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amid [387]
3 years ago
11

Dado os intervalos A=(x e r/-4< x <2 )e b =(x e r/-2< x <4 )

Mathematics
1 answer:
ch4aika [34]3 years ago
6 0

A) AUB = ao conjunto B ou seja {-2,-1,0,1,2,3,4}
B) AUB = {-3,-2,-1,0,1,2,3,4,5,6,7}
? · 7 anos atrás
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Answer:


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Please answer all parts of the question and all work shown.
faust18 [17]

Answer:

a. 0.4931

b. 0.2695

Step-by-step explanation:

Given

Let BG represents Boston Globe

NYT represents New York Times

P(BG) = 0.55

P(BG') = 1 - 0.55 = 0.45

P(NYT) = 0.6

P(NYT') = 1 -0.6 = 0.4

Number of headlines = 5

Number of depressed articles = 3 (at most)

a.

Let P(Read) = Probability that he reads the news the first day

P(Read) = P(He reads BG) and P(He reads NYT)

For the professor to read BG, then there must be at most 3 depressing news

i.e P(0) + P(1) + P(2) + P(3)

But P(0) + P(1) + .... + P(5) = 1 (this is the sample space)

So,

P(0) + P(1) + P(2) + P(3) = 1 - P(4) - P(5)

P(4) or P(BG = 4) is given as the binomial below

(BG + BG')^n where n = 5, r = 4

So, P(BG = 4) = C(5,4) * 0.55⁴ * 0.45¹

P(BG = 5). = (BG + BG')^n where n = 5, r = 5

So, P(BG = 5) = C(5,5) * 0.55^5 * 0.45°

P(0) + P(1) + P(2) + P(3)= 1 - P(BG = 4) - P(BG = 5)

P(0) + P(1) + P(2) + P(3) = 1 - C(5,4) * 0.55⁴ * 0.45¹ - C(5,5) * 0.55^5 * 0.45°

P(0) + P(1) + P(2) + P(3) = 0.7438

For the professor to read NYT, then there must be at most 3 depressing news

i.e P(0) + P(1) + P(2) + P(3)

But P(0) + P(1) + .... + P(5) = 1 (this is the sample space)

So,

P(0) + P(1) + P(2) + P(3) = 1 - P(4) - P(5)

P(4) or P(NYT = 4) is given as the binomial below

(NYT+ NYT')^n where n = 5, r = 4

So, P(NYT = 4) = C(5,4) * 0.6⁴ * 0.4¹

P(NYT = 5). = (NYT + NYT')^n where n = 5, r = 5

So, P(NYT = 5) = C(5,5) * 0.6^5 * 0.4°

P(0) + P(1) + P(2) + P(3)= 1 - P(NYT = 4) - P(NYT = 5)

P(0) + P(1) + P(2) + P(3) = 1 - C(5,4) * 0.6⁴ * 0.4¹ - C(5,5) * 0.6^5 * 0.4°

P(0) + P(1) + P(2) + P(3) = 0.6630

P(Read) = P(He reads BG) and P(He reads NYT)

P(Read) = 0.7438 * 0.6630

P(Read) = 0.4931

b.

Given

n = Number of week = 7

P(Read) = 0.4931

R(Read') = 1 - 0.4931 =

He needs to read at least half the time means he reads for 4 days a week

So,

P(Well-informed) = (Read + Read')^n where n = 7, r = 4

P(Well-informed) = C(7,4) * (0.4931)⁴ * (1-0.4931)³ = 0.2695

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3 years ago
You roll two six-sided dice. Match each event with its probability.
Nina [5.8K]

Answer:

A) probability the sum is 8 or 11= 4/21

B) probability that sum is 12 or less than 10 = 6/7

C) Probability that the sum is 3 or less than 3 = 2/21

D) Probability that the sum is 2 or 10 = 1/7

Step-by-step explanation:

Since we have the same probability of each event in each dice, the answer would be just to check the different outcomes, two dices, each with 1 to 6;

(1,1);(1,2);(1,3);(1,4);(1,5);(1,6);(2,2);(2,3);(2,4);(2,5);(2,6);(3,3);(3,4);(3,5);(3,6);(4,4);(4,5);(4,6);(5,5);(5,6);(6,6)

Thus, there are 21 possible outcomes.

Now,

A) probability that the sum is 8 or 11;

From the outcomes above, the number of outcomes that have a sum as 8 or 11 are;

(2,6) ; (3,5) ; (4,4) ; (5,6)

So,probability = 4/21

B) From the outcomes above, the number of outcomes that are 12 or less than 10 are;

(1,1);(1,2);(1,3);(1,4);(1,5);(1,6);(2,2);(2,3);(2,4);(2,5);(2,6);(3,3);(3,4);(3,5);(3,6);(4,4);(4,5);(6,6).

There are 18 possible outcomes.

So, probability that sum is 12 or less than 10 = 18/21 = 6/7

C)From the initial 21 outcomes, the number of outcomes that the sum is 3 or less than 3 are;(1,1);(1,2)

Thus,

Probability that the sum is 3 or less than 3 = 2/21

D) From the initial 21 outcomes, the number of outcomes that the sum is 2 or 10 are;

(1,1); (4,6) ; (5,5)

Thus,

Probability that the sum is 2 or 10 = 3/21 = 1/7

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