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3 years ago

Question 5 Multiple Choice Worth 1 points)

Mathematics

1 answer:

lara [203]3 years ago

7 0

Its key to copying a specified line is to open your beacon to the length of both the section, then use that amount of entrance to label off another section of the same length.
To bisect a segment or angle means to divide it into confocal scanning parts. A line segment's bisector would then pass thru the midpoint of its segment.
Once copying a segment as well as bisecting a section, its Draw a ray including one end-point step will be the same.
So if copying a line and bisecting a line segment, this same needle of both the compass is placed mostly on-line segment's end.
Therefore, the "Draw a ray with one endpoint" choices are correct, and the other choices were incorrect.

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Suppose that 75% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou

RSB [31]

Answer:

b) 0.2961

c) 0.2251

d) Mean = 11.25, Standard deviation = 1.667

Step-by-step explanation:

We are given the following information:

We treat trucks undergoing a brake inspection passin as a success.

P( trucks undergoing a brake inspection passes the test) = 75% = 0.75

a) Conditions for binomial probability distribution

There are n independent trial.
Each trial have a success probability p
The probability of success is same for all trials.

Then the number of trucks undergoing a brake inspection follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 15

b) P(proportion of groups will between 8 and 10 trucks pass the inspection)

We have to evaluate:

$P(8\leq x \leq 10) = P(x = 8) + P(x = 9) + P(x = 10)\\= \binom{15}{8}(0.75)^8(1-0.75)^7 + \binom{15}{9}(0.75)^9(1-0.75)^6 + \binom{15}{10}(0.75)^{10}(1-0.75)^5\\= 0.0393 + 0.0917 + 0.1651\\= 0.2961$

c) P( exactly 3 trucks fail the inspection)

p = 0.25

$P(x = 3)\\= \binom{15}{3}(0.25)^3(1-0.25)^{12}\\=0.2251$

d) Mean and standard deviation

$\mu = np = 15(0.75) = 11.25\\\sigma = \sqrt{np(1-p)} = \sqrt{15(0.75)(1-0.75)} = 1.667$

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shtirl [24]

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