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3 years ago

Anthony has a sink that is shaped like a half-sphere

1 answer:

3 0

I think you are referring to this:
The volume of the sink shaped like a half sphere is [2000/3] π in^2.

A. The volume of a cylinder =area of the base * height
For this part the cup has a diameter, d, of 4 in and height, h, of 8 in.
[2000 / 3]π / [32π] = 20.83
Answer: 21 cups.

B. For this part the diameter of the cup is 8 in and the height is 8 in
And the number of cups needed is [2000π/3] / (128π) = 5.2
Answer: 5 cups

You might be interested in

A 95% confidence interval for the mean number of television per American household is (1.15, 4.20). For each of the following st

liq [111]

Answer:

a. False

b. True

c. False

d. False

e.True

f. True

Step-by-step explanation:

The 95% is confidence interval its not a probability estimate. The probability will be different from the confidence interval. Confidence interval is about the population mean and is not calculated based on sample mean. Every confidence interval contains the sample mean. There is 95% confidence that the number of televisions per American household is between 1.15 to 4.20.

7 0

3 years ago

(02.04 LC)

Georgia [21]

Answer:

x = - 1

Step-by-step explanation:

the equation of a vertical line is

x = c

where c is the value of the x- coordinates the line passes through

the line passes through the x- intercept of - 1 , then

x = - 1 ← equation of vertical line

7 0

1 year ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa

blagie [28]

Answer:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

$p_v =P(z>1.9)=0.0287$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=12.95$ represent the sample mean

$\sigma=3$ represent the population standard deviation for the sample

$n=36$ sample size

$\mu_o =12$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:

Null hypothesis: $\mu \leq 12$

Alternative hypothesis: $\mu > 12$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

P-value

Since is a right tailed test the p value would be:

$p_v =P(z>1.9)=0.0287$

Conclusion

4 0

3 years ago

Read 2 more answers

Write 83 as a sum of tens and ones

Travka [436]

8 tens and 3 ones.

hope that helped

6 0

3 years ago

Read 2 more answers