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Sloan [31]
2 years ago
6

Find the measure of angle A.​

Mathematics
1 answer:
Dmitriy789 [7]2 years ago
3 0

Answer:

30 degrees

Step-by-step explanation:

Keep this in mind, ALL triangles must equal 180 degrees.

This is an acute triangle, hence one of the angles is 90 degrees. To find the other two angles, we must solve for x. Since we know the sum of all angles in this triangle must equal 180 degrees, we know for setting up our equation, all the angles should equal 180.

90 + (x+37) + (x+67) = 180\\

Now solve for x by isolating x

2x + 194 = 180\\2x=-14\\x=-7

Our x is -7, now that we know that the numerical value of x is -7, replace x with -7 for angle A.

-7 + 37 = 30

<A is 30 degrees.

<u>Check your work</u>

Plug in -7 where you see x.

90 + (x+37) + (x+67) = 180\\90 + (-7+37) + (-7+67) = 180\\90 + 30 + 60 = 180\\180 = 180

Turns out this was a 30-60-90 triangle. ✅

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Answer:

a) 0.8

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c) 0.9

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Step-by-step explanation:

A probability always adds up to 1. It is between the scale of 0 (never likely) to 1 (very likely).

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1 - 0.2 = 0.8

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Answer:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

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

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

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Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent 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

Assume the following dataset:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      32   18    50

Medium                                 68     7    75

Large                                     89    11    100

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Total                                     189    36   225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1:  NO independence between heath insurance coverage and size of the company

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{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*189}{225}=42

E_{2} =\frac{50*36}{225}=8

E_{3} =\frac{75*189}{225}=63

E_{4} =\frac{75*36}{225}=12

E_{5} =\frac{100*189}{225}=84

E_{6} =\frac{100*36}{225}=16

And the expected values are given by:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      42    8    50

Medium                                 63     12    75

Large                                     84    16    100

_____________________________________

Total                                     189    36   225

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\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

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p_v = P(\chi^2_{2} >19.221)=0.000067

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"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

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