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2 years ago

Please help! can you explain it to me?

Mathematics

1 answer:

Semenov [28]2 years ago

7 0

Answer:

It is increasing before x = -2 and from x = 0 to x = 2

Step-by-step explanation:

Option 1 - you can see the line going down starting at -2 and that means the line is decreasing but the choice says "It is increasing before x = 0". This contradicts the actual graph.

Option 2 - The line isn't increasing before x= -1, it's actually decreasing as seen in the graph.

Option 3 - The line doesn't even touch the point before x = -3 so this choice makes no sense.

Option 4 - Since all the other choices were eliminated this is the only choice standing. Looking at the line it is increasing before x = -2.

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Answer:

a) 0.778

b) 0.9222

c) 0.6826

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Step-by-step explanation:

Given:

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a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

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b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

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c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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