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3 years ago

6

On January 1st, the high temperature was 81℉ in Kona, Hawaii. The low temperature was -29℉ in Barrow, Alaska. What was the diffe

rence between the two temperatures?

1 answer:

Alborosie3 years ago

3 0

Answer:

110

Step-by-step explanation:

put 89-110= -29 therefore u have an answer. difference= subtraction.

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Which graph best represents an exponential function?

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The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.08

kvv77 [185]

Answer:

a) 44.93% probability that there are no surface flaws in an auto's interior

b) 0.03% probability that none of the 10 cars has any surface flaws

c) 0.44% probability that at most 1 car has any surface flaws

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the binomial probability distributions.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

So $\mu = 10*0.08 = 0.8$

(a) What is the probability that there are no surface flaws in an auto's interior?

Single car, so Poisson distribution. This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

44.93% probability that there are no surface flaws in an auto's interior

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?

For each car, there is a $p = 0.4493$ probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003$

0.03% probability that none of the 10 cars has any surface flaws

(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?

At least 9 cars without surface flaws. So

$P(X \geq 9) = P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.4493)^{9}.(0.5507)^{1} = 0.0041$

$P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0041 + 0.0003 = 0.0044$

0.44% probability that at most 1 car has any surface flaws

5 0

3 years ago