What is the value of x ?!! Please HELP !!!!!! Will mark BRIANLIEST !!!!!!!!!

he price of oil recently went from $6.60 to $8.80 per case of 12 quarts. Find the ratio of the increase in price to the origin

Artist 52 [7]

It increased by $2.20, so the increase is 2.20. But out of ten dollars, it should be a 24% increase.

8 0

3 years ago

A ball is thrown with an initial speed of 17 mph in a direction that makes an angle of 22 degrees with the positive x-axis. Expr

Agata [3.3K]

Answer:

The speed of the ball in terms of i and j is (15.76 i, 6.37 j)

Step-by-step explanation:

Given;

velocity of the ball, v = 17 mph

angle of inclination, θ = 22⁰

The horizontal component of the velocity is calculated as;

$V_x = Vcos\theta\\\\V_x = 17cos(22)\\\\V_x = 15.76 \ i$

The vertical component of the velocity is calculated as;

$V_y = Vsin\theta\\\\V_y = 17sin(22)\\\\V_y = 6.37 \ j$

Therefore, the speed of the ball in terms of i and j is (15.76 i, 6.37 j)

7 0

2 years ago

I need some help! i will try to give brainliest!!!

o-na [289]

Answer:

its A because the Y intercept is at 3

7 0

2 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

3 years ago

Marisol is painting on a canvas that has an area of 180 square inches. The length of the painting is 11/4 times the width. What

Alona [7]

Notice the picture below, see the length "l"?

well $\bf Area=l\times w\qquad 
\begin{cases}
l=\frac{11}{4}w\\
Area=180
\end{cases}\\\\
-----------------------------\\\\
thus
\\\\
180=\cfrac{11}{4}w\times w$

solve for "w", to find the width,

what's the length? well, l = 11/4 w

3 0

3 years ago