Find the distance if mr.Nichols had been able to drive straight from the entrance to his campsite

Factor 5(x^2n-1)·(2x^3n 1)^2

nikdorinn [45]

I think I may be wrong check

5(x^2n-1)×(2x^3n-1) ^2

=20n3 x8 -20n2 x5 +20n x3 +20n x3 +5n x2-5

3 0

3 years ago

PLs Help! Pls help! Pls Help!

LiRa [457]

70% of 60,000,000 is 42,000,000

5 0

3 years ago

According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating

Natali [406]

Answer:

a) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

$H_0: \mu=32.48\\\\H_a:\mu< 32.48$

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565$

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Critical value approach

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

6 0

3 years ago

Need help please :( What is the second term of (s+v)^5?

Citrus2011 [14]

First compute the coefficient like this:
$C_5^1=\frac{5!}{1!(5-1)!}=\frac{5!}{4!}=\frac{5\times4!}{4!}$
Simplifying the fraction over 4! we get:
$C_5^1=5$
and the variables are $s^4v$ . So answer $5s^4v$ .

The correct answer is C then.

3 0

3 years ago

If there are 400 cubic centimeters of a chemical in 1 liter of solution, how many cubic centimeters of water must be added to di

mars1129 [50]

$1l=1dm^3\\1dm=10cm\to1dm^3=(10cm)^3=1000cm^3\\\\1l=1000cm^3\\-------------\\ \begin{array}{ccc}400cm^3&-&25\%\\1000cm^3+x&-&100\%\end{array}\\cross\ multiply\\25(1000+x)=400\times100\\25000+25x=40000\ \ \ \ \ |subtract\ 25000\ from\ both\ sides\\25x=15000\ \ \ \ |divide\ both\ sides\ by\ 25\\\boxed{x=600\ (cm^3)}\\\\Answer:Must\ be\ added\ 600cm^3\ of\ water.$

5 0

3 years ago