4 years ago

A bear fun weighed 64 oz in March. After three

Mathematics

1 answer:

Alex Ar [27]4 years ago

3 0

What are u asking lol?

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What is the simplified form of <img src="https://tex.z-dn.net/?f=3%5Csqrt%7B135%7D" id="TexFormula1" title="3\sqrt{135}" alt="3\

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The first term of a geometric sequence is 300 and the common ratio is 2 What is the 7th term of the sequence

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4 years ago

Suppose that 7 in every 10 auto accidents involve a single vehicle. If 15 auto accidents are randomly selected, compute the prob

kobusy [5.1K]

Answer:

$\mathbf{P(X \le 4 ) \simeq 0.0006722}$

Step-by-step explanation:

From the information given:

p = x/n

p = 7/10

p = 0.7

sample size n = 15

Suppose X be the number of accidents involved by a single-vehicle.

Then;

$X \sim Binom (15,0.7)$

Thus, the required probability that at most 4 involve in a single-vehicle is

$P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)$

$P(X \le 4 ) = (^{15}_0) *0.7^0 *0.3^{15-0} + (^{15}_1) *0.7^1 *0.3^{15-1} + (^{15}_2) *0.7^2 *0.3^{15-2} + (^{15}_3) *0.7^3 *0.3^{15-3} + (^{15}_4) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) = (\dfrac{15!}{0!(15-0)!}) *0.7^0 *0.3^{15-0} + (\dfrac{15!}{1!(15-1)!}) *0.7^1 *0.3^{15-1} + (\dfrac{15!}{2!(15-2)!}) *0.7^2 *0.3^{15-2} + (\dfrac{15!}{3!(15-3)!}) *0.7^3 *0.3^{15-3} + (\dfrac{15!}{4!(15-4)!}) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) =1.4348907 \times 10^{-8} +5.02211745 \times 10^{-7} + 8.20279183 \times 10^{-6} + 8.29393397 \times 10^{-5} + 5.80575378 \times 10^{-4}$