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dem82 [27]
2 years ago
13

If a person tosses a coin 23 times, how many ways can he get 11 heads

Mathematics
1 answer:
EastWind [94]2 years ago
3 0

Tossing a coin is a binomial experiment.

Now lets say there are 'n' repeated trials to get heads. Each of the trials can result in either a head or a tail.

All of these trials are independent since the result of one trial does not affect the result of the next trial.

Now, for 'n' repeated trials the total number of successes is given by

_{r}^{n}\textrm{C}

where 'r' denotes the number of successful results.

In our case n=23 and r=11,

Substituting the values we get,

_{11}^{23}\textrm{C}=\frac{23!}{11!\times 12!}

\frac{23!}{11!\times 12!}=1352078

Therefore, there are 1352078 ways to get heads if a person tosses a coin 23 times.


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Answer:

D. All three chose a valid first step toward solving the equation.

Step-by-step explanation:

Aaron, Blaine, and Cruz are solving the equation 4/7 (7 − n) = −1. Aaron started his solution by multiplying both sides of the equation by 7/4 . Blaine started by using the distributive property to multiply 4/7 by both 7 and −n. Cruz started by dividing both sides of the equation by 4/7 .

Which of the following is true?

A. Blaine and Cruz made an error in picking their first steps.

B. Cruz made an error in picking his first step.

C. All three made an error because the right side equals -1.

D. All three chose a valid first step toward solving the equation.

Given:

4/7(7 - n) = -1

Aaron:

4/7(7 - n) = -1

Multiple both sides by 7/4

4/7(7 - n) * 7/4 = -1 * 7/4

7 - n = -7/4

- n = -7/4 - 7

- n = (-7-28)/4

- n = -35/4

n = 35/4

Blaine:

4/7(7 - n) = -1

4/7(7 - n) × 7 = -1 × 7

4(7 - n) = -7

28 - 4n = -7

-4n = -7 - 28

- 4n = - 35

n = -35/-4

n = 35/4

Cruz:

4/7(7 - n) = -1

Divide both sides by 4/7

4/7(7 - n) ÷ 4/7 = -1 ÷ 4/7

4/7(7 - n) × 7/4 = -1 × 7/4

7 - n = -7/4

- n = (-7-28)/4

- n = -35/4

n = 35/4

D. All three chose a valid first step toward solving the equation.

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2 years ago
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Hi there!

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(A)

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Seems like it would be

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