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jolli1 [7]
3 years ago
14

How do you find the circumference or a circle with only a diameter?​

Mathematics
2 answers:
goblinko [34]3 years ago
6 0

Answer:

pi * d

Step-by-step explanation:

Multiply 3.14 by the Diameter.

bogdanovich [222]3 years ago
3 0

To find the circumference of a circle that lists only as a diameter, you would do pi*d so lets say the diameter is 6, you would do pi*6 which would end up giving you 18.84. I hope this helps a bit! (:

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Find the perimeter of a square with a side length of 4x
julia-pushkina [17]

Answer:16x^{2}

Step-by-step explanation:

4 0
3 years ago
6. Pleassss helppp!
yaroslaw [1]

Answer:

Step-by-step explanation:

First expand the expression so you know the coefficients of x and y.

4(2x+y+2y) = 4(2x+3y) = 8x+12y

8x+12y  =  8x+12y

6x+7y  ≠  6x+12y

8x+y+2y  = 8x + 3y ≠  8x+12y

8x+4y+8y = 8x+12y = 8x+12y

3 0
3 years ago
Each gallon of shingle stain covers 120 square feet. How many gallons should you buy to cover 658 square feet?
hjlf
You need to divide 658 by 120 because you know one gallon covers 120 square feet.
6 0
3 years ago
A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed i
Hunter-Best [27]

Answer:

a)  σ/√n= 1.43 min

c) Margin of error 2.8028min

d) [30.1972; 35.8028]min

e) n=62 customers

Step-by-step explanation:

Hello!

The variable of interest is

X: Time a customer stays at a restaurant. (min)

A sample of 49 lunch customers was taken at a restaurant obtaining

X[bar]= 33 mi

The population standard deviation is known to be δ= 10min

a) and b)

There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;σ²/n)

Where μ is the population mean and σ²/n is the population variance of the sampling distribution.

The standard deviation of the mean is the square root of its variance:

√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min

c)

The CI for the population mean has the general structure "Point estimator" ± "Margin of error"

Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

The formula for the interval is:

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

The margin of error of the 95% interval is:

Z_{1-\alpha /2}= Z_{1-0.025}= Z_{0.975}= 1.96

d= Z_{1-\alpha /2}*(σ/√n)= 1.96* 1.43= 2.8028

d)

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

[33±2.8028]

[30.1972; 35.8028]min

Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.

e)

Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:

d= Z_{1-\alpha /2}*(σ/√n)

\frac{d}{Z_{1-\alpha /2}}= σ/√n

√n*(\frac{d}{Z_{1-\alpha /2}})= σ

√n= σ* (\frac{Z_{1-\alpha /2}}{d})

n=( σ* (\frac{Z_{1-\alpha /2}}{d}))²

n= (10*\frac{1.96}{2.5})²= 61.47≅ 62 customers

I hope this helps!

3 0
3 years ago
Does anyone know this?
maksim [4K]
I think its 24% because .24 x 50 is 12 but I could be doing it wrong, sorry I can't say for sure.
4 0
3 years ago
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