The probability of getting 3 or more who were involved in a car accident last year is 0.126.
Given 9% of the drivers were involved in a car accident last year.
We have to find the probability of getting 3 or more who were involved in a car accident last year if 14 were selected randomly.
We have to use binomial theorem which is as under:
n
where p is the probability an r is the number of trials.
Probability that 3 or more involved in a car accident last year if 14 are randomly selected=1-[P(X=0)+P(X=1)+P(X=2)]
=1-{
}
=1-{1*0.2670+14!/13!*0.9*0.29+14!/2!12!*0.0081*0.2358}
=1-{0.2670+0.3654+0.2358}
=1-0.8682
=0.1318
Among the options given the nearest is 0.126.
Hence the probability that 3 or more are involved in the accident is 0.126.
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The measurements are 3 times as big.
Answer:
£29.75.
Step-by-step explanation:
15% = 0.15 as a decimal fraction.
Sale price = 35 - 0.15*35
= £29.75.
Answer:
a. 10
Step-by-step explanation:
Profit = Total Cost - Total Revenue
The profits are charted on the chart attached.
Profit is maximized when the quantity is 6. At this quantity, the difference between marginal cost and marginal revenue is $10.