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3 years ago

Number 3

Mathematics

1 answer:

sdas [7]3 years ago

3 0

Answer: Choice A) 1990 to 1991

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Explanation:

Let's go through the answer choices one by one.

For choice A, we go from the year 1990 to 1991 which is a change of 1 year. The wage goes from 3.80 to 4.25 over this timespan, which is an increase of 4.25-3.80 = 0.45 dollars. Divide the change in wage (0.45 dollars) over the change in change in time (1 year) to get 0.45/1 = 0.45; this indicates that the wage increased by 0.45 dollars per year over the timespan 1990 to 1991

Now onto choice B. We have a wage increase of 3.80-3.35 = 0.45 over a course of 9 years (since 1990-1981 = 9) so 0.45/9 = 0.05 is the rate of wage growth, meaning that the wage bumps up by a nickel each year. So far choice A is the winner.

Moving onto choice C, we have a wage increase of 0.25 dollars (3.35-3.10 = 0.25) over an 1 year period (1981-1980 = 1) so the rate of change for this slice of time is 0.25/1 = 0.25 dollars per year. Choice A is still the winner.

Finally, for choice D, this is over a year as well (1980-1979 = 1) and the wage increases by 0.20 dollars (3.10-2.90 = 0.20) leading to a rate of change to be 0.20/1 = 0.20

So choice A has the largest rate of change which is the same as saying it has the largest rate of wage increase. This shows why choice A is the answer.

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Solve for r write the answer in simplified form<br><br> 7 = - 7r

alex41 [277]

Answer:

r = -1

Step-by-step explanation:

To get r isolated, you have to divide each side by -7

7 divided by -7 is -1, therefore r = -1

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3 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$