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s344n2d4d5 [400]
2 years ago
12

What is the solution to 5(3-x)?

Mathematics
1 answer:
Anit [1.1K]2 years ago
8 0
The answer is 15 - 5 x

Multiply 5 by 3 ( 15 )
Multiply 5 by -x ( -5x )

I hope this helps and please mark brainliest!
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koban [17]
9/3 is already made a fraction.

Why?

Because it is not a whole numer.

Well, this improper fraction can be simplified.

<u>9 </u>  ÷ 3<u>
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4 0
3 years ago
Read 2 more answers
Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0
3 years ago
A university will add fruit juice vending machines to its classroom buildings if the student body president is convinced that mo
Artemon [7]

Answer:

E) The sample size is n = 1,000, and 50 percent of all students use the vending machines.

Step-by-step explanation:

CollegeBoard

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Shipping rates for Company A and Company B are shown in the tables below. Which company has shipping rates that you can represen
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Answer:

OMG I KNOW THE ANSWER!!!!!!!!

Step-by-step explanation:

its b.

5 0
3 years ago
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Jun 07, 9:22:26 AM
german

Answer:

i dont know

Step-by-step explanation:

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