Jace recorded the grade-level and instrument of everyone in the middle school School of Rock below.

The height of a cylinder is 10 and the area of a base is 36 pi square units. What

ollegr [7]

Answer:

1130.97336 units^3

Step-by-step explanation:

The volume of a cylinder can be found using:

$v=\pi r^2h$

We have the area of the base, but not the radius

$a=\pi r^2$

We know the area is $36\pi$ , so we can substitute that in for a

$36\pi =\pi r^2$

We want to find r, so we need to isolate it

Divide both sides by pi

36=r^2

Take the square root of both sides

6=r

Now we know the radius, and can substitute it into the volume formula, and we can substitute the height (10) in

$v=\pi r^2h$

$v=\pi 6^210$

Solve the exponent

$v=\pi 36(10)$

$v=\pi 360$

v=1130.97336

The volume is 1130.97336 units^3

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

2 years ago

the line y=2x-4 is dilated bt a scale factor of 3/2 and centered at the origin. Write an equation that represent that image of t

Veseljchak [2.6K]

Answer: $y = 2x -6$

Step-by-step explanation:

Since, If (x,y) represents the coordinates of a line, then after dilation about the origin by the scale factor k, the coordinates of the resultant line will get by the rule,

$(x,y)\rightarrow (kx,ky)$

Here, The equation of the given line is,

y = 2x - 4

x-intercept and y-intercept of the line are (2,0) and (0,-4) respectively,

Hence, the line y = 2x - 4 will be passes through the points (2,0) and (0,-4),

By the above property,

The dilated line about the origin will be passes through the points (3,0) and (0,-6), ( Because (3/2× 2, 3/2×0) = (3,0) and (3/2×0, 3/2×-4) = (0,-6) )

Hence, the equation of transformed line after dilation,

$y-0 = \frac{-6-0}{0-3}(x-3)$