Write in simplest form: -8(1.2c - 3.3d) 16.8cd -9.6c+26.4d 16.8(c + d)

TC= 0.0001Q^2-0.02Q+5+500Q

Leona [35]

Answer: 0

.

0

1

2

−

0

.

0

2

+

5

+

5

0

Step-by-step explanation:

5 0

2 years ago

What is the difference between secant and tangent line?

Naily [24]

Answer:

A secant line is a straight line joining two points on a function

A tangent line is a straight line that touches a function at only one point.

Step-by-step explanation:

6 0

4 years ago

Rebecca and dan are biking in a national park for three days they rode 5 3/4 hours the first day and 6 4/5 hours the second day

likoan [24]

Answer:

Rebecca and Dan need to ride $7\frac{9}{20}\ hrs.$ on the third day in order to achieve goal of biking.

Step-by-step explanation:

Given:

Goal of Total number of hours of biking in park =20 hours.

Number of hours rode on first day = $5\frac34 \ hrs.$

So we will convert mixed fraction into Improper fraction.

Now we can say that;

To Convert mixed fraction into Improper fraction multiply the whole number part by the fraction's denominator and then add that to the numerator,then write the result on top of the denominator.

$5\frac34 \ hrs.$ can be Rewritten as $\frac{23}{4}\ hrs$

Number of hours rode on first day = $\frac{23}{4}\ hrs$

Also Given:

Number of hours rode on second day = $6\frac45 \ hrs$

$6\frac45 \ hrs$ can be Rewritten as $\frac{34}{5}\ hrs.$

Number of hours rode on second day = $\frac{34}{5}\ hrs.$

We need to find Number of hours she need to ride on third day in order to achieve the goal.

Solution:

Now we can say that;

Number of hours she need to ride on third day can be calculated by subtracting Number of hours rode on first day and Number of hours rode on second day from the Goal of Total number of hours of biking in park.

framing in equation form we get;

Number of hours she need to ride on third day = $20-\frac{23}{4}-\frac{34}{5}$

Now we will use LCM to make the denominators common we get;

Number of hours she need to ride on third day = $\frac{20\times20}{20}-\frac{23\times5}{4\times5}-\frac{34\times4}{5\times4}= \frac{400}{20}-\frac{115}{20}-\frac{136}{20}$

Now denominators are common so we will solve the numerator we get;

Number of hours she need to ride on third day = $\frac{400-115-136}{20}=\frac{149}{20}\ hrs \ \ Or \ \ 7\frac{9}{20}\ hrs.$

Hence Rebecca and Dan need to ride $7\frac{9}{20}\ hrs.$ on the third day in order to achieve goal of biking.

8 0

3 years ago

PLEASE ANSWER FAST WILL DO BRAINLIESTThe coordinate grid shows points A through K. What point is a solution to the system of ine

Alja [10]

Answer: points B (4,7) and I (9,3)

If the inequalities are

y > −2x + 10 and y> (½)x -2

Step-by-step explanation: If I interpreted the inequalities correctly, the attached graph shows them. It is possible that you meant y > 1/(2x-2) for the second inequality. If so, we start over!

You can test the values for all the points, but it appears that (4,7) and (9,3) both work.

The other coordinates appear to be outside the solution -- the dark-shaded area.

I hope this is your Brainliest answer. It was a lot of work!

4 0

3 years ago

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0

4 years ago