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Eddi Din [679]
3 years ago
13

If the area of a rectangle is 50m^2 what is the perimeter?

Mathematics
2 answers:
elixir [45]3 years ago
4 0

Answer:

Step-by-step explanation:

2l+2w= 50

l= 4w-8

2 (4W - 8) + 2W = 50.

2 (4W - 8) + 2W = 50

8W - 16 + 2W = 50

10W - 16 = 50

10W = 66

W = 6.6 meters

Solve for L

L = 4W - 8

L = 4 (6.6) -8

L = 26.4 - 8

L = 18.4

Dimensions are 6.6 and 18.4

I hope this is right

nika2105 [10]3 years ago
3 0

It's most likely 30 m

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a. The positive difference between Nelson's height and the population mean is: \\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

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d. Nelson's height is <em>usual</em> since \\ -2 < -1.174 < 2.

Step-by-step explanation:

The key concept to answer this question is the z-score. A <em>z-score</em> "tells us" the distance from the population's mean of a raw score in <em>standard deviation</em> units. A <em>positive value</em> for a z-score indicates that the raw score is <em>above</em> the population mean, whereas a <em>negative value</em> tells us that the raw score is <em>below</em> the population mean. The formula to obtain this <em>z-score</em> is as follows:

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Where

\\ z is the <em>z-score</em>.

\\ \mu is the <em>population mean</em>.

\\ \sigma is the <em>population standard deviation</em>.

From the question, we have that:

  • Nelson's height is 68 in. In this case, the raw score is 68 in \\ x = 68 in.
  • \\ \mu = 70.7in.
  • \\ \sigma = 2.3in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson​'s height and the​ mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

That is, <em>the positive difference is 2.7 in</em>.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

To find how many <em>standard deviations</em> is that, we need to divide that difference by the <em>population standard deviation</em>. That is:

\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174

In words, the difference found in part (a) is 1.174 <em>standard deviations</em> from the mean. Notice that we are not taking into account here if the raw score, <em>x,</em> is <em>below</em> or <em>above</em> the mean.

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Using formula [1], we have

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d. If we consider​ "usual" heights to be those that convert to z scores between minus2 and​ 2, is Nelson​'s height usual or​ unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

\\ -2 < z_{Nelson} < 2

\\ -2 < -1.174 < 2

Then, Nelson's height is <em>usual</em> according to that statement.  

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