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Nitella [24]
3 years ago
10

Paul can read 15 pages in 5 minutes. how long does it take paul to read 75 at the same rate?

Mathematics
1 answer:
Leno4ka [110]3 years ago
6 0

I think 150 hope this helps


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Find the slope of the line that passes through<br> (10, 1) and (5,2)<br> -5<br> -1/5<br> 1/5<br> 5
rusak2 [61]

Answer:

-1/5 I believe, since you go down one and to the right by 5.

7 0
3 years ago
A prism is filled with 70 cubes with a side length of 1 2 unit. What is the volume of the prism in cubic units? Enter answer as
galina1969 [7]

Answer:

The volume of the cube is;

8\dfrac{3}{4} \ cube \, units

Step-by-step explanation:

The parameters of the prism are;

The number of cubes used to fill the prism, n = 70 cubes

The side length of each cube, s = 1/2 unit

The volume of the cube, V = n × s³

∴ Therefore, the volume of the cube, V = 70 × (1/2)³ = 8\dfrac{3}{4} \ cube \, units = 8.75 cube unit

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3 years ago
X &lt; -10/3 or x &gt; 2/3
Fittoniya [83]

Hello! :)

Here is a graph for x < -10/3 or x > 2/3.

THEDIPER

6 0
4 years ago
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
3 years ago
The basket of golf balls at a miniature golf course contains 18 golf balls, of which 9 are red. What is the probability that a r
alexira [117]
It will be 2 because 19/9 is 2
5 0
3 years ago
Read 2 more answers
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