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2 years ago

Given:circle O, with tangent segments AC and AB . What is the measure of angle A

Mathematics

1 answer:

tino4ka555 [31]2 years ago

3 0

Answer:

B 84

Step-by-step explanation:

Angle COB = 96 (Angle CDB x2)

Lets split Quadrileteral ABOC into 2 equal parts.

Angle ABO = 90 (tangent segment of circle)

Angle AOB = 96/2 = 48

Angle OAB = 180 - 90 - 48 = 42

Angle A = Angle OAB x 2 = 42 x 2 = 84

Therefore Answer is B.

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4) −4b + 3 = 35 <br> solve equation

rosijanka [135]

Answer:

-8

Step-by-step explanation:

when you substitute -8 for b it is -4(-8)+3=35

-4(-8) is 32 and 32+3=35

Boom.

also btw need help give brainliest

5 0

3 years ago

Read 2 more answers

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

2 years ago

A taxicab starts at (1,-2) on the grid. It goes 3 blocks east and 1 block south to pick up a passenger. Then it goes 6 blocks we

Jobisdone [24]

The coordinates of the drop off would be (-2,2)

4 0