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antoniya [11.8K]
3 years ago
15

I need an equation of the line that passes through the given point and is parallel to the given line (6,-3);y=-2/3 x+5

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

<em>y = -2/3 x + 1</em>

Step-by-step explanation:

The line y = -2/3 x + 5 is in y = mx + b form, where m is the slope, so the slope of the given line is -2/3. Parallel lines have equal slopes, so the equation you are looking for also has slope -2/3.

The parallel line has equation

y = -2/3 x + b

We need to find b.

Since we are given a point, we substitute x and y with the x- and y-coordinates of the given point and solve for b. x = 6, and y = -3.

y = -2/3 x + b

-3 = (-2/3)(6) + b

-3 = -4 + b

1 = b

Now that we know b = 1, we can write the equation.

y = -2/3 x + b

y = -2/3 x + 1

You might be interested in
There are 7 ushers and 11 technicians helping at the Harper Middle School fall play.
bearhunter [10]

Answer:

<h3>The ratio of technicians to all helpers  is 11 : 7, or \frac{11}{7} or 11 to 7.</h3>

Step-by-step explanation:

  • Given that there are 7 ushers and 11 technicians helping at the Harper Middle School fall play.
  • Let x be the number of ushers ( or helpers ).
  • Therefore x=7 helpers.
  • Let y be the number of technicians.
  • Therefore y=11 technicians.
<h3>To find the ratio of technicians to all helpers :</h3>

That is to find the ratio of y to x.

We can write the ratio of technicians to all ushers(helpers) as y : x

Which implies that 11 : 7,  (since y=11 and x=7)

Or \frac{11}{7} or 11 to 7

<h3>The ratio of technicians to all helpers  is 11 : 7, or \frac{11}{7} or 11 to 7</h3>
4 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
How can I do this? someone help me out please,, nn<br>​
Lelechka [254]

Answer:

didn't give clear instructions on what to do. I just assumed you add them. answer is down below. :)

Step-by-step explanation:

if ur supposed to add them then

1. 1/8 + 8/8 + 3/8 = 12/8 = 1 1/2

2. 1/4 + 2/4 + 1/4 = 4/4 = 1

3. 1/8 + 8/8 + 4/8 = 1 5/8

4. 1/4 + 1/4 + 2/4 = 4/4 = 1

5. 1/4 + 1/4 + 1/4 + 1/4 = 1

5 0
2 years ago
A ball, with an initial position of x = 25.89 meters, undergoes a displacement of 32.2 meters. What is it’s final position?
Snowcat [4.5K]

Answer:

The answer is 58.09

Step-by-step explanation:

To get the final position, you add the displacement to the initial position.

Final position = 25.89 + 32.2

Final position = 58.09metres

4 0
3 years ago
What part of a relation is the set of all first components from each ordered pair?
Orlov [11]
The mathematical term for a set<span> of </span>ordered pairs<span> is a </span>relation<span>. A </span>relation<span> is any </span>set <span>of </span>ordered pairs<span>. The </span>set of all<span> first </span>components<span> of the </span>ordered pairs<span> is called the domain of the </span>relation<span>, and the </span>set of all second components<span> is called the range of the </span>relation<span>.

I hope my answer helped you.</span>
5 0
3 years ago
Read 2 more answers
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