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storchak [24]
2 years ago
8

Write a system of two equations in two variables to solve the problem. when fully extended, a ladder is 28 feet in length. if th

e extension is 4 feet shorter than the base, how long is each part of the ladder? base
Mathematics
1 answer:
Alisiya [41]2 years ago
5 0
Let the base of the ladder be x ft and the extension be y ft.

"when fully extended, a ladder is 28 feet in length."

means: x+y=28

"<span>the extension is 4 feet shorter than the base</span>"

means: y=x-4,   that is x-y = 4, (taking y to the side of x, and -4 to the other side of x and y)

thus, we have the system of equations:

1) x+y= 28 

2) x-y=4

adding the 2 equations, we have:

(x+y)+(x-y)=28+4

2x=32

x=32/2=16


x+y=28, so 16+y=28, thus y=12


Answer: Base: 16, Extension 12
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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 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 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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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