1² x 1/4m² x (-6)ml find the product<br>

If f(x)=4x^3 - 6x^2 + 2x - 5, then f(-2) = ___________

Art [367]

f(x)=4x^3 - 6x^2 + 2x - 5

f(-2)

Replace X with -2:

4(-2)^3 - 6(-2)^2 + 2(-2) - 5 =

Calulate the powers:

4(-8) - 6(4) +(-4) -5=

Multiply:

-32 - 24 -4 -5=

Subtract

-65

7 0

3 years ago

A circular lawn has a row of bricks around the edge. The diameter of the lawn is about 40 feet.

Murrr4er [49]

Answer:

1250

Step-by-step explanation:

please tell me if I am incorrect

7 0

3 years ago

Read 2 more answers

Four friends are out for a walk at night and come to an old bridge. it isn’t safe to cross the bridge without a flashlight, and

Anna71 [15]

First the person with a flashlight and one of his friends cross the bridge. After that same person with a flashlight comes back and leads another friend to the other side. Finally he comes back again and leads the last of them.
Number of times that he will have to cross the bridge is 5 ( five ).

8 0

3 years ago

If AA and BB are two disjoint events with P(A)=0.15P(A)=0.15 and P(B)=0.75P(B)=0.75, find the following probabilities:

zubka84 [21]

Answer:

a 0

b) 0.9

c) 0.85

d) 0.25

e) 0.1

f) 0.0375

Step-by-step explanation:

a)

As A and B are two disjoints events, then P(A and B) is

P(A and B)=0

b)

As A and B are two disjoints events, then P(A or B) is

P(A or B)=P(A)+P(B)

P(A or B)=0.15+0.75=0.9

c)

P(not A)=1-P(A)

P(not A)=1-0.15=0.85

d)

P(not B)=1-P(B)

P(not B)=1-0.75=0.25

e)

P(not (A or B))=1-P(A or B)

P(not (A or B))=1-0.9=0.1

f)

P(A and (not B))=P(A)*P(not B)

P(A and (not B))=0.15*0.25=0.0375

5 0

3 years ago

According to the norms established for a reading comprehension test, eighth-graders should average 83.2 with a standard deviatio

joja [24]

Answer:

$t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73$

$p_v =P(t_{44}>2.73)=0.0045$

If we compare the p value and a significance level given $\alpha=0.01$ 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 population mean is signficantly higher than 83.2 at 1% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=86.7$ represent the average life for the sample

$s=8.6$ represent the population standard deviation

$n=45$ sample size

$\mu_o =83.2$ represent the value that we want to test

$\alpha=0.01$ 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.

We need to apply a two tailed test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \leq 83.2$

Alternative hypothesis : $\mu > 83.2$

Compute the test statistic

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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:

$t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73$

Give the appropriate conclusion for the test

First we need to calculate the degrees of freedom given by:

$df=n-1=45-1=44$

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

$p_v =P(t_{44}>2.73)=0.0045$

Conclusion

If we compare the p value and a significance level given $\alpha=0.01$ 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 population mean is signficantly higher than 83.2 at 1% of significance.

8 0

2 years ago

1² x 1/4m² x (-6)ml find the product​

1² x 1/4m² x (-6)ml find the product