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Citrus2011 [14]
3 years ago
7

For this problem what I did was add all the measurements and I got 48 m. However, it is wrong. How do I go about solving the per

imeter then?

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
8 0

9514 1404 393

Answer:

  66 m

Step-by-step explanation:

The perimeter is the sum of the measures of <em>all</em> of the sides. There are two side measures that are missing from the diagram.

The missing horizontal measure is ...

  17 m - 8 m = 9 m

The missing vertical measure is ...

  16m -7 m = 9 m.

If you add these to the sum you already calculated, you will get the correct answer:

  48 m + 9 m + 9 m = 66 m . . . perimeter of the figure

_____

If you're paying attention, you see that the sum of the measures of the two shorter horizontal segments is the same as the measure of the longer horizontal segment. Likewise, the sum of the measurements of the two shorter vertical segments is the same as that of the longer vertical segment.

In other words, the perimeter of this (and any) L-shaped figure is the same as the perimeter of a rectangle having the same horizontal and vertical dimensions as the long sides of the figure.

  P = 2(17 m +16 m) = 2(33 m) = 66 m

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Step-by-step explanation:

So we get 5 hrs.

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4 0
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.

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

\\ 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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Step-by-step explanation:

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<em>Comment on the formula</em>

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