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DiKsa [7]
4 years ago
13

Find the volume of a right circular cone that has a height of 9.1 m and a base with a circumference of 12.1 m. Round your answer

to the nearest tenth of a cubic meter.
Mathematics
1 answer:
Orlov [11]4 years ago
7 0

Answer:

V=35.3\ m^3

Step-by-step explanation:

-Given the cone's height is 9.1m and the Circumference is 12.1

#We use the circumference formula to calculate the radius:

C=2\pi r\\\\12.1=2\pi r\\\\r=\frac{12.1}{2\pi}=\frac{6.05}{\pi}\ cm

-The volume of the cone using the radius above can then be calculated as follows:

V=\pi r^2\frac{h}{3}\\\\=\pi\times(\frac{6.05}{\pi})^2\times \frac{9.1}{3}\\\\\\=35.3\ m^3

Hence, the cone's volume is 35.3\ m^3

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_____ have congruent corresponding parts.
aniked [119]

Answer:

Two congruent figures

Step-by-step explanation:

Here we have to use the definition of congruent.

Congruent means equal to each other.

Example.

If the two triangles said be congruent, their corresponding angles and corresponding sides are equal.

They will have to same size of the sides and same measure of angles.

Two equal polygons, that is two congruent figures.

Hope this will help you.

Thank you.

4 0
3 years ago
The volume of the rectangular prism is 240 cubic centimeters. A rectangular pyramid has the same length, width,and height as a p
LuckyWell [14K]
The volume of the pyramid would be 80 cubic centimeters. The reason for this is because the formula for volume of a pyramid is 1/3bh, so if all the dimensions are the same for both figures, just divide the volume of the prism by three to find the volume of the pyramid. If you do this 240÷3=80 or the volume of the pyramid.
4 0
3 years ago
Read 2 more answers
Find x, round your answer to the nearest integer.
Makovka662 [10]

Answer:

B. 8

Step-by-step explanation:

cos37° = 4/5 = x/10

=> 4(10) = 5x

=> x = 40/5

=> x = 8

The correct option is B.

8 0
2 years ago
At a raffle, 1500 tickets are sold at $2 each. There are four prizes given for $500, $250, $150, and $75. You buy one ticket. Us
wolverine [178]

Answer:

<u>The expected value of every ticket is a loss of $ 1.35</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of tickets sold at the raffle = 1,500

Price of each ticket = $ 2

Total prizes = 1 * $ 500 + 1 * $ 250 + 1 * $ 150 + 1 * $ 75

2. Use a probability distribution table to calculate the expected value of your gain. Your expected value is_____ ?

Let's answer the question using a probability distribution for the gains, this way:

  • Probability of 1st prize of $ 498 (500 - ticket) = 1/1,500
  • Probability of 2nd prize of $ 248 (250 - ticket) = 1/1,500
  • Probability of 3rd prize of $ 148 (150 - ticket) = 1/1,500
  • Probability of 4th prize of $ 73 (75 - ticket) = 1/1,500
  • Probability of losing $ 2 (ticket) = 1,496/1,500

Now, we calculate the mean for all the tickets (winners and non-winners), this way:

Expected value = [(1,496 * -2) + (1 * 498) + (1 * 248) + ( 1 * 148) + ( 1 * 73)]/1,500

Expected value = [- 2,992 +498 + 248 + 148 + 73)/1,500

Expected value = -2,025/1,500 = - 1,35

<u>The expected value of every ticket is a loss of $ 1.35</u>

8 0
4 years ago
What the recursive formula for the sequence
klemol [59]

\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})


so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}


just a quick note on notation:

\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}

8 0
4 years ago
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