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2 years ago

HELP GEO QUESTION THX... GIVE ME EXPLANATION PLZZZZZ THXXXXX SOO MUCCHHHH

Mathematics

1 answer:

xxMikexx [17]2 years ago

7 0

E is the midpoint of DA, and F is the midpoint of DB. This means EF is a midsegment of the triangle. It is half as long as AB, and EF is also parallel to AB.

When we extend EF to form segment EC, it is now twice as long compared to EF. So that shows EC is the same length as AB, in other words, EC = AB. Also, EC is parallel to AB. This works due to points E, F, and C being on the same line.

Since EC || AB, we know that angle EBA = angle BEC. These are alternate interior angles.

From the reflective property, we can say EB = EB

---------------

Using these three facts

EC = AB
angle EBA = angle BEC
EB = EB

to prove that triangle EBA is congruent to triangle BEC. The next step is to use CPCTC (corresponding parts of congruent triangles are congruent) to lead to

angle BEA = angle EBC

we can use the alternate interior angle theorem converse to conclude that EA is parallel to BC.

-----------------

In summary, we have proven:

EC is parallel to AB
EA is parallel to BC

By the definition of what it means to be a parallelogram, we have proven quadrilateral EABC is a parallelogram. In short, we've shown that the opposite sides are parallel.

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It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

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Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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 life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

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 using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

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:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

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:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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i will tell you how,
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Now solve use ∠BDE,
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Answer:

y = -0.25x - 6

Step-by-step explanation:

You can either just read off the slope m = -0.25 and the intersection with the vertical n = -6 or you form a system of equations from points you choose on the graph. For example:

(0,-6);(4,-7)

The general equation for a line:

y = m*x + n

The two points on the graph form a system of two equations:

1. -6 = m*0 + n

2. -7 = m*4 + n

You can read off n right away fro equation 1:

n = -6

Plugging n into equation 2:

-7 = m*4 - 6 => -1 = m*4 => m = -1/4

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2 years ago

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