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STALIN [3.7K]
3 years ago
10

Find the total length of fencing required to fence the outer edges of a rectangular park that is 35 meters by 25 meters

Mathematics
1 answer:
kykrilka [37]3 years ago
7 0

Perimeter = 2(length) + 2(width)

                =   2(35)    +   2(25)

                =     70      +     50

                =             120

Answer: 120 meters

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You spend $77 shopping. You buy some shirts at $12 each, and spend $5 on food. How many shirts did you buy?
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You buy 6 shirts. So $77 minus $5=$72
Then you divide $72 by $12 and you get 6 shirts
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Gina has been present at school for 40 of the first 45 days of the school year. The school year is 180 days long. If Gina contin
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The answer is going to be 160.

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A circle has its center at the origin, and (5, -12) is a point on the circle. How long is the radius of the circle?
aivan3 [116]
Hello here is a solution :
<span> the radius of the circle is OA        A(5;-12) and O (0;0)
OA= </span><span>  square root ((5-0)² + (-12-0)²) 
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7 0
3 years ago
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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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Find the length of side a.
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The length of side A is 194
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