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Lena [83]
3 years ago
5

A poster is to have an area of 240 in² with 1-inch margins at the bottom and sides and a 2-inch margin at the top. Find the exa

ct dimensions that will give the largest printed area.
Mathematics
1 answer:
Rus_ich [418]3 years ago
3 0

Answer: 41 inches by 6 inches

Step-by-step explanation:

A poster is in form of a rectangle and the area is length × breadth.

If the area of the poster is 240in² with 1 inch margins at the bottom and sides and 2in margin at the top.

The extra length of the poster will be 1-in at the bottom plus 2-in margin at the top making additional height of 3-inches.

Also, the extra breadth the 1-in at the sides to give breadth of extra 2-inches in total.

The area of the extra length and breath will be 3inches × 2inches to give 6inches.

The area that will give the exact dimension will be 240inches+6inches = 246inches

The dimension an be 41inches by 6inches

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Answer:

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Step-by-step explanation:

For the first one just subtract

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2 years ago
The ideal size of a first-year class at a particular college is 150 students. The college, knowing from past experiences that on
katen-ka-za [31]

Answer:

6.18% probability that more than 150 first-year students attend this college.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

For each item selected, there are only two possible outcomes. Either it is defective, or it is not. This means that we use concepts of the binomial probability distribution to solve this problem.

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:

\sqrt{V(X)} = \sqrt{np(1-p)}

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.

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In this problem, we have that:

n = 450, p = 0.3

So

\mu = E(X) = np = 450*0.3 = 135

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{450*0.3*0.7} = 9.72

Approximate the probability that more than 150 first-year students attend this college.

This is 1 subtracted by the pvalue of Z when X = 150. So

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

Z = \frac{150 - 135}{9.72}

Z = 1.54

Z = 1.54 has a pvalue of 0.9382

1 - 0.9382 = 0.0618

6.18% probability that more than 150 first-year students attend this college.

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