PLEASE HELP I WILL MARK YOU BRAINLIEST. FIND THE AREA

Mathematics

1 answer:

Mandarinka [93]2 years ago

7 0

Answer:

<em>The family will need to purchase 187 square ft carpeting.</em>

Step-by-step explanation:

<u>Area of Rectangles</u>

Given a rectangle with width W and length H, the area is calculated as follows:

A = W * H

The new carpet of the family has the shape provided in the figure. We have completed the shape to add some missing dimensions by adding and subtracting the given lengths. Please refer to the attached image.

The total area of the carpet is the sum of the areas of three rectangles.

The rectangle at the bottom has dimensions 3 ft x 20 ft, thus its area is:

A1 = 3 ft * 20 ft = 60 square ft

The rectangle at the middle has dimensions 3 ft x 17 ft, thus its area is:

A2 = 3 ft * 17 ft = 51 square ft

The rectangle at the top has dimensions 4 ft x 19 ft, thus its area is:

A3 = 4 ft * 19 ft = 76 square ft

The total area of the carpet is:

A = 60 square ft + 51 square ft + 76 square ft

A = 187 square ft

The family will need to purchase 187 square ft carpeting.

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velikii [3]

Answer:

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .