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3 years ago

If f(x) = 5x – 2 and g(x) = x+1, find (f = g)(x).

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

5 0

Answer: Choice A) 5x^3 - x - 3

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Work Shown:

We subtract the two functions like so

(f - g)(x) = f(x) - g(x)

(f - g)(x) = ( 5x^3-2 ) - ( x+1 )

(f - g)(x) = 5x^3 - 2 - x - 1

(f - g)(x) = 5x^3 - x - 3 ..... choice A

Note: be sure to remember to distribute the negative to every term inside (x+1), and not just to the x only.

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Step-by-step explanation:

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Consider the equality xy k. Write the following inverse proportion: y is inversely proportional to x. When y = 12, x=5.

skelet666 [1.2K]

Answer:

$y=\dfrac {60} {x}$ or $xy=60$ (depending on your teacher's format preference)

Step-by-step explanation:

<h3><u>Proportionality background</u></h3>

Proportionality is sometimes called "variation". (ex. " 'y' varies inversely as 'x' ")

There are two main types of proportionality/variation:

Direct
Inverse.

Every proportionality, regardless of whether it is direct or inverse, will have a constant of proportionality (I'm going to call it "k").

Below are several different examples of both types of proportionality, and how they might be stated in words:

$y=kx$ y is directly proportional to x
$y=kx^2$ y is directly proportional to x squared
$y=kx^3$ y is directly proportional to x cubed
$y=k\sqrt{x}}$ y is directly proportional to the square root of x
$y=\dfrac {k} {x}$ y is inversely proportional to x
$y=\dfrac {k} {x^2}$ y is inversely proportional to x squared

From these examples, we see that two things:

things that are <u>directly proportional</u> -- the thing is <u>multipli</u>ed to the constant of proportionality "k"
things that are <u>inversely proportional</u> -- the thing is <u>divide</u>d from the constant of proportionality "k".

<h3><u>Looking at our question</u></h3>

In our question, y is inversely proportional to x, so the equation we're looking at is the following $y=\dfrac {k} {x}$ .

It isn't yet clear what the constant of proportionality "k" is for this situation, but we are given enough information to solve for it: "When y=12, x=5."

We can substitute this known relationship pair, and find the "k" that relates this pair of numbers:

<h3><u>Solving for k, and finding the general equation</u></h3>

General Inverse variation equation...

$y=\dfrac {k} {x}$

Substituting known values...

$(12)=\dfrac {k} {(5)}$

Multiplying both sides by 5...

$(12)*5= \left ( \dfrac {k} {5} \right ) *5$

Simplifying/arithmetic...

$60=k$

So, for our situation, k=60. So the inverse proportionality relationship equation for this situation is $y=\dfrac {60} {x}$ .

The way your question is phrased, they may prefer the form: $xy=60$

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2 years ago

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

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4 years ago