18. Solve the equation.<br> m- 10/3

How does the volume of a cylinder change if the radius is quadrupled and the height is reduced to a third of its original size?

KIM [24]

Answer:

the answer is 1/3 pie r2h

Step-by-step explanation:

The volume of a cylinder is given by πr²h where, r is the radius of the cylinder and h is the height of the cylinder.

Also r=d/2 , where d is the diameter of the cylinder.

Therefore if the diameter is halved, the radius also gets halved ,i.e., it becomes r/2. Therefore the new volume = π(r/2)²h

=π(r²/4)h

=(1/4) πr²h

Therefore the volume becomes one-fourth of the initial volume.

8 0

2 years ago

Write an expression to represent the perimeter of:<br> 2a-3<br> 2a<br> 3a+1

Marina86 [1]

I am assuming this is a triangle.....P = a + b + c

P = 2a - 3 + 2a + 3a + 1....combine like terms
P = 7a - 2 <==

5 0

2 years ago

A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2

Yuliya22 [10]

A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2 and 5.1.

Answer: The correct option is B) about 34%

Proof:

We have to find $P(4.2$

To find $P(4.2$ , we need to use z score formula:

When x = 4.2, we have:

$z = \frac{x-\mu}{\sigma}$

$=\frac{4.2-5.1}{0.9}=\frac{-0.9}{0.9}=-1$

When x = 5.1, we have:

$z = \frac{x-\mu}{\sigma}$

$=\frac{5.1-5.1}{0.9}=0$

Therefore, we have to find $P(-1$

Using the standard normal table, we have:

$P(-1$ = $P(z$

$=0.50-0.1587$

$=0.3413$ or 34.13%

= 34% approximately

Therefore, the percent of data between 4.2 and 5.1 is about 34%

7 0

3 years ago

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$