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Rasek [7]
3 years ago
14

59​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.
Mathematics
1 answer:
Bumek [7]3 years ago
8 0

Answer:

(a) 20.9%

(b) 60.8%

(c) 6.3%

Step-by-step explanation:

Use binomial probability.

P = nCr pʳ qⁿ⁻ʳ

(a)

P = ₁₀C₅ (0.59)⁵ (1−0.59)¹⁰⁻⁵

P = 0.209

(b)

P = ₁₀C₆ (0.59)⁶ (1−0.59)¹⁰⁻⁶

+ ₁₀C₇ (0.59)⁷ (1−0.59)¹⁰⁻⁷

+ ₁₀C₈ (0.59)⁸ (1−0.59)¹⁰⁻⁸

+ ₁₀C₉ (0.59)⁹ (1−0.59)¹⁰⁻⁹

+ ₁₀C₁₀ (0.59)¹⁰ (1−0.59)¹⁰⁻¹⁰

P = 0.608

(c)

P = ₁₀C₀ (0.59)⁰ (1−0.59)¹⁰⁻⁰

+ ₁₀C₁ (0.59)¹ (1−0.59)¹⁰⁻¹

+ ₁₀C₂ (0.59)² (1−0.59)¹⁰⁻²

+ ₁₀C₃ (0.59)³ (1−0.59)¹⁰⁻³

P = 0.063

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Answer:

v = -3

Step-by-step explanation:

-8(v + 7) = 3v - 23

Expand.

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Add -3v and 56 on both sides.

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-8v - 3v = -23 + 56

Combine like terms.

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Divide -11 into both sides.

-11v/-11 = 33/-11

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{\bold{\red{\huge{\mathbb{QUESTION}}}}}

The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has length 2. What is the area of the shaded sector formed by obtuse angle WXY?

\bold{ \red{\star{\blue{GIVEN }}}}

RADIUS = 2

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\bold{\blue{\star{\red{TO \:  \: FIND}}}}

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\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

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Area Under Unshaded Part .

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my bad I couldn't put the line under the sign

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