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2 years ago

Quick please , find the volume of the shape

Mathematics

1 answer:

Blizzard [7]2 years ago

3 0

Answer:

7068.58 Cubic inch

Step-by-step explanation:

$V=\frac{2}{3} \pi r^{3} =\frac{2}{3} \pi 15^{3}=7068.58 in^{3}$

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Write the equation in slope-intercept form for a line that has a y-intercept of -3 and a slope of 6. Use no spaces in your equat

Lilit [14]

This is slope-intercept form y=mx+b
when you substitute it, it becomes y=6x-3

7 0

3 years ago

How do you do 6/3 (improper fraction) on a number line?

ss7ja [257]

Answer:

The number ins't simplified

Step-by-step explanation:

If you convert it into a mixed number it will be a lot easier to see where it goes. In this case tho 6 divided by 3 just equals 2, so the two is what you put on a number line.

8 0

3 years ago

Store ABC sells 8 lbs. of oranges for $6.00. Store XYZ sells 14 lbs. or oranges for $11.20. Which store is less expensive, and b

Fed [463]

Answer : 3. ABC is less expensive by $0.05 per pound

So let's find the price for one pound first:
ABC = $6/8pounds = $0.75/ 1 pound
XYZ = $11.20/14 pounds = $0.8/ 1 pound

So we can see ABC cost less for 1 pound and we can find the difference by subtracting:
0/8=0/75 = 0.05

Hope this helps! Please give me the brainliest answer if you like it! If you have further questions, please leave a comment or add me as a friend!

5 0

2 years ago

In a clinical trial of 268 subjects treated with a drug, 11% of the subjects reported dizziness. The margin of error is plus o

Mariulka [41]

Answer:

$\hat p =0.11$ represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

$\hat q = 1-\hat p = 1-0.11= 0.89$ represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

$\alpha= 1-0.90 =0.1$ and this value represent the significance level of the test or the probability of error type I

Step-by-step explanation:

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

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

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The margin of error is given by:

$ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

$Lower= 0.11-0.04 = 0.07$

$Upper= 0.11+0.04 = 0.15$

And for this case we have the following info:

$\hat p =0.11$ represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

$\hat q = 1-\hat p = 1-0.11= 0.89$ represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

$\alpha= 1-0.90 =0.1$ and this value represent the significance level of the test or the probability of error type I

4 0

3 years ago

For the given function, find the vertical and horizontal asymptote(s) (if there are any).

Doss [256]

There are three rules of finding the horizontal asymptote depending on the orders of the numerator and denominator. If the degrees are equal for the numerator and the denominator, then the horizontal asymptote is equal to y = the ratio of the coefficients of the highest order from the numerator and the denominator. If the degree in the numerator is less than the degree in the denominator, then there the x axis is the horizontal asymptote. If on the other hand, the order in the numerator is greater than that of the denominator, then there is no horizontal asymptote.

8 0

2 years ago