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Naya [18.7K]
4 years ago
7

Use the figure shown.

Mathematics
1 answer:
Olegator [25]4 years ago
6 0
You didn’t attach the picture to show the figure
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A cone has a diameter of 7.5 ft and a height of 4.25 ft. What is the volume of the cone to the nearest tenth? Use 3.14 for π.
lubasha [3.4K]

Answer:

V = ⅓πr²h = ⅓(3.14)(7.5/2)²(4.25) = 62.5546875 ≈ 62.6 ft³

Step-by-step explanation:

4 0
3 years ago
Do you agree? explain why or why not
castortr0y [4]
No. when plugging 4 into the equation, 0 is not the final result. instead it is 6. 

square root of 2(4)+1 = square root of 9 = 3 

3+3 = 6
and 6 is not 0 so x cannot = 4

3 0
3 years ago
Read 2 more answers
A salesperson sold a car for $18,200 and their commission is $700. What percentage of the sale price is their commission?
marshall27 [118]
The percentage of their commission is 3.85%
700/18200= 0.03846
0.03846 x 100 = about 3.85
7 0
3 years ago
Please help I need help I
Crazy boy [7]

Answer:

5 runners will be needed

Step-by-step explation:

To solve this problem you have to find how many 1/25ths it takes to get 1/5. You can do that by dividing 25 by 5 which eauals 5. In other words 5/25 is equal to 1/5. This means you will need 5 runners to complete the race.

4 0
3 years ago
You flip a coin and then roll a 6-sided number cube (a die).
expeople1 [14]

Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

6 0
2 years ago
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