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Nataly [62]
3 years ago
10

Q-The general solution of inequality cos 2 x≤- sin x is

Mathematics
1 answer:
frozen [14]3 years ago
3 0

Answer:

x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I

Step-by-step explanation:

1−2sin2  x≤−sin x    ⇒    (2sin x+1)(sin x−1)≥0

sin x≤−1/2    or    sin x≥1

−5π/6+2nπ≤x≤−π/6+2nπ    or    , n ϵ I x=(4n+1)π/2, n ϵ I⇒    -5π6+2nπ≤x≤-π6+2nπ    or    , n ϵ I x=4n+1π2, n ϵ I     (as sin x = 1 is valid only)

In general⇒    In general    x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I

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Answer:

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Step-by-step explanation:

Given:

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a. No. of donors that had type B blood = \frac{11}{100}*145 = \frac{1,595}{100} = 15.95

Type B blood donors ≈ 16

b. No. of blood donors without blood type O = \frac{40 + 11 + 4}{100}*145 = \frac{55}{100}*145 = \frac{7,975}{100} = 79.75

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Type AB = \frac{4}{100}*145 = 5.8 ≈ 6. This is less than 10 donors.

Therefore, blood type AB had less than 10 donors.

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