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sveticcg [70]
3 years ago
14

Pauline is building a fence around her vegetable garden, what length of fence will she need to build?

Mathematics
1 answer:
Nuetrik [128]3 years ago
6 0
I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
 SEE ATTACHED IMAGE. 
 First we look for the hypotenuse of both triangles.
 Left triangle:
 Sine (68.1) = (1.75) / (h)
 h = (1.75) / Sine (68.1)
 h = 1.886108667
 h = 1.9m
 Right triangle:
 Sine (49.4) = (1.75) / (h)
 h = (1.75) / Sine (49.4)
 h = 2.304841475
 h = 2.3m
 Finally adding the perimeter:
 P = 5 + 1.9 + 2.75 + 2.3
 P = 11.95 m
 Answer: 
 she will need to build 11.95 m of fence

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Answer:

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Step-by-step explanation:

<u>Use the Pythagorean Theorem</u>

The equation for the Pythagorean Theorem is \tt a^2+b^2=c^2

* a and b are the lengths of the legs, and c is the hypotenuse.*

(To know which side is the hypotenuse look at longest side which is across the right angle).

a= x, b=8, c=9

\sf a^2+b^2=c^2

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\sf x^2+8^2=9^2

Evaluate 8^2 and 9^2:

\sf x^2+64=81

Subtract 64 from both sides:

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\sf x^2=17

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\sf x=\sqrt{17}

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2. The time between engine failures for a 2-1/2-ton truck used by the military is
OLEGan [10]

Answer:

A truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

For a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

Step-by-step explanation:

We have here a <em>random variable</em> <em>normally distributed</em> (the time between engine failures). According to this, most values are around the mean of the distribution and less are far from it considering both extremes of the distribution.

The <em>normal distribution</em> is defined by two parameters: the population mean and the population standard deviation, and we have each of them:

\\ \mu = 6000 miles.

\\ \sigma = 800 miles.

To find the probabilities asked in the question, we need to follow the next concepts and steps:

  1. We will use the concept of the <em>standard normal distribution</em>, which has a mean = 0, and a standard deviation = 1. Why? With this distribution, we can easily find the probabilities of any normally distributed data, after obtaining the corresponding <em>z-score</em>.
  2. A z-score is a kind of <em>standardized value</em> which tells us the <em>distance of a raw score from the mean in standard deviation units</em>. The formula for it is: \\ z = \frac{x - \mu}{\sigma}. Where <em>x</em> is the value for the raw score (in this case x = 5000 miles).
  3. The values for probabilities for the standard normal distribution are tabulated in the <em>standard normal table</em> (available in Statistics books and on the Internet). We will use the <em>cumulative standard normal table</em> (see below).

With this information, we can solve the first part of the question.

The chance that a truck will be able to travel a total distance of over 5000 miles without an engine failure

We can "translate" the former mathematically as:

\\ P(x>5000) miles.

The z-score for x = 5000 miles is:

\\ z = \frac{5000 - 6000}{800}

\\ z = \frac{-1000}{800}

\\ z = -1.25

This value of z is negative, and it tells us that the raw score is 1.25 standard deviations <em>below</em> the population mean. Most standard normal tables are made using positive values for z. However, since the normal distribution is symmetrical, we can use the following formula to overcome this:

\\ P(z

So

\\ P(z

Consulting a standard normal table available on the Internet, we have

\\ P(z

Then

\\ P(z1.25)

\\ P(z1.25)

However, this value is for P(z<-1.25), and we need to find the probability P(z>-1.25) = P(x>5000) (Remember that we standardized x to z, but the probabilities are the same).

In this way, we have

\\ P(z>-1.25) = 1 - P(z

That is, the complement of P(z<-1.25) is P(z>-1.25) = P(x>5000). Thus:

\\ P(z>-1.25) = 1 - 0.10565

\\ P(z>-1.25) = 0.89435  

In words, a truck "<em>will be able to travel a total distance of over 5000 miles without an engine failure</em>" with a probability of 0.89435 or about 89.435%.

We can see the former probability in the graph below.  

The chance that a fleet of a dozen trucks will have an average time-between-failures of 5000 miles or more

We are asked here for a sample of <em>12 trucks</em>, and this is a problem of <em>the sampling distribution of the means</em>.

In this case, we have samples from a <em>normally distributed data</em>, then, the sample means are also normally distributed. Mathematically:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the samples means are normally distributed with the same mean of the population mean \\ \mu, but with a standard deviation \\ \frac{\sigma}{\sqrt{n}}.

We have also a standardized variable that follows a standard normal distribution (mean = 0, standard deviation = 1), and we use it to find the probability in question. That is

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z \sim N(0, 1)

Then

The "average time-between-failures of 5000" is \\ \overline{x} = 5000. In other words, this is the mean of the sample of the 12 trucks.

Thus

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z = \frac{5000 - 6000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{\frac{800}{\sqrt{12}}}

\\ z = \frac{-1000}{230.940148}

\\ z = -4.330126

This value is so low for z, that it tells us that P(z>-4.33) is almost 1, in other words it is almost certain that for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is almost 1.

\\ P(z

\\ P(z

\\ P(z

The complement of P(z<-4.33) is:

\\ P(z>-4.33) = 1 - P(z or practically 1.

In conclusion, for a sample of 12 trucks, its average time-between-failures of 5000 miles or more is 0.9999925 or practically 1.

7 0
2 years ago
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