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3 years ago

HELP! <3 I have other questions as well not yet posted, if you'd like more points! :3

Mathematics

1 answer:

ratelena [41]3 years ago

3 0

Option B: $\frac{AB}{BC}=\frac{AE}{ED}$ is sufficient to prove that $BE$ ll $CD$

Explanation:

Given that ACD is a triangle.

The line EB intersects the sides AC and AD of the triangle.

The point E intersect the side AD and B intersect the side AC.

We need to prove that $BE$ ll $CD$

Then, the side splitter theorem states that "If a line intersects two sides of a triangle and divides the sides proportionally, the line is parallel to the third side of the triangle".

From the side splitter theorem, the line EB intersects the two sides of the triangle AC and AD and divides the sides proportionally.

Thus, the proportion of the sides is given by

$\frac{AB}{BC}=\frac{AE}{ED}$

This proportionality shows that the line is parallel to the third side of the triangle.

Hence, $\frac{AB}{BC}=\frac{AE}{ED}$ is sufficient to prove that $BE$ ll $CD$

Therefore, Option B is the correct answer.

You might be interested in

Twice the sum of a number and 3 times a second number is 4. The difference of ten times the seco

34kurt

Two numbers that gives twice the sum of a number and 3 times a second number is 4. The difference of ten times the second number and five times the first is 90 are -10 and 4

Given :

Twice the sum of a number and 3 times a second number is 4. The difference of ten times the second number and five times the first is 90.

Let a and b be the two unknown numbers

Lets frame equation using the given statements

Twice the sum of a number and 3 times a second number is 4.

$2(a+3b)=4\\a+3b=2$

the difference of ten times the second number and five times the first is 90

$10b -5a=90\\Divide \; by \;5 \\2b-a=18$

Now use these two equations to solve a and b

$a+3b=2\\-a+2b=18$

Add both the equations

$a+3b=2\\-a+2b=18\\--------------------------\\5b=20\\b=4$

Now find out 'a'

$a+3b=2\\a+3(4)=2\\a+12=2\\a=2-12\\a=-10$

The two numbers are -10 and 4

Learn more : brainly.com/question/13856304

4 0

3 years ago

den301095 [7]

X+2y
Plug in the variables
4+2(7)
4+14
x+2y=18

8 0

3 years ago

Help againnn<br><br> i hate matg

ser-zykov [4K]

Hello User!

Correction "Math"

Answer: 12 4/7

6 0

3 years ago

(20 points) please give me motivation to study and study tips or test taking tips for finals. it would be greatly appreciated.

Daniel [21]

Answer: Think about graduating. Think about never having to take the courses again. You're almost at the finish line! It'll be worth it. You've worked hard all year for this. You can do it!

Study tips: I would recommend Quizlet! They have a section that generates study games. It's a lot more fun than normal studying. It's also a good idea to make a goal for yourself. Try to make a challenge of achieving a certain score! By the time you accomplish said score, you'll find that you've learned a lot. Another tip is to make sure you take breaks. If you work too long without giving yourself a break, it will become harder to focus and your brain will become tired. Just don't get too distracted! set yourself an alarm during break times to help you stay on task. If you become frustrated with a certain subject or task, take a break from that task. Use this time as an opportunity to work on another subject. You can begin working on the first subject again once you feel refreshed. A lot of this may sound redundant, but hopefully it will help at least a little bit. Good luck!

6 0

3 years ago

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

scoray [572]

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.

5 0

3 years ago