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azamat
2 years ago
7

The diagram below shows the dimensions of a rectangular house with a rectangular backyard. What is the total area, in square fee

t, of the house and backyard?

Mathematics
1 answer:
nignag [31]2 years ago
7 0

Answer:

D) 5,200

Step-by-step explanation:

The formula to find the area of a r rectangle: l x w, where l is the length and w is width.

Area of the backyard: 50 x 80

= 4,000 ft

Area of the house: 40 x 30

= 1,200 ft

Total = 4,000 + 1,200

= 5,200 ft²

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a and d

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The factored form of a quadratic equation is y=(2x+1)(x-5), and the standard form is y=2x²-9x-5. Which of the following statemen
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Step-by-step explanation:

As you can see as explained in the pictures

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Write an equivalent expression<br><br> -3 + 2/3y - 4 - 1/3y
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Vera_Pavlovna [14]

Answer:

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

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 z-score 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 p-value, 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.

Mean of 20 hours, standard deviation of 6:

This means that \mu = 20, \sigma = 6

Sample of 150:

This means that n = 150, s = \frac{6}{\sqrt{150}}

What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So

X = 21

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

By the Central Limit Theorem

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

Z = \frac{21 - 20}{\frac{6}{\sqrt{150}}}

Z = 2.04

Z = 2.04 has a p-value of 0.9793

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

Z = \frac{19.5 - 20}{\frac{6}{\sqrt{150}}}

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