Line PQ is parallel to line RS<br> ∠MOQ=65°<br><br><br> What is ∠RTN=

(60) POINTS AND 5 STAR AND THANKS AND BRIAN-LIST!!!!!

katovenus [111]

Answer:

1. Send me the chart and ill get you the answer.

2. unit cost is $0.30

Step-by-step explanation:

3 0

2 years ago

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Sierra’s ice cream shop made $45 before she had to close early when the power went out. She realized she lost Two-thirds of a do

lozanna [386]

Answer:

Sierra's profit was $41.

Step-by-step explanation:

Sierra realized she lost two-third of a dollar each hour when the shop was closed.

The expression that models this situation is, ............. (1)

Now, we have to calculate Sierra's profit if the shop was closed for 6 hours.

So, putting h = 6 hours in the equation (1) we get,

dollars.

So, Sierra's profit was $41. (Answer)

Step-by-step explanation:

3 0

3 years ago

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Please answer these two questions precisely :)

GuDViN [60]

1. A)
7x= 3(35)
7x= 105
105/7= 15
Therefore x=15

2. C)

4 0

3 years ago

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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

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

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago

Please come answer this, I need help asap

Vadim26 [7]

Answer:

26

Step-by-step explanation:

To evaluate this expression, all we have to do is substitute the given values of the variables into the expression. That is, replace every $m$ in the expression with a $1$ and every $p$ in the expression with a 5.