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Effectus [21]
3 years ago
12

4. In the accompanying diagram, AABC – ADEF, DE = x + 2, DF = x+8, AB = x-3, and AC = x. Determine, algebraically, the value of

x. E X-3 tX = X+2+x+8 2X-3=2x+10 2x+B - 2x +3 B A 스 x+2 - 13 x-3 ol C X48 А x Х D​

Mathematics
1 answer:
alexira [117]3 years ago
3 0
Danggg that is sooo longgg I feel bad for you
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Answer:

a) The Ohio Bureau of Motor Vehicles states that 7 out of 8 people pass the written driver’s test.

Let X be the number of test given by the test taker to pass out.

So X~Geometric(p) where p=Probability that for a particular test anyone will pass the written test=7/8

and here support of X be equal to 1,2,3,..... i.e X is a Natural number

So ,probability that he will pass the written test in fewer than 4 tries

=P(X<4)

=\sum _{x=0}^{3}p(1-p)^{x-1}

b) LAIMO manufacturing company makes parts for the auto industry. Approximately 3% of the parts it makes are defective.

So let X=number of non defective parts sampled before the 3rd defective part is sampled

then X~Negative Binomial(r,p) where here r=3 and p=Probability that a randomly selected part is defective= 0.03

where support of X is {0,1,2,3,...}

So the probability that the third defective part is the 20th one sampled.

P(X=20-3=17)

=\binom{r+16}{17}p^r(1-p)^{16}

c) A BigMart store is going to hire 3 new cashiers. It has 18 applicants (10 male, 8 female) for these 3 cashier jobs.

So let X be number of female cashier appointed.

Here X~Hypergeometric(3,8,18) where

f(x)=P(X=x)

=\left\{\begin{matrix} \frac{\binom{8}{x}.\binom{10}{3-x}}{\binom{18}{3}} & ,x=0,1,2,3\\ 0 & ,otherwise \end{matrix}\right.

So the probability that none of the positions are filled by females

=P(X=0)

d) A gardener is inspecting the fall flowers in her garden. She notices, on average, 4 bugs on a flower. She randomly picks one flower from her garden.

Let X be the numbers of bugs on that flower

So X follows Poisson distribution with mean 4 where support of X is {0,1,2,3.....}

So the probability that the flower she picked has at least one bug on it

=P(X\geq 1)=1-P(X=0)

=1-e^{-4}\frac{4^x}{x!}|_{x=0}

e) A student is taking a true/false test that consists of 15 questions. Based on past performance the student has approximately a 70% chance of getting any individual question correct.

So let X be the number of questions that are correct among those 15 questions.

so X~Binomial(n,p) where n=15 and p=Probability that he get an individual question correct =0.7

where support of X be {0,1,2,3,...,15}

So the probability that the student gets at least 60% of the questions on the test correct or 15x60%=9 questions are correct

=P(X\geq9)

=\sum _{x=9}^{15}\binom{15}{x}p^x(1-p)^{15-x}

f) A certain radio station’s phone lines are busy approximately 95% of the time when trying to call during a contest.

Let X denotes the number of calls to get into the contest.

So X~Geometric(p) where p=Probability that in a call I get through into the contest=1-0.95=0.05

support of x={1,2,3,....}

So the probability that the 4 th time you call is the 1st time you get through during a contest.

=P(X=4)

=p(1-p)^4

4 0
2 years ago
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