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Pani-rosa [81]
3 years ago
5

Points A, B, and C are collinear. Point B is between A and C. Solve for x if AC = 3x + 3, BC = 3, and AB = 2x + 2.

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

x=2.

Step-by-step explanation:

It is given that points A, B, and C are collinear. Point B is between A and C.

Using segment addition property, we get

AC=AB+BC

It is given that AC = 3x + 3, BC = 3, and AB = 2x + 2.

3x+3=(2x+2)+3

3x+3=2x+5

Isolate variable terms.

3x-2x=5-3

x=2

Therefore, the value of x is 2.

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Please help!<br><br> Thank you!
Dafna1 [17]

Answer:

(-16,17)

Step-by-step explanation:

The answer for the and point D is negative 16, 17. how you will solve this is you would take them in point and the endpoint c. to find that x of D you will do negative 4 minus 8 this equals negative 12. what you will do next is you will take the negative 12 and added to -4. this is how you get negative 16. you will then find the y of D. to do this you will take negative 7 and subtract 5. this equals negative 12. you will then take the midpoint and do 5 + 12. this equals 17. Hope this helps!

6 0
2 years ago
Convert the degrees measure 180° to radian measure.
Mandarinka [93]
180° = 3.1415926536 rad
4 0
3 years ago
Read 2 more answers
KN is perpendicular bisector of MQ identify the value of x
ExtremeBDS [4]

Answer:

x = 6

Step-by-step explanation:

Since KN is the perpendicular bisector, that means ∠KNM = ∠KNQ = 90° and MN = NQ so therefore, since they are right triangles, ΔKNM ≅ ΔKNQ because of HL. Therefore, KM = KQ by CPCTC so:

5x - 3 = 3x + 9

2x = 12

x = 6

4 0
3 years ago
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
3 years ago
PLS HELP ASAP THIS IS DUE IN 5 MINUTES
MA_775_DIABLO [31]

Answer:

A and B

Step-by-step explanation:

6 0
3 years ago
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