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2 years ago

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Liam needed to get his computer fixed. He took it to the repair store. The technician at the store worked on the computer for 5.

25 hours and charged him $111 for parts. The total was $767.25. What was the cost of labor, per hour?

1 answer:

Luden [163]2 years ago

7 0

Answer:

$767.25 - $111 = $656.25

$656.25 ÷ 5.25 = $125

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Anyone help me solve this? And show work please

inessss [21]

Answer:

A) 2a + 6

Step-by-step explanation:

(5a + 3) + (2a + 3) + x = 9a + 12

Add both side lengths plus the unknown side to equal the perimeter.

7a + 6 + x = 9a + 12

-7a -6 -7a -6 Subtract 7a + 6 on both sides

x = 2a + 6 This is the missing side length.

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2 years ago

Luis purchased 24 purple plants and 8 pink plants. He wants to plant them in equal groups in his garden. What is the largest num

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3 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

alexandr402 [8]

Answer:

1) H0: There is significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

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

3) $\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $p_v = P(\chi^2_{2} >19.72)=0.00005222$

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated 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:

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 significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

Part 2

The statistic to check the hypothesis is given by:

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

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:

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

Part 3

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

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

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

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=0.00005222$

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.

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2 years ago

A triangular sail has a base length of 2.5 centimeters. The area of the sail is 3.75 square meters. How tall is the sail?

seraphim [82]

The area of any triangle = (1/2) x (length of the base) x (height)

2.5 centimeters = 0.025 meter

The area of the sail is (3.75) = (0.5) x (0.025) x (height)

Multiply each side by 2 : (7.5) = (0.025) x (height)

Divide each side by 0.025 : (7.5)/(0.025) = height =<em> 300 meters</em>

The situation is stupid, but the math is bullet-proof.

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3 years ago

How to substitute 3x+3y=-3 and y=5x-1?

schepotkina [342]

Since y = 5x-1, we can fill it into 3x + 3y = -3. First, let's look at relating to a simpler equation. Let's say that x + y = 9 and y = 3 + 5. Now, we can fill it in to get that x + (3 + 5) = 9. Now, we know that 3+5 is 8, so x + 8 = 9. Now, x = 1. Likewise, we can do the same. For 3x + 3y = -3, all we need to do is to switch the y in 3x + 3y = -3 with 5x - 1. So it would become 3x + 3(5x - 1) = -3. Now we distribute to get 3x + 15x - 3 = -3. Now add three to both sides to get 3x + 15x = 0. Now simplify to get 18x = 0. Now we know that x = 0. Now fill x into y = 5x - 1. So y = 5(0) - 1. Now we know that y = -1.

To check fill in the answer to 3x + 3y = -3.
3(0) + 3(-1) = -3
0 + (-3) = -3
0 - 3 = -3
-3 = -3
Now that our check is completed we now know that x is 0 and y is -1.

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3 years ago