equation:

2w + 2L = P

Use the information given in the word problem to fill in the variables.

2(59) + 2L = 238

118 + 2L = 238

move 118 to the other side of the equation.

2L = 238 - 118

2L = 120

move 2 to the other side of the equation.

L = \frac{120}{2}

L = 60

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Answer:

4.01% probability that the cable company will keep the shopping channel.

Step-by-step explanation:

For each subscriber, there are only two possible outcomes. Either they watch the shopping channel at least once a week, or they do not. This means that we can solve this problem using the binomial probability distribution.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 100, p = 0.20

So

E(X) = 100*0.20 = 20

\sqrt{V(X)} = \sqrt{100*0.20*0.80} = 4

The cable company has decided to keep the shopping channel if the sample proportion is greater than 0.27. What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only 0.20?

This probability is 1 subtracted by the pvalue of Z when X = 0.27*100 = 27. So

Z = \frac{X - \mu}{\sigma}

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Z = 1.75 has a pvalue of 0.9599.

So there is a 1-0.9599 = 0.0401 = 4.01% probability that the cable company will keep the shopping channel.

5 0
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