All categories

2 years ago

How do you find direct variation. I’m having trouble understanding some of these examples.

1 answer:

5 0

Direct variation is a relation that has the form

y = kx

where k is the constant of proportionality.

If you are told that a relation is a direct proportion, and you are given one data point, you can find k. The you can write the equation of the direct relation.

Here is an example.
The price of gasoline follows a direct variation.
John bought 5 gallons of gas and paid $15.
a) Write an equation for the relation.
b) Using the relation you found, how much do 13.8 gallons cost?

Solution:
Since the relation is a direct variation, it follows the general equation of a direct variation:

y = kx

We are given one data point, 5 gallons cost $15.
We plug in 5 for x and 15 for y and we find k.

y = kx
15 = k * 5
k = 3

Now that we know that k = 3, we rewrite the relation using our value of k.

y = 3x

This is the answer to part a).

Part b)

We use our relation, y = 3x, and we plug in 13.8 into x and find y.

y = 3x

y = 3 * 13.8

y = 41.4

The price of 15 gallons of gas is $41.40.

You might be interested in

1+4=5<br> 2+5=12<br> 3+6=21<br> 8+11=?

MariettaO [177]

Oh, I SEE what da pattern is! =D
You see, like in this problem:
2 + 5 = 12
They did 5 times 2 = 10, plus da 2 in da problem = 12.
Same with 3 + 6 = 21, 6 times 3 = 18, 18 plus da 3 given in da problem = 21.
Just like 1 + 4 = 5, 4 times 1 = 4, plus da 1 in da problem = 5.
SO,
with that strategy, 8 +11 = 96, because 8 times 11 = 88, plus da 8 given in da problem = 96.
I hope I helped! =D

6 0

2 years ago

PLEASE HELP PLEASE IM TIMED!!!

inysia [295]

Answer:

9316000000000

Simply move the decimal 12 places to the left.

If this helped please mark as brainiest <3

8 0

2 years ago

What do the following two equations represent? -3x-y = 3−3x−y=3minus, 3, x, minus, y, equals, 3 -3x-y = 3−3x−y=3

allochka39001 [22]

3 x minus y equals 3

4 0

2 years ago

Please help and show your work

ElenaW [278]

#6 (you should download an app call “Photomath “)

8 0

2 years ago

A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. The average nicotine content from a

tatuchka [14]

Answer:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

$df=n-1=15-1=14$

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

Step-by-step explanation:

Information provided

$\bar X=42.6$ represent the average nicotine content

$s=3.7$ represent the sample standard deviation

$n=15$ sample size

$\mu_o =40$ represent the value to check

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:

Null hypothesis: $\mu \leq 40$

Alternative hypothesis: $\mu > 40$

The statistic for this case would be:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replcing we got:

$t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722$

The degrees of freedom are:

$df=n-1=15-1=14$

The p value would be:

$p_v =P(t_{(14)}>2.722)=0.0083$

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

6 0

2 years ago