All categories

3 years ago

12. (2v^3 - v + 8) + (-v^3 + v - 3) 14. (4h^3 + 3h + 1) - (-5h^3 + 6h - 2)

1 answer:

7 0

12. On addition, you can just combine like terms.
2v^3+(-v^3)=v^3
-v+v cancels each other out
8+(-3)=5
So you have v^3+5

14 On subtraction, you have to remember to distribute the negative sign so after you do that you have:
4h^3+3h+1+5h^3-6h+2
Then you can combine like terms
4h^3+5h^3=9h^3
3h-6h=-3h
1+2=3
So you end up with:
9h^3-3h+3

Hope that helps and feel free to ask any questions.

You might be interested in

PLEASE HELP!!!!!!!!!!!!!!!

kotykmax [81]

What's the question?

7 0

3 years ago

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa

Norma-Jean [14]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation

$\bar X=18.3$ represent the sample mean

$s=3.72$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

If we analyze the size for the sample is < 30 and we don't 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".

Part b: P-value and conclusion

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

Conclusion

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0

3 years ago

Simplify the difference quotient f ( x + h ) â f ( x ) h for the function f ( x ) = 6 x â x 2

ladessa [460]

Answer:

final answer is 6

Step-by-step explanation:

Given $f(x)=6x-2$