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MAXImum [283]
2 years ago
13

Find the circumfrance and the area of a circle with a 5 ft radius. Use the value 3.14

Mathematics
1 answer:
Crazy boy [7]2 years ago
6 0

Answer:

Circumference is 2piR =31.4

Area is pir^2=78.5

Step-by-step explanation:

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Evaluate the series.<br><br> A)125<br> B) 38<br> C) 210<br> D) 165
Reika [66]
165 is answer we also use a formula
5 ( n/2(2a + ( n-1)d)
in that n=6 ,a=3 , d= 1

4 0
3 years ago
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Safiya deposited money into a bank account that earned 6.5% simple interest each year. After 3 1/2 years, she had earned $36.40
N76 [4]

Answer:

$1,792

Step-by-step explanation:

3 1/2×36.40=116.48

116.48÷6.5%=1,792

$1,792

5 0
3 years ago
Find f if f(x)=23; 23= x²+7
ioda
Don't you mean "x?"
8 0
3 years ago
How does a person use credit to make a
Black_prince [1.1K]

Answer:

makes a promise to pay at some point in the future

Step-by-step explanation:

Using credit means you borrow money to buy something, which is called a "loan", then you can buy the things you want, and you can pay it back later (with interest).

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
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