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4 years ago

Find the difference. Simplify completely. Please show work.1. 7/9 - 3/8 2. 5 4/5 - 2 1/4

Mathematics

1 answer:

fenix001 [56]4 years ago

4 0

1. 7/9 - 3/8 = (7·8 - 9·3)/(9·8) = (56 -27)/72 = 29/72

2. (5 4/5) - (2 1/4) = (5 - 2) + (4/5 - 1/4) = 3 + (4·4 -5·1)/(5·4) = 3 11/20

_____

(a/b) -(c/d) = (ad -bc)/(bd) . . . . a useful formula to remember when working problems like this. It is equivalent to multiplying the first fraction by d/d and the second one by b/b to get a common denominator of bd. Of course if the second fraction is added instead of subtracted, the sign changes in both places.

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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

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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.