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lorasvet [3.4K]
3 years ago
5

Could you please work out this and show answers if possible :)

Mathematics
1 answer:
g100num [7]3 years ago
5 0

Answer:‍ here !!

Step-by-step explanation:

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PLEASE HELP ASAP!!<br><br> Solve the equation<br><br> |x−7|=8
steposvetlana [31]

here you go -brainliest

6 0
3 years ago
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
Which expression have the same value as the product of 2.7 and 4
Maurinko [17]

Answer:

Step-by-step explanation:

2.7*4=10.80

27*0.4=10.80 we multiplied 27 by 10 but divide 4 by 10

0.27*40=10.80

3 0
3 years ago
Piz help me fast asap thx​
Effectus [21]

Answer:

y= 7703.5

Step-by-step explanation:

multiply 7100 by 1.085

5 0
3 years ago
-2 4/5 + 1/4 equals what ?
vlada-n [284]
Hey there! :)

Since 2 4/5 is greater than 1/4, the answer is negative.

You can just subtract 2 4/5 and 1/4 and add the (-) sign

2 4/5 - 1/4

Change the fractions so they can have a common denominator: 20

2 4/5 = 2 16/20

1/4 = 5/20

2 16/20 - 5/20 = 2 11/20

Add the (-) sign → -2 11/20

Answer ⇒ -2 11/20

:)
5 0
3 years ago
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