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makvit [3.9K]
3 years ago
8

Enter the equation of the circle with the given center and radius.

Mathematics
1 answer:
Montano1993 [528]3 years ago
7 0

HOPE IT HELPS!!!!!!!!!!

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CAN SOMEONE HELP ME WITH THESE QUESTIONS PLEASE
Lady_Fox [76]

Answer:

17. A

18. B

Step-by-step explanation:

<u>17.</u> We need to find the longest side length. Remember that in a triangle, the smallest angle corresponds to the shortest side length and the largest angle corresponds to the longest side length.

Here, we already know two angles are 75 and 60, which means that the third angle is 180 - 75 - 60 = 45 (because all angles of a triangle add up to 180). Since 45 is the smallest angle, it corresponds to the shortest side, which we know is 10 units. Since 75 is the largest angle, it corresponds to the longest side, which we want to find.

We now use the Law of Sines, which states that for a triangle with side lengths a, b, and c and angles A, B, and C:

\frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}

Here, let's say that a = 10, A = 45, and B = 75. Then:

\frac{a}{sinA} =\frac{b}{sinB}

\frac{10}{sin(45)} =\frac{b}{sin(75)}

Remember that sin(45) = √2/2 and sin(75) = (√2 + √6) / 4. Plug these in and solve for b, which is the longest side we want to find:

b = [(√2 + √6) / 4] * [10 / (√2/2)] = 5 + 5√3 = 5(1 + √3)

The answer is thus A.

<u>18.</u> We need to use trigonometry for this. Notice that in triangle ABC, we can write the tangent (which is opposite divided by adjacent) expression:

tan(30) = BC / AB = h / (4 + x)

Remember that tan(30) = √3/3. So:

h/(4 + x) = √3/3

Now we can also write a tangent expression for triangle BDC:

tan(60) = BC / DB = h/x

Remember that tan(60) = √3, so:

h/x = √3

We're looking for x, so write h in terms of x from the second equation:

h/x = √3

h = x√3

Plug this in for h in the first equation:

h/(4 + x) = √3/3

(x√3) / (4 + x) = √3/3

x√3 = (4√3 / 3) + x√3/3

x√3 - x√3/3 = 4√3/3

2x√3/3 = 4√3/3

x = 2

The answer is thus B.

7 0
3 years ago
Which lines are parallel? Justify your answer.
svlad2 [7]

Answer:

It's C

Step-by-step explanation:

I just took the test :)

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ANSWER ASAP WILL GIVE BRAINLIEST PLS HELP
jeka57 [31]

Answer:

Step-by-step explanation:

11) Probability of Black is  1/4

12) P of  not orange is  3/4

13) P of blue or black is 1/2

this should be pretty easy for you, why is it tough?  don't say .. " it just is" say what  makes it tough.  I'm curious why this is difficult.   :?

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Ms Feheley buys 9 bottles of eye of Newt for the same price, and also a new couldron for $9.45. She spent $118.89. How much did
scoundrel [369]

See picture for answer and solution steps.

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3 years ago
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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
4 years ago
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