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4 years ago

Which of the following must be true? The diagram is not to scale.

Mathematics

2 answers:

satela [25.4K]4 years ago

7 0

I think it would be D) AC<FH

seraphim [82]4 years ago

3 0

Answer:

D. AC < FH

Step-by-step explanation:

We know that,

Hinge Theorem states that 'if two sides of a triangle are congruent to the two sides of an another triangle and the included angle of the first triangle is larger than the angle of the second triangle, then the third side of the first triangle is greater than the third side of the second triangle'.

Now, according to our figures, we see that,

In ΔABC and ΔHGF, sides AB is congruent to HG and BC is congruent to GF. Also, ∠HGF = 74° and ∠ABC = 31°.

Since, 74 > 31. Therefore, ∠HGF > ∠ABC.

So, according to the 'Hinge Theorem', side FH is greater than AC or AC is less than FH.

Hence, AC < FH.

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Step-by-step explanation:

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3 years ago

8. A rectangular prism measures 5 cm by

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Your answer is 817 cm3.

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3 years ago

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Which is an example of a statistical question?

tiny-mole [99]

Answer:

Question 4

Step-by-step explanation:

Question 4 is the correct answer because you are asking numerous students and are getting more than one answer. Then you can find the average or whatever.

The other questions only get 1 answer.

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4 years ago

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1 point

lana [24]

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3 years ago

2. MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of

antiseptic1488 [7]

Answer:

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

$X \sim N(75,15)$

Where $\mu=75$ and $\sigma=25$

The distribution for the sample mean $\bar X$ is given by:

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

b) $\mu$ represent the true average for the scores of the aptitude test

c) We can calculate the z scores and we got:

$z = \frac{70.14-75}{\frac{15}{\sqrt{25}}}= -1.62$

$z = \frac{82.14-75}{\frac{15}{\sqrt{25}}}= 2.38$

And we can calculate the probability with this difference:

$P(-1.62$

d) We can calculate the z scores and we got:

$z = \frac{82.68-75}{\frac{15}{\sqrt{25}}}= 2.56$

And we can calculate the probability with this difference:

$P(Z$

e) We can calculate the z scores and we got:

$z = \frac{78.69-75}{\frac{15}{\sqrt{25}}}= 1.23$

And we can calculate the probability with this difference:

$P(Z$

Step-by-step explanation:

a. What are the expected value, the standard deviation, and the shape of the sampling distribution of $\bar X$ ?

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".

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

$X \sim N(75,15)$

Where $\mu=75$ and $\sigma=25$

The distribution for the sample mean $\bar X$ is given by:

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

b. What is the random variable in this problem? Define it in words.

$\mu$ represent the true average for the scores of the aptitude test

c. What is the probability that the average aptitude test score in the sample will be between 70.14 and 82.14?

We can calculate the z scores and we got:

$z = \frac{70.14-75}{\frac{15}{\sqrt{25}}}= -1.62$

$z = \frac{82.14-75}{\frac{15}{\sqrt{25}}}= 2.38$

And we can calculate the probability with this difference:

$P(-1.62$

d. What is the probability that the average aptitude test score in the sample will be greater than 82.68?

We can calculate the z scores and we got:

$z = \frac{82.68-75}{\frac{15}{\sqrt{25}}}= 2.56$

And we can calculate the probability with this difference:

$P(Z$

e. What is the probability that the average aptitude test score in the sample will be less than 78.69?

We can calculate the z scores and we got:

$z = \frac{78.69-75}{\frac{15}{\sqrt{25}}}= 1.23$

And we can calculate the probability with this difference:

$P(Z$

7 0

3 years ago