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3 years ago

Combine and simplify the radical below. Sqrt8 × sqrt20

Mathematics

2 answers:

Colt1911 [192]3 years ago

8 0

Once multiplied it would be Square Root of 160. 160 once simplified it would be 4 Root 10.

Serjik [45]3 years ago

7 0

I got the answer t^14

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Doug is 2 more than twice the age of Joey. The sum of their ages is 74. How old is Doug?

Citrus2011 [14]

The answer would be 37

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3 years ago

The value of 7 in 67908 is blank times the value of 7 in 74702

Anastasy [175]

The value of 7 in 67098 is 7000 and the value of 7 in 74702 is 70000 or 700 so it is either 10 times or 1/10th of the the value.

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3 years ago

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I need this immediately!! Put the equation in standard form with a positive X coefficient. Y + 6 = 3/2 ( x - 4 ) HELP!!

Readme [11.4K]

Answer:

see explanation

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

given

y + 6 = $\frac{3}{2}$ (x - 4)

Multiply through by 2 to eliminate the fraction

2y + 12 = 3(x - 4) ← distribute

2y + 12 = 3x - 12 ( subtract 2y from both sides )

12 = 3x - 2y - 12 ( add 12 to both sides )

24 = 3x - 2y, that is

3x - 2y = 24 ← in standard form

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3 years ago

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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 distribution of sample means. 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 normally 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 normal standard distribution, 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 distribution of sample means. 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 standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then