Which expression is equivalent to the given expression? Assume the denominator does not equal zero.

12n^3 + 16n^3 can someone please help me with this?

vodka [1.7K]

Answer:

$4n^{3} (3n^{2} + 4)$

Step-by-step explanation:

Assuming the problem is asking you to factor out the formula, it can be done easily by finding a common number with the two givens.

4 goes into both 12 and 16 evenly, so we will use this to factor. But that's not all, we have the n-variable to worry about.

Look at the n-variable exponents in the problem, take the highest power that can go be subtracted from both and use that. We have a $n^{5}$ and $n^{3}$ . Since 5 can't go into 3, we must use 3 to factor.

Our factoring variable will therefore be $4n^{3}$ . Now, use this to simply divide and get the answer:

$( 12n^{5} + 16n^{3} ) / 4n^{3} = 4n^{3}(3n^{2} + 4)$

So how did we get this? For the whole numbers, we have to divide: (12/4, 16/4). Next, we have to minus the exponents since we are dividing.

When dividing the 12 and 4, minus the $n^{5}$ and $n^{3}$ to get $n^{2}$ . Get rid of the $n^{3}$ term for the 16 and leave the 4. And there's your answer!

7 0

3 years ago

Vail Resorts pays part-time seasonal employees at ski resorts on an hourly basis. At a certain mountain, the hourly rates have a

notsponge [240]

Answer:

$z=0.842$

And if we solve for a we got

$\mu=13.16 +0.842*3=15.69$

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".

Solution to the problem

Let X the random variable that represent the hourly rates of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,3)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>13.16)=0.20$ (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 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.20

If we use condition (b) from previous we have this:

$P(X$