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3 years ago

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A library building is in the shape of a rectangle. Its floor has a length of (3x + 5) meters and a width of (5x − 1) meters. The

expression below represents the area of the floor of the building in square meters: (3x + 5)(5x − 1) Which of the following simplified expressions represents the area of the floor of the library building in square meters? (5 points)

1 answer:

Alecsey [184]3 years ago

4 0

Answer:

(16x + 8)m²

Step-by-step explanation:

area of a rectangle = 2length + 2width

=2(3x + 5) + 2(5x-1)

=6x + 10 + 10x - 2

=(16x + 8)m²

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Suppose that in a random selection of 100 colored candies, 26% of them are blue. The candy company claims that the percentage

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Answer: a) -0.2252, b) 0.8219

Step-by-step explanation:

Since we have given that

Sample size n = 100

Probability that candies are blue = p= 0.26

Probability that company claims that it is blue candy = P = 0.27

So, Q = 1-P= 1-0.27 = 0.73

So, Null hypothesis : $H_0:p=P$

Alternate hypothesis : $H_1:p\neq P$

So, the test statistic would be

$z=\dfrac{p-P}{\sqrt{\dfrac{P.Q}{n}}}\\\\z=\dfrac{0.26-0.27}{\sqrt{\dfrac{0.27\times 0.73}{100}}}\\\\z=-0.2252$

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3 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

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$\\ P(z$

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

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