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masha68 [24]
3 years ago
10

Help, with this multiple choice question?

Mathematics
1 answer:
lbvjy [14]3 years ago
8 0

he would hove to work 20 hours to meet the second job I think

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andre [41]

Yes because volume has 3 dimensions, hence the power of 3

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What two products would you add to find 713​x48
umka21 [38]

9514 1404 393

Answer:

  40·713 and 8·713

Step-by-step explanation:

When this multiplication is carried out "by hand", the usual sum of partial products is ...

  8·713 + 40·713

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Step-by-step explanation:

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3 years ago
If country A has 161,500 million more computers than country B and the total number of computers for both countries is 321,050 m
Vladimir [108]

Answer:

b has 79,775 million

a has 241,275 million

Step-by-step explanation:

a has more than b

b+161,500 is a

b=b

b+b+161,500=321,050

2b=321,050-161,500

2b=159,550

b=159,550/2

b=79,775

a=b+161,500

a=79,775+161,500

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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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