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AleksandrR [38]
2 years ago
13

An isosceles triangle has two equal sides. If two of the sides are 5 inches and 2 inches how long is the perimeter?

Mathematics
1 answer:
Jlenok [28]2 years ago
4 0
Perimeter is the sum of the lengths of all sides on any given shape. In this case, we know that since the triangle is isosceles, two sides must be the same (in a triangle, two of the same angles will produce two of the same sides opposite of the angles). However, there can not be two 2 inch long sides. Think of this logically; If you had two sides of 2 inches connected to a 5 inch long side, no matter what angle you pull those 2 inch sides down, the maximum it could reach is close to their cumulative sum, or around 4 inches. Since 4 inches isn't long enough to reach above 5 inches, then the only possibility for the side lengths is 5 inches + 5 inches + 2 inches. Thus, the perimeter is 12 inches.
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a) \bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b) Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c) P(\bar X >31.25)=0.006=0.6\%

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e) P(28.25

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =2)

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})

b. What is the standard error of the mean time?

The standard error is given by this formula:

Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >31.25)=1-P(\bar X

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(\bar X >28.25)=1-P(\bar X

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

P(28.25

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2 years ago
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