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Serga [27]
3 years ago
10

Find the perimeter and area of the figure:

Mathematics
2 answers:
blagie [28]3 years ago
6 0

Answer:

The primeter is 12 inches

The area is 9 square inches

Step-by-step explanation:

The area of a square is calculated by multiplying side length by itself (all four side lengths would be same for square)

so the area is 3 × 3 = 9 square inches

The primeter is calculated by multiplying side length by 4

so the primeter is 4 × 3 = 12 square inches

erastova [34]3 years ago
3 0
The perimeter is 12 and the area is 9
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Step-by-step explanation:

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What is 124 decreased by 67%
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40.92

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What is the solution for f(n)=1.25n+6.25; 25???<br> Please answer this quickly
gogolik [260]
It would be f= 1.25n+6.25/n
                

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3 years ago
The shape of the distribution of the time required to get an oil change at a 20 minute oil change facility. However, records ind
Katyanochek1 [597]

Answer:

For a mean oil change time of 20.51 minutes there would be a​ 10% chance of being at or​ below

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 normally distributed samples are 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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 21.3, \sigma = 3.9, n = 40, s = \frac{3.9}{\sqrt{40}} = 0.6166

Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below?

This is the 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

By the Central Limit Theorem

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

-1.28 = \frac{X - 21.3}{0.6166}

X - 21.3 = -1.28*0.6166

X = 20.51

For a mean oil change time of 20.51 minutes there would be a​ 10% chance of being at or​ below

6 0
3 years ago
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