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igor_vitrenko [27]

2 years ago

7

Solve the system of equations below by graphing both equations with a pencil and paper. What is the solution? y = -x + 21 y = -2

x + 30

1 answer:

Kamila [148]2 years ago

4 0

-x+21=-2x+30
-x+2x=-21+30
x=9
-(9)+21=12

y=12

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Simplify 5(x - 2) + 6<br><br> 5x + 4<br> 5x - 4<br> 5x + 20

Akimi4 [234]

5x-10+6 (then you sum the like terms)
5x-4
the answer is B 5x-4

6 0

3 years ago

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1. The perimeter of a piece of paper is 18 cm.

OlgaM077 [116]

Answer:

a) length = 3 and width = 6

Area of the rectangle = 3 × 6 = 18 cm²

b) length = 7 and width = 2

Area of the rectangle = 7 × 2 = 14 cm²

c) length = 5 and width = 4

Area of the rectangle = 5× 4 = 20 cm²

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given perimeter of the paper = 18 cm

we know that the paper has a rectangle shape

The perimeter of the rectangle = 18 cm

2( l + w ) = 18

a)

we choose length = 3 and width = 6

The perimeter of the rectangle = 2( 3+6) = 18 cm

Area of the rectangle = 3 × 6 = 18 cm²

b)

we choose length = 7 and width = 2

The perimeter of the rectangle = 2( 7+2) = 18 cm

Area of the rectangle = 7 × 2 = 14 cm²

c)

we choose length = 5 and width = 4

The perimeter of the rectangle = 2( 5+4) = 18 cm

Area of the rectangle = 5× 4 = 20 cm²

3 0

2 years ago

Please help! I will mark as brainliest IF answer is right. <3

EastWind [94]

Answer:

b^2 + 3b - 4 + 4/(b-2)

Step-by-step explanation:

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7 0

2 years ago

Read 2 more answers

If x = 4, then 5x = 20

Alexandra [31]

Answer:

Step-by-step explanation:

3 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

3 years ago