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finlep [7]
3 years ago
6

Write an equation of the line with the given slope and y-intercept. m equals one fourth, b equals negative three fourths

Mathematics
2 answers:
victus00 [196]3 years ago
7 0
Plug in m and then b and you get y=x/4 -3/4
pentagon [3]3 years ago
5 0
We are going to use mx+b=y.
B= the y intercept=-0.75
M= slope=0.25
0.25x-0.75=y
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Cody works at a car dealership where he earns $300 each week plus a commission equal to 1% of his total sales. This week his goa
Dmitrij [34]

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$20,000 = sales

Step-by-step explanation:

Giving the following information:

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5 0
3 years ago
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
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