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3 years ago

What is the answer to 2+1/3

Mathematics

2 answers:

Strike441 [17]3 years ago

3 0

the answer is 2 1/3 -------------------------------------------------------------------------------------------------

Levart [38]3 years ago

3 0

Yea its 2 1/3 bro...................

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M is directly proportional to r2 <br> When r=2, m=14<br> Work out the value of m when r=12

Genrish500 [490]

we have M is durectly porpotional to r^2

so M=(k)r^2

and when r=2, m=14

so 14=(k)(2)^2

k=14/4 =7/2

so when r=12

m= (7/2)(12)^2 =(7/2)(144) = 504

8 0

2 years ago

Can someone help me

mafiozo [28]

If you need to you can write it out. the answer is X=14

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2 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.

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Answer:

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