What is the solution to the equation below?Round your answer to two decimal places.3*In x=9.9

Evaluate the function at the indicated values. f(x)= 6x^2+ 5x -12

seropon [69]

Step-by-step explanation:

We need to find the value of the function at the indicated value.

$f(x)= 6x^2+ 5x -12$

Indicated values are : f(-7), f(0), f(6), f(7)

To find f(-7), put x = -7 in the given function

$f(-7)= 6(-7)^2+ 5(-7) -12=247$

To find f(0), put x = 0 in the given function

$f(0)= 6(0)^2+ 5(0) -12=-12$

To find f(6), put x = 6 in the given function

$f(6)= 6(6)^2+ 5(6) -12=234$

To find f(7), put x = 7 in the given function

$f(7)= 6(7)^2+ 5(7) -12=317$

Hence, this is the required solution.

8 0

3 years ago

Help with this solution

Juli2301 [7.4K]

Literally y is across from 112, so y=112

7 0

3 years ago

In a random sample of 75 American women age 18 to 30, 26 agreed with the statement that a woman should have the right to a legal

ddd [48]

Answer:

a) $z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

$p_v =2*P(Z>0.236)=0.813$

If we compare the p value and using any significance level for example $\alpha=0.01$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.

b) We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

Step-by-step explanation:

Previous concepts and data given

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$p_A$ represent the real population proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_A =\frac{26}{75}=0.347$ represent the estimated proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_A=75$ is the sample size for A

$p_B$ represent the real population proportion for women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_B =\frac{21}{64}=0.328$ represent the estimated proportion of women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_B=64$ is the sample size required for B

$z$ represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:

Null hypothesis: $p_{A} = p_{B}$

Alternative hypothesis: $p_{A} \neq p_{B}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}$ (1)

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

Statistical decision

Since is a two sided test the p value would be:

$p_v =2*P(Z>0.236)=0.813$

If we compare the p value and using any significance level for example $\alpha=0.01$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.

Part b

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$(0.347-0.328) - 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=-0.188$

$(0.347-0.328) + 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=0.226$

And the 99% confidence interval for the difference of proportions would be given (-0.188;0.226).

We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

5 0

3 years ago

The ordered pair (4, -8) is in the 4th quadrant if we took the opposite of the x- value and the opposite of the y-value, which q

harina [27]

9514 1404 393

Answer:

2nd quadrant

Step-by-step explanation:

Reversing the signs of both coordinates reflects the point across the origin. The quadrant diagonally opposite quadrant 4 is quadrant 2.

_____

You recall that quadrants are numbered 1 to 4 counterclockwise, starting from upper right.

8 0

3 years ago

How many different right triangles are there with a hypotenuse of lenght 5 cm

Sveta_85 [38]

Answer:

Draw a horizontal line across a page of paper, somewhere in the middle. Mark a point A on the line, toward the left margin.

Spread open a compass to make good sized circle, but so that if the point of the compass is on the point you drew, the pencil fits on the page, both below the top edge and to the left of the right edge of the page. Draw the arc, from roughly the 12 O’Clock position over and down to the intersection of the line segment, at the 3 O’Clock position.

Call the opening of your compass, the radius of the arc you just drew, “5 units”.

Pick any point on the arc between the 12 and 3 positions, B. Drop a line down from that point B, perpendicular to the original horizontal line. Label the point that it intersects the horizontal line, C.

ABC is a right triangle with hypotenuse 5. Do this again with a point a little closer to the 3 O’Clock position. It’s another right triangle with hypotenuse 5.

Indeed, as you get closer to the right, along the arc, the height of the triangle declines, but the width of the triangle increases. The hypotenuse remains 5 in all cases. We picked points on the circle. We could have picked points on the horizontal line first, and raised perpendicular lines until they intersected the circle. Each point forms a distinct right triangle.

There are as many possible right triangles as there are possible points in a line segment.

Step-by-step explanation:

6 0

2 years ago