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3 years ago

The formula for finding the perimeter of a rectangle with length, l, and width, w, is P = 2l + 2w.

2 answers:

7 0

P1=2(l+w)---->(equation 1)
after the length and the width are doubled
p2=2(2l+2w)
we will take 2 common factor out of the bracket
p2=2(2(l+w))
then from equation1
p2=2p1
then the answer is the choice number c

creativ13 [48]3 years ago

6 0

The perimeter will grow by a factor of 2, so C is the correct answer.

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What is the truth value for the following conditional statement?

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True I think is the answer

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3 years ago

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The width of a rectangle is 7 meters greater than its length. If the area of the rectangle is 170 square meters, write the quadr

soldier1979 [14.2K]

Since x is the length, then x+7 has to be the width.
The area of a rectangle can be found via the formula :
A = length * width
So if we replace the area with the given number (170) and the length, and width, with what we assumed, we get..
170 = x * (x+7)
[which is.. ]
170 = x^2 +7x
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And that is the answer.

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3 years ago

Can someone help me please!!?? Eleven and seven-hundredths added to four and twenty-five hundredth the sum is

zloy xaker [14]

Answer:

$15.32$

Step-by-step explanation:

Eleven and seven-hundredths $=11+\frac{7}{100}$

(Here, $\frac{1}{100}$ denotes the hundredth)

Four and twenty-five hundredth $=4+\frac{25}{100}$

Therefore,

Sum of Eleven and seven-hundredths added to four and twenty-five hundredth = $11+\frac{7}{100}+4+\frac{25}{100}$

$=15+\frac{32}{100}$

$=15+0.32$

$=15.32$

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3 years ago

A book sold 31,200 copies in its first month of release. Suppose this represents 8.4% of the number of copies sold to date. How

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4 years ago

The United States Coast Guard assumes the mean weight of passengers in commercial boats is 185 pounds. The previous value was lo

Valentin [98]

Answer:

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Step-by-step explanation:

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

Normal probability distribution

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 $\frac{\sigma}{\sqrt{n}}$ .

In this problem, we have that:

$\mu = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85$

The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

The probability of an extreme value below the mean.

This is the pvalue of Z when X = 177.6.

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

By the Central Limit Theorem

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

$Z = \frac{177.6 - 185}{3.85}$

$Z = -1.92$

$Z = -1.92$ has a pvalue of 0.0274.

So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.

The probability of an extrema value above the mean.

Measures above the mean have a positive z score.

So this probability is 1 subtracted by the pvalue of $Z = 1.92$

$Z = 1.92$ has a pvalue of 0.9726.

So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.

Total:

2*0.0274 = 0.0548 = 0.055

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

4 0

3 years ago