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9966 [12]
3 years ago
9

The radius of a circular rug is 4 feet. How much ribbing will you need to buy to go around the rug? Use 3.14 for ?. A) 13 feet B

) 26 feet C) 51 feet D) 102 feet D) 628 feet
Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer:

C =25.12 ft

I would get 26 ft

B) 26 feet

Step-by-step explanation:

To go around the rug, we are looking for circumference

C = pi*d = 2*pi*r

C = 2*pi*4

C = 8pi

We are approximating pi by 3.14

C = 8*3.14

C =25.12 ft

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A rectangular deck has an area of 320ft^2. the length of the deck is 4 feet longer than the width. find the dimensions of the de
Digiron [165]
A = L * W
A = 320
L = W + 4

320 = W(W + 4)
320 = W^2 + 4W
W^2 + 4W = 320
W^2 + 4W + 4 = 320 + 4
(W + 2)^2 = 324
W + 2 = (+-) sqrt 324
W = -2 (+-) 18

W = -2 + 18 = 16 ft <== this is the width
W = -2 - 18 = -20....not this one because it is negative

L = W + 4
L = 16 + 4
L = 20 ft <=== this is the length

in summary...the width is 16 ft and the length is 20 ft 

7 0
3 years ago
Find the average value of f over region
yan [13]
The area of D is given by:

\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx  \\  \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}

The average value of f over D is given by:

\frac{1}{ \frac{343}{3} }  \int\limits^7_0  \int\limits^{x^2}_0 {4x\sin(y)} \, dydx  = -\frac{3}{343}  \int\limits^7_0 {4x\cos(x^2)} \, dx  \\  \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49
3 0
3 years ago
Find the radius of a sphere whose surface area is 154cm².​with stepss
worty [1.4K]

Answer:

THIS IS YOUR ANSWER:

area \: of \: spere \:  = 4\pi \: r {}^{2}  \\ 154cm {}^{2} = 4 \times  \frac{22}{7}  \times r {}^{2}

\frac{154 \times 7}{4 \times 22} = r {}^{2} \\ r {}^{2} = 12.25 \\

r =  \sqrt{12.25} \\ r = 3.5cm

<em>✍️</em><em>HOPE</em><em> </em><em>IT HELPS</em><em> </em><em>YOU</em><em> </em><em>✍️</em>

7 0
2 years ago
Read 2 more answers
Explain why a two-dimensional net is useful for finding the surface area of a three-dimensional figure. Answer as soon as possib
Allushta [10]
A two-dimensional net is useful because it flattens the 3-d figure so that you can see all the faces of the figure.  You can calculate the area of each face to find the surface area. 
5 0
3 years ago
"A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean
Arada [10]

Answer:

85.31% probability that their mean rebuild time exceeds 8.1 hours.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 8.4, \sigma = 1.8, n = 40, s = \frac{1.8}{\sqrt{40}} = 0.2846

If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours.

This is 1 subtracted by the pvalue of Z when X = 8.1. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.1 - 8.4}{0.2846}

Z = -1.05

Z = -1.05 has a pvalue of 0.1469

1 - 0.1469 = 0.8531

85.31% probability that their mean rebuild time exceeds 8.1 hours.

4 0
3 years ago
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