17.

Evaluate the expression \dfrac{5\cdot x^3}{x^2} x 2 5⋅x 3 start fraction, 5, dot, x, cubed, divided by, x, squared, end fracti

Fantom [35]

Answer:

10

Step-by-step explanation:

Given the expression:

(5 * x³) / x² ; for x = 2

Simplifying (5 * x³) / x²

5 * x^(3-2)

5 * x^1

= 5x

For x = 2

5x = 5*2 = 10

8 0

3 years ago

Which graph shows a system of equations with no solutions?

podryga [215]

Graph of Parallel lines shows a system of equations with no solutions

Step-by-step explanation:

Consider a set of equations

$7x - 2y = 16\\21x + 6y =24$

If we solve this both equations using any one of the solving method, (Substitution method) then we will get

$7x-2y=16\\7x=16+2y\\x=\frac{16+2y}{7}$

substituting the following x in 2nd equation (21x + 6y = 24) We get

$21(\frac{16+2y}{7} )+6y=24\\3(16+2y)+6y=24\\48+6y+6y=24\\12y=24-48\\y=-\frac{24}{12} \\y=-2$

Put y= -2 in x equation

$x=\frac{16+2(-2)}{7}\\ x=\frac{16-4}{7}\\\x=\frac{12}{7} \\x=1.71$

Comparing these (x,y) values we can understand that they never meet at a point

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

The area of a rectangle is greater than or equal to 115 square cm. The width of the rectangle is 5 cm and the Length of the rect

sattari [20]

Area= length*width
115= x * 5
115/5 = x
23 = x
So the length will be 23 cm.

4 0

3 years ago

What is 4/50 as a decimal?

Pavel [41]

0.08 is your answer to the question

5 0

3 years ago