ASAP!!! PLEASE help me with this question!

Which linear equations does the graph show the solution to? Select all that apply. y = 2x + 13 y = –x – 2 y = 3x – 5 y = negativ

salantis [7]

Answer:

y = 2x + 13

y = –x – 2

Step-by-step explanation:

The options are:

y = 2x + 13

y = -x - 2

y = 3x - 5

y= -(1/2)x + 6

y = -2x - 2

The graph is shown in the figure attached. There, point (-5, 3) is shown. Replacing it into the equations we get:

y = 2(-5) + 13 = 3 (so, it is a solution)

y = -(-5) - 2 = 3 (so, it is a solution)

y = 3(-5) - 5 = -20 ≠ 3 (so, it isn't a solution)

y= -(1/2)(-5) + 6 = 8.5 ≠ 3 (so, it isn't a solution)

y = -2(-5) - 2 = 8 ≠ 3 (so, it isn't a solution)

6 0

3 years ago

Read 2 more answers

Please help me thank you so much i will give brainliest to the right answer

AnnyKZ [126]

Is is the first one. 1600=.45x

3 0

3 years ago

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How many terms are in the expression shown? 2n + 5 – 3p + 4q 1 2 3 4

MA_775_DIABLO [31]

Step-by-step explanation: A term can be a number, a variable, or a number times one or more variables.

So in this expression, the terms are +2n, +5, -3p, and +4q.

This means that there are 4 terms.

3 0

3 years ago

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Jess owns 4 red shirts, 5 blue shirts, and 3 white shirts. She chooses a shirt to wear at random. What is the probability that s

xxTIMURxx [149]

D is the answer because in total she has 12 shirts and 3 out of the twelve shirts are white. so find the percentage of 3/12 which i belive is 1/4

7 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