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alexandr402 [8]
2 years ago
15

Select three ratios that are equivalent to 11:111:111, colon, 1. Choose 3 answers:

Mathematics
1 answer:
kirill115 [55]2 years ago
4 0

The three ratios that are equivalent to the given ratio are options B, C and E.

<h3>How to determine the equivalent ratios?</h3>

In order to determine the ratios that are equivalent to the given ratio, we would evaluate by dividing each of them by their respective denominator value:

(11:111:111)/1 = 11:111:111.

For choice A, we have:

1:111:111/11 = 1/11:111:111 (not equivalent).

For choice B, we have:

22:222:222/2 = 11:111:111. (equivalent).

For choice C, we have:

55:555:555/5 = 11:111:111. (equivalent).

For choice D, we have:

9:999:999/99 = 1/11:111:111 (not equivalent).

For choice E, we have:

110:1110:1110/10 = 11:111:111. (equivalent).

Read more on ratio here: brainly.com/question/13513438

#SPJ1

<u>Complete Question:</u>

Select three ratios that are equivalent to 11:111:111, colon, 1. Choose 3 answers: Choose 3 answers: (Choice A) A 1:111:111, colon, 11 (Choice B) B 22:222:222, colon, 2 (Choice C) C 55:555:555, colon, 5 (Choice D) D 9:999:999, colon, 99 (Choice E) E 110:1110:1110, colon, 10.

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Consider a binomial experiment with n = 20 and p = .70.
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Complete question:

Consider a binomial experiment with n = 20 and p = .70.

A. Compute f(12).

B. Compute f(16).

C. Compute P(x≥ 16).

D. Compute P(x≤15).

E. Compute E(x).

F. Compute Var(x).

Answer:

a) 0.1144

b) 0.1304

c) 0.2375

d) 0.7625

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Step-by-step explanation:

Given:

n = 20

p = 0.70

q = 1 - p ==>  1 - 0.70 = 0.30

a) Use the formula:

P(x) = CC\left(\begin{array}{ccc}n\\x\end{array}\right) p^x q^(^n^-^x^)

Thus,

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b) P(16) = C\left(\begin{array}{ccc}20\\16\end{array}\right) 0.7^1^6 (0.3^(^2^0^-^1^6^))

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c) Compute P(x≥16):

P(x ≥ 16) = P(16) + P(17) + P(18) + P(19) + P(20)

= C\left(\begin{array}{ccc}20\\16\end{array}\right) 0.7^1^6 (0.3^(^2^0^-^1^6^)) + C\left(\begin{array}{ccc}20\\17\end{array}\right) 0.7^1^7 (0.3^(^2^0^-^1^7^) ) + C\left(\begin{array}{ccc}20\\18\end{array}\right) 0.7^1^8 (0.3^(^2^0^-^1^8^)) + C\left(\begin{array}{ccc}20\\19\end{array}\right) 0.7^1^9 (0.3^(^2^0^-^1^9^)) + C\left(\begin{array}{ccc}20\\20\end{array}\right) 0.7^2^0 (0.3^(^2^0^-^2^0^))

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= n*p

= 20 * 0.7

= 14

f) Var(x)

Use the formula: npq

npq = 20 * 0.7 * 0.3

= 4.2

σ

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