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masha68 [24]
3 years ago
5

PLEASE HELP!

Mathematics
2 answers:
vodka [1.7K]3 years ago
7 0

Volume of a cone is pi*r^2*h/3

Volume of cone is approx. 20.9.

 

Volume of a sphere is 4/3pi*r^3

Volume of sphere is approx. 113.1


113.1 - 20.9 = 92.2 approx.

pshichka [43]3 years ago
4 0

Answer: 92 cubic meters

Step-by-step explanation:

To find the volume of shaded region we need to subtract the volume of cone from volume of sphere.

radius of cone = 2 m

height of cone = 5 m

Volume of cone is given by :_

\text{Volume}=\frac{1}{3}\pi\times r^2h\\\\\Rightarrow\text{Volume}=\frac{1}{3}\times3.14\times(2)^2(5)\\\\\Rightarrow\text{Volume}=20.9333333333\approx20.93cubic meters

radius of sphere = 3 m

Volume of sphere is given by :-

\text{Volume}=\frac{4}{3}\pi\times r^3\\\\\Rightarrow\text{Volume}=\frac{4}{3}\times3.14\times(3)^3)\\\\Rightarrow\text{Volume}=113.04cubic meters

Volume of shaded region=Volume of sphere-Volume of cone=113.04-20.93=92.11≈92 cubic meters

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vladimir1956 [14]

The hypothesis test shows that we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%

<h3>What is the claim that the return rate is less than 20% by using a statistical hypothesis method?</h3>

The claim that the return rate is less than 20% is p < 0.2. From the given information, we can compute our null hypothesis and alternative hypothesis as:

\mathbf{H_o :p =0.2}

\mathbf{H_i:p < 0.2}

Given that:
Sample size (n) = 6965

Sample proportion \mathbf{\hat p = \dfrac{x}{n} = \dfrac{1302}{6965} \sim0.1869}

The test statistics for this data can be computed as:

\mathbf{z = \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n}}}}

\mathbf{z = \dfrac{0.1869 -0.2}{\sqrt{\dfrac{0.2(1-0.2)}{6965}}}}

\mathbf{z = \dfrac{-0.0131}{0.0047929}}

z = -2.73

From the hypothesis testing, since the p < alternative hypothesis, then our test is a left-tailed test(one-tailed.

Hence, the p-value for the test statistics can be computed as:

P-value = P(Z ≤ z)

P-value = P(Z ≤ - 2.73)

By using the Excel function =NORMDIST (-2.73)

P-value = 0.00317

P-value ≅ 0.003

Therefore, we can conclude that since P-value is less than the significance level at ∝ = 0.01, we reject the null hypothesis and there is sufficient evidence to support the claim that the return rate is less than 20%

Learn more about hypothesis testing here:

brainly.com/question/15980493

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3 0
2 years ago
Calculate the volume, in cubic centimeters, of a box which is 125 cm long, 37 cm wide, and 68 cm high. Report your answer with c
viva [34]

Answer:

Volume = 314500cm^3

Step-by-step explanation:

Given

Length = 125cm

Width = 37cm

Height = 68cm

Required

Determine the volume

Volume is calculated as:

Volume = Length * Width * Height

Substitute values for Length, Width and Height

Volume = 125cm * 37cm * 68cm

Volume = 314500cm^3

<em>Hence, the volume of the box is </em>314500cm^3<em></em>

6 0
3 years ago
2.5 hours is equal to 2 hours and 30 minutes <br> True or false<br>​
Feliz [49]

Answer:

True because 1/2 an hour is 30 minutes.

Step-by-step explanation:

4 0
3 years ago
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Help ASAP! I’m confused. Thanks!!
KATRIN_1 [288]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
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I’m having a lot of trouble, can someone guide me, step by step?
shutvik [7]

Answer:

Hi hopefully this helps you!

Step-by-step explanation:

To find the area of a circle you can use the formula A = πr^2

The radius of a circle is just the diameter divided by 2. In this case we know the diameter is 3, so the radius is 1.5

A = π(1.5)^2

   = 7.07

Because this is a semicircle, divide this area by 2

   = 3.53429 in^2

Add up the area of this semi circle with the area of the rectangle

A = (3.53429) + (3x4)

   = 15.53429 in^2

To find the circumference/ perimeter of a circle use this formula C = 2πR

C = 2π(1.5)

   = 9.42478 inches

Again because this is a semicircle, divide by 2

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To find the perimeter of this entire shape add up the circumference of the semicircle and the rectangle's sides and bottom

P = 4.71239 + 4 + 4 + 3

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So the final answer would be

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