URGENT. Find f(3) if<br> f(x)=x^3 + 2x^2-x-1

If two sides of a triangle are 5 and 7 inches longer than the third side and the perimeter measures 21, find the length of each

aleksandr82 [10.1K]

Hello

x
x +5
x + 7

x + x + 5 + x + 7 = 21
3 x = 21 - 12
3 x = 9
x = 3

3 , 5 et 7 pouces

7 0

3 years ago

Help me now if i get this one wrong i wont be free

jok3333 [9.3K]

Answer: its the first one "opposite sides with equal length"

5 0

2 years ago

If x = 0 and y< 0 where is the point (x, y) located with

maria [59]

Answer:

on the negative y- axis

7 0

3 years ago

What has my life come to

Verizon [17]

Step-by-step explanation:

whaaaaaat ........how do we know that

8 0

2 years ago

Read 2 more answers

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

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

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ 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)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

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

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0

3 years ago

URGENT. Find f(3) if f(x)=x^3 + 2x^2-x-1