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Leokris [45]
3 years ago
15

A green rectangle tile and a yellow rectangular tile are similar. The green tile has a length of 24 centimeters and a perimeter

of 80 centimeters. The yellow tile has a length of 72 centimeters. What is the perimeter of the yellow tile?
Mathematics
1 answer:
sesenic [268]3 years ago
8 0
"The perimeter of yellow tile is 240 cm."

Green tile and yellow tile are similar.
Perimeter of green tile = 80 cm
length of green tile = 24 cm
length of yellow tile = 72 cm
perimeter of yellow tile = ?
now you see if you multiply the length of green tile with 3 you get the length of yellow tile;
24 x 3 = 72 cm
As given that both tiles are similar so, if you multiply the perimeter of green tile with 3, you will get the perimeter of yellow tile;
80 x 3 = 240 cm
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By 3 cubic feet since 3 cubes is 9
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Please help!
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Because
\frac{f(x)}{x-3} = 2x^{2} + 10x - 1
therefore
f(x) = (x-3)(2x² + 10x - 1) + k, where k =  constant.

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6 0
3 years ago
A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th
Burka [1]

Answer:

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = \bar{x}=\frac{\sum x}{n}

Sample mean = \bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}

Sample mean = \bar{x}=23

Sample standard deviation = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}

Sample standard deviation = \sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table t_{\frac{\alpha}{2} = 3.883

Formula of confidence interval : \bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}

Substitute the values in the formula

Confidence interval : 23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 23 -3.883 \times \frac{1.72}{\sqrt{20}} to 23 + 3.883 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

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3 years ago
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I hope this helps you

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