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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:
m∠2 = 53°
Step-by-step explanation:
We will use two properties of a rhombus to solve this problem.
1). Opposite angles of a rhombus are equal.
2). Diagonals bisect the angles.
Since ∠JFG and ∠JHG are the opposite angles of the rhombus,
m∠JFG = m∠JHG
Since, diagonal FH bisect ∠JHG,
m∠FHJ = m∠GHF = m∠JFH = m∠GFH = 37°
In triangle JFH,
m∠FHJ + m∠JFH + m∠HJF = 180°
37° + 37° + m∠HJF = 180°
m∠HJF = 180 - 74
= 106°
Since, diagonal GJ bisects angle HJF,
m∠FJG = = 53°
Therefore, m∠2 = 53°.