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3 years ago

ILL GIVE BRAINLIEST Given the diagram, which of the following must be true?

Mathematics

2 answers:

Alex787 [66]3 years ago

8 0

Answer:

Choice D

Step-by-step explanation:

QS ≈ TR

they have same direction, sense and magnitude

max2010maxim [7]3 years ago

7 0

Answer:

B. QR ≅ TS

Step-by-step explanation:

Angle TR shows us that it is a right angle. Flip it to the other side as the inverse and we get the angle QR.

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2 years ago

Can u tell me the answers? please quick

stepan [7]

Answer:

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Step-by-step explanation:

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8 0

2 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

2 years ago

V-9/10v+6=11(hlp)this problem is confusing.

Valentin [98]

$\frac{v-9}{10v+6}=11 \\\\ 11(10v+6)=v-9 \\\\ 110v+66=v-9 \\\\ 110v-v=-9-66 \\\\ 109v= -75 \\\\ \boxed{v=-\frac{75}{109}}$

4 0

3 years ago

Read 2 more answers

PLEASE PLEASE PLEASE HELP ME. IM STUCK IN A SUMMER SCHOOL AND IF I DONT GET GOOD GRADES ILL GET IN BIG TROUBLE AND IM TOTALLY LO

Mrrafil [7]

Answer:

We know that in the box there are:

4 twix

3 kit-kat

Then the total number of candy in the box is:

4 +3 = 7

Here we want to find the probability that we draw two twix.

All the candy has the same probability of being drawn from the box.

So, the probability of getting a twix in the first drawn, is equal to the quotient between the number of twix and the total number of candy in the box, this is:

p = 4/7

Now for the second draw, we do the same, but because we have already drawn one twix before, now the number of twix in the box is 3, and the total number of candy in the box is 6.

this time the probability is:

q = 3/6 = 1/2

The joint probability is the product of the individual probabilities, so here we have

P = p*q = (4/7)*(1/2) = 2/7

b) same reasoning than in the previous case:

For the first bar, the probability is:

p = 3/7

for the second bar, the probability is:

q = 2/6 = 1/3

The joint probability is:

P = p*q = (3/7)*(1/3) = 1/7

c) Suppose that first we draw a twix.

The probability we already know that is:

p = 4/7

Now we want another type, so we need to draw a kit-kat, the probability will be equal to the quotient between the remaining kit-kat bars (3) and the total number of candy in the box (6)

q = 3/6

The joint probability is:

P = p*q = (4/7)*(3/6) = 2/7

But, we also have the case where we first draw a kit-kat and after a twix, so we have a permutation of two, then the probability in this case is:

Probability = 2*P = 2*2/7 = 4/7

3 0

3 years ago