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evablogger [386]
3 years ago
13

Can someone help me?​

Mathematics
2 answers:
Soloha48 [4]3 years ago
8 0
PLS I NEED HELP ASAP I DONT HAVE TIME IT ALSO DETECTS IF IT RIGHT OR WRONG.
kirill [66]3 years ago
6 0

Answer:

X=25

Step-by-step explanation:

180-5 =

175 divided by 7 = 25

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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
What is the measure of angle CAD?​
frez [133]
<h3>Answer: 128 degrees</h3>

Work Shown:

(angle CAD) + (angle CAB) = 180

angle CAD = 180 - (angle CAB)

angle CAD = 180 - 52

angle CAD = 128 degrees

7 0
3 years ago
Read 2 more answers
By dividing one quantity by the other
Step2247 [10]

Answer:

The method of comparing two quantities by dividing one quantity by the other is called ratio.

Step-by-step explanation:

4 0
2 years ago
Please help for test due today
krek1111 [17]
I don’t really know, but it could be b or c. If none then just pick a question :D
3 0
3 years ago
Thelma and louise drove 341 miles and used 15.5 gallons of gas how many miles did their car get per gallon of gas
poizon [28]

Answer:

22 MPG

Step-by-step explanation:

The miles and the gallons result in the formula Miles per gallon. Which can be substituted as (341 miles) per (15.5 gallons). If you simplify it, the result is  22miles per gallon.

4 0
2 years ago
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