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mafiozo [28]
3 years ago
11

A rectangle has a length that is 2 meters more than the width. The area of the rectangle is 288 square meters. Find the dimensio

ns of the rectangle.
Mathematics
1 answer:
Rufina [12.5K]3 years ago
7 0

Answer:

L = 18 and w = 16

Step-by-step explanation:

The area of a rectangle is found by A = l*w. Since the length here is 2 more than the width or 2 + w and the width is w, substitute these values and A = 288 to solve for w.

A = l*w\\288 = w(2+w)\\288 = w^2 + 2w

To solve for w, move 288 to the other side by subtraction. Then factor and solve.

w^2 + 2w  - 288 = 0 \\(w +18)(w-16) = 0\\

Set each factor equal to 0 and solve.

w - 16 = 0 so w = 16

w + 18 = 0 so w = -18

Since w is a side length and length/distance cannot be negative, then w = 16 is the width of the rectangle.

This means the length is 16 + 2 = 18.

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Answer:

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Step-by-step explanation:

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3 years ago
A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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 question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

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

X = 205

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

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

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

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

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

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

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

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

7 0
3 years ago
Will be giving brainliest to the correct one, no links or account will be reported:)
Y_Kistochka [10]

Answer:

63

Step-by-step explanation:

64+53+ (ㅣ) = 180

117 + (ㅣ) = 180

(ㅣ) = 63

4 0
2 years ago
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