The article "Should You Report That Fender-Bender?"† reported that 7 in 10 auto accidents involve a single vehicle. Suppose 11 accidents are randomly selected. (Round your answers to five decimal places.)

What is the probability that exactly nine involve multiple vehicles?

Using the principle of binomial probability, the probability that exactly 9 involve multiple vehicle is 0.00053

Probability of single vehicle = 7/10

Probability of multiple vehicle = 1 - (7/10) = 0.3

From the concept of binomial probability :

P(x = x) = nCx * p^x * q^(n-x)

x = 9

n = number of trials = 11

p = probability of success = 0.3

q = 1 - p = 0.7

Therefore, the probability that exactly 9 involve multiple vehicles can be defined as :

P(x = 9) = 11C9 × 0.3^9 × 0.7^2

P(x = 9) = 55 × 0.00000964467

P(x = 9) = 0.00053045685

Therefore, the probability that exactly 9 involve multiple vehicle is 0.00053

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0.19975

Step-by-step explanation:

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

Given:

n = 11

r = 9

p = 0.7

q = 0.3

Plugging in values:

P = ₁₁C₉ (0.7⁹) (0.3¹¹⁻⁹)

P = 0.19975

fwms@lil_ben06

step-by-step explanation:

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.