Howard pays $327 for one dozen collector’s edition baseball cards. About how much does he pay for each baseball card?

Instructions: Simplify the algebraic expression.<br><br> −12(32x−10y)

allochka39001 [22]

Answer:

-384x - 120y

Step-by-step explanation:

General Concepts and Formulas:

Brackets

Parentheses

Exponents

Multiplication

Division
Addition
Subtraction

Steps:

1) Distribute

$-12(32x-10y)$

We must distribute the expression above by multiplying -12 by the coefficients and then adding the variables.

So first, multiple -12 by 32 and add the variable 'x' when getting the product:

-12(-384x - 10y)

Now that we have one part, let's solve the other part.

We now must multiple -12 by -10, which gets us 120, now we must plug in:

(-384x - 120y)

(Remove parenthesis)

⇒ -384x - 120y

4 0

1 year ago

Drag each relation to the correct location on the table.

asambeis [7]

So what is the question exactly I’m confused on how this is asked

5 0

2 years ago

Read 2 more answers

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

Cscx/secx = cotx<br><br> Verify with steps please :)

Dovator [93]

Csc(x) = 1/sin(x)
sec(x) = 1/cos(x)
cot(x) = [1/sin(x)] / [1/cos(x)]
cot(x) = 1/sin(x) * cos(x)/1
cot(x) = cos(x) / sin(x)
cot(x) = cot(x)

6 0

3 years ago

Simplify; 5a+3b-7a+4b

serg [7]

Combine like terms:

5a-7a=-2a

3b+4b=7b

-2a+7b

Hope this helps!!

7 0

2 years ago

Read 2 more answers