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3 years ago

In the diagram to the right what is the length of N

Mathematics

1 answer:

svetoff [14.1K]3 years ago

8 0

Answer:

Step-by-step explanation:

Using the sine ratio in the right triangle and the exact value

sin30° = $\frac{1}{2}$

sin30° = $\frac{opposite}{hypotenuse}$ = $\frac{8}{n}$ = $\frac{1}{2}$ ( cross- multiply )

n = 16 → C

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Mr smith had 4 daughters each daughter had a brother.. how many kids mr. Smith has

blsea [12.9K]

Mr. Smith has 5 children, since each of the 4 daughters share the same 1 brother.

3 0

3 years ago

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K (x) = 6x + 100 <br> k(-5) = ?

siniylev [52]

Answer:

im not super sure but i think its 70

Step-by-step explanation:

6 0

2 years ago

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Factor each completely JEED THIS ASAP

Maurinko [17]

Answer: (m-1)(m-8)

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2 years ago

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Please help!

kirill115 [55]

Answer:

73 and 113 / 144 ft^2

Step-by-step explanation:

Lets convert everything into pure feet.

10ft 5in = 10 and 5/12 feet = 125/12 feet

14ft 2in = 14 and 1/6 feet = 85/6

Now lets calculate area:

1/2 * 125/12 * 85/6 = 125*85 / 144 = 73 and 113/144

3 0

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.

7 0

2 years ago