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lana66690 [7]
3 years ago
6

Mrs. Martin wants to place a ribbon around the outer edge of a rectangular mirror. The area of the mirror is 324 square inches.

The width of the mirror is 12 inches. How many inches of ribbon does Mrs. Martin need?
Mathematics
2 answers:
lesya692 [45]3 years ago
6 0
78, because 324 divided by 12 is 27. this means that the width is 12 and the length is 27. 27 times 2 equals 54 and 12 times 2 equals 24. 54+24= 78

Ostrovityanka [42]3 years ago
5 0

Answer:

78 inches of ribbon does Mrs. Martin need

Step-by-step explanation:

Area of a rectangle (A)  is given by:

A = lw                          .....[1]

where,

l is the length of the rectangle and w is the width of the rectangle.

As per the statement:

The area of the mirror is 324 square inches. The width of the mirror is 12 inches.

⇒A = 324 square inches and w = 12 inches.

By [1] we have;

324 = 12l

Divide both sides by 12 we have;

27 = l

or

l = 27 inches

We have to find the inches of ribbon does Mrs. Martin need.

Perimeter of the rectangle(P) = 2(l+w)

Substitute the given values we have;

P = 2 \cdot (27+12) = 2 \cdot 39 = 78 in

Therefore, 78 inches of ribbon does Mrs. Martin need

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Answer:

54-1.64\frac{21}{\sqrt{16}}=45.39    

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So on this case the 90% confidence interval would be given by (45.39;62.61)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=54 represent the sample mean  

\mu population mean (variable of interest)

\sigma=21 represent the sample standard deviation

n=16 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.90 or 90%, the value of \alpha=1-0.9=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

Now we have everything in order to replace into formula (1):

54-1.64\frac{21}{\sqrt{16}}=45.39    

54 +1.64\frac{21}{\sqrt{16}}=62.61    

So on this case the 90% confidence interval would be given by (45.39;62.61)    

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4 years ago
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