<span>(r^2+7r+10/3) * (3r-30/r^2-5r-50)
1/3(3r^2+21r+10) x 3(r-10)/(r^2-5r-50)
(3r^2+21r+10)x(r-10)/(r^2-10r+5r-50)
(3r^2+21r+10)x(r-10)/(r-10)(r+5)
3r^2+21r+10/(r+5)
</span>
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.
Learn more about probability at brainly.com/question/24756209
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Answer:
y = -1/2x + 2
Step-by-step explanation:
Answer:
19 days
Step-by-step explanation:
805 = 7x
x = 115 dollars per day
115x = 2185
divide by 115
x=19
he needs to work 19 days