IO.7(4.2)+15.9=<br><br> 51.25+43.85+14=

Perform the following mathematical operation, and report the answer to the correct umber of significant figures 7.125*8.00

creativ13 [48]

Answer:23☺️

Step-by-step explanation:

4 0

3 years ago

Write a real world situation for 15x-20<130 (greater than or equal to is the symbol)

sammy [17]

15x<130+20, 15x<150 divided both sides by 15 ,the answer is 10

6 0

3 years ago

19 POINTS AND ILL GIVE BRAINLIEST!!

Archy [21]

Answer:

3.104 (option c)

Step-by-step explanation:

3 + 0.1 + 0.004 = 3.104

6 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

What is the slope of the line that passes through the points (-3, 5) and (1, 7)?

Stells [14]

Answer: The slope of the line is $\frac{1}{2}$

Step-by-step explanation:

The slope of a line can be calculated with the following formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Then, given the points $(-3, 5)$ and $(1, 7)$ , you can identify that:

$y_2=5\\y_1=7\\\\x_2=-3\\x_1=1$

Knowing these values you can substitute them into the formula for calculate the slope:

$m=\frac{5-7}{-3-1}$

Finally, evaluating, you get that the slope of that line is:

$m=\frac{-2}{-4}\\\\m=\frac{1}{2}$

8 0

3 years ago

IO.7(4.2)+15.9= 51.25+43.85+14=