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tia_tia [17]
4 years ago
14

Mrs. Long keeps a garden in her backyard. The garden is in the shape of a rectangle. The garden is 27 feet wide. If the perimete

r of the garden is 200 feet, what is the length of Mrs. Long’s garden
Mathematics
1 answer:
uranmaximum [27]4 years ago
7 0

Answer:

so Area = length x width

Area = 450,  width = 9

450 = 9 x length

divide 450 by 9, and you have the length.

Perimeter = length + length+ 9 + 9

and you got your perimeter!

Step-by-step explanation:

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3 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 111.4-cm and a standard dev
Kipish [7]

Answer:

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

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 = 111.4, \sigma = 0.5, n = 23, s = \frac{0.5}{\sqrt{23}} = 0.1043

Find the probability that the average length of a randomly selected bundle of steel rods is between 111.2-cm and 111.4-cm.

This is the pvalue of Z when X = 111.4 subtracted by the pvalue of Z when X = 111.2. So

X = 111.4

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

By the Central Limit Theorem

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

Z = \frac{111.4 - 111.4}{0.1043}

Z = 0

Z = 0 has a pvalue of 0.5.

X = 111.2

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

Z = \frac{111.2 - 111.4}{0.1043}

Z = -1.92

Z = -1.92 has a pvalue of 0.0274.

0.5 - 0.0274 = 0.4726

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

5 0
3 years ago
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