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Andre45 [30]
1 year ago
13

A power line stretches down a 400-foot country road. A pole is to be put at each end of the road, and 1 in the midpoint of the w

ire. How far apart is the center pole from the left-most pole?

Mathematics
1 answer:
Usimov [2.4K]1 year ago
4 0

Answer:

The distance between the center pole and the left most pole = <em>200 ft</em>

<em></em>

Step-by-step explanation:

Given that a power line has a length of 400 ft.

Three poles are to be put, two on each end and one at the midpoint of the wire.

Let us represent this using the number line as shown in the attached diagram.

To find:

The distance between the center pole and the left most pole ?

Solution:

Please refer to the attached diagram for the number line representation of the given situation.

Point A refers to the starting of the road.

Point B refers to the mid point of the road and

Point C refers to the other end point of the road.

So, point A is a at 0.

Point C is at 400.

The mid point will be at:

AB = \dfrac{A's\ position+C's\ position}{2}\\\Rightarrow AB = \dfrac{0+400}{2}\\\Rightarrow AB = 200\ ft

The distance between the center pole and the left most pole = <em>200 ft</em>

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Step-by-step explanation:

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2 years ago
A probability calculator is required on this problem; answer to six decimal places. Suppose we will spin the wheel pictured 400
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Answer:

P(90< X< 110)= P(\frac{90-80}{8}

And we can find this probability with this difference:

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"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"

And we got 0.118 a very close value from the value obtained using the normal approximation

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=400, p=0.2)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

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nCx=\frac{n!}{(n-x)! x!}

We need to check the conditions in order to use the normal approximation.

np=400*0.2=80 \geq 10

n(1-p)=400*(1-0.2)=320 \geq 10

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And we want this probability:

P(90< X< 110)

We can use the z score formula given by:

z = \frac{x -\mu}{\sigma}

And replacing we got:

P(90< X< 110)= P(\frac{90-80}{8}

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P(90< X< 110)=P(z

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"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"

And we got 0.118 a very close value from the value obtained using the normal approximation

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