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4 years ago

Find lateral area of square pyramid

Mathematics

2 answers:

77julia77 [94]4 years ago

5 0

Answer: 352 m²

Step-by-step explanation:

You need to use this formula to calculate the lateral area of the square pyramid:

$LA=\frac{1}{2}pl$

Where "p" is the perimeter of the base and "l" is the slant height.

You need to calculate the perimeter of the square base with this formula:

$P=4s$

Where "s" is the lenght of any side of the square.

Then, this is:

$p=4(8m)=32m$

Substitute the slant height ( $l=22m$ ) and the perimeter into the formula $LA=\frac{1}{2}pl$ . Then you get:

$LA=\frac{1}{2}(32m)(22m)=352m^2$

AURORKA [14]4 years ago

4 0

Answer:

The lateral area of pyramid = 352 m²

Step-by-step explanation:

Points to remember

Lateral area of square pyramid = 2bh

Where b - Base of pyramid and

h - Slant height of pyramid

From the figure we can see a square pyramid

To find the lateral area of pyramid

Here b = 8 m and h = 22 m

Area = 2bh = 2 * 8 * 22

= 352 m²

Therefore the lateral area of pyramid = 352 m²

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Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

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

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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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

In this problem, we have that:

$\mu = 432, \sigma = 722, n = 40, s = \frac{722}{\sqrt{40}} = 114.16$

What is the probability that the mean gain for the sample was between 250 and 500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.

X = 500

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

By the Central Limit Theorem

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

$Z = \frac{500 - 432}{114.16}$

$Z = 0.6$

$Z = 0.6$ has a pvalue of 0.7257.

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

By the Central Limit Theorem

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

$Z = \frac{250 - 432}{114.16}$

$Z = -1.59$

$Z = -1.59$ has a pvalue of 0.0559.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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width = W

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