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Slav-nsk [51]
2 years ago
10

Triangle EFG is dilated by a scale factor of 1/4 ​ to form triangle E'F'G'. Side E'F' measures 12.512.5. What is the measure of

side EF?
Mathematics
1 answer:
WINSTONCH [101]2 years ago
6 0

Answer: 3.125

Step-by-step explanation:

Given

Triangle is dilated by a factor of \frac{1}{4} i.e. each side multiplies to 0.25.

Side E'F' becomes 0.25 times the original length

\Rightarrow E'F'=\dfrac{1}{4}\times 12.5=3.125

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The distribution of weights of United States pennies is approximately normal with a mean of 2.5 grams and a standard deviation o
stich3 [128]

Answer:

a) P(X

b) \bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)

c) P(\bar X

And using a calculator, excel or the normal standard table we have that:

P(Z

d) Figure attached

e) If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:

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

And we can't estimate the probabilities on a easy way.

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(2.5,0.03)  

Where \mu=2.5 and \sigma=0.03

We are interested on this probability

P(X

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

P(X

And we can find this probability on this way:

P(-0.50

Part b

Since the distribution for X is normal then the distribution for the sample mean is:

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

\bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)

Part c

P(\bar X

And using a calculator, excel or the normal standard table we have that:

P(Z

Part d

See the figure attached the deviation for the sample mean is lower for this reason we have the pattern in the graph attached.

Part e

If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:

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

And we can't estimate the probabilities on a easy way.

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3 years ago
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vesna_86 [32]

Answer:

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Step-by-step explanation:

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2 years ago
An example of an abelian group which is not cyclic
Leokris [45]

G=Z6×Z2G=Z6×Z2 will do (where ZnZn denotes the cyclic group of order nn). As a direct product of cyclic (so abelian) groups, GG is again abelian. Given any element (x,y)∈G(x,y)∈G, the order of (x,y)(x,y) will be the least common multiple of the orders of x,y.x,y.  

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Given ΔRST : ΔLMN, which of the following is true<br> 1.
kvasek [131]
I hope this helps you



m (R)=m (L)


m (S)=m (M)


m (T)=m (N)
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gregori [183]
Let the number be x.
9(1/4)+x=x/3+6
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