1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sav [38]
3 years ago
10

Point M is the midpoint of AB. The coordinates of point A are (-7, 1) and the coordinates of M are (-4,1). What are the coordina

tes of point B? The coordinates of point B are :
Mathematics
1 answer:
I am Lyosha [343]3 years ago
3 0

Answer:

The coordinates of point B are : (-1, 1)

Step-by-step explanation:

\frac{-7+x^{2} }{2} = -4\\\frac{1+y^{2} }{2} =1 You multiply -4 by 2 which gives you -8 then you multiply 1 by 2 which gives 2.

Then you create the equations -7 + x^{2} = -8\\1 + y^{2} = 2 you substitute -7 for +7 and add 7 to -8 which gives -1 (x^{2}) and subtract 1 from 2 which gives you 1 (y^{2})

To check your answer use the midpoint formula \frac{-7+-1}{2} =\frac{-8}{2} =-4\\\frac{1+1}{2} =\frac{2}{2} =1 which gives you your midpoint (-4,1)

You might be interested in
g A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are he
stira [4]

Answer:

Step-by-step explanation:

Corresponding heights of presidents and height of their main opponents form matched pairs.

The data for the test are the differences between the heights.

μd = the​ president's height minus their main​ opponent's height.

President's height. main opp diff

191. 166. 25

180. 179. 1

180. 168. 12

182. 183. - 1

197. 194. 3

180. 186. - 6

Sample mean, xd

= (25 + 1 + 12 - 1 + 3 + 6)/6 = 5.67

xd = 5.67

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (25 - 5.67)^2 + (1 - 5.67)^2 + (12 - 5.67)^2+ (- 1 - 5.67)^2 + (3 - 5.67)^2 + (- 6 - 5.67)^2 = 623.3334

Standard deviation = √(623.3334/6 sd = 10.19

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 6 - 1 = 5

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (5.67 - 0)/(10.19/√6)

t = 1.36

We would determine the probability value by using the t test calculator.

p = 0.12

Since alpha, 0.05 < than the p value, 0.12, then we would fail to reject the null hypothesis.

Therefore, at 5% significance level, we can conclude that for the population of heights for presidents and their main​ opponents, the differences have a mean greater than 0 cm.

5 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
I will give brainliest for the best awnser
hodyreva [135]

I worked it out and I got south Africa

7 0
3 years ago
Read 2 more answers
Help me answer this question please
Klio2033 [76]

Answer:

The answer is C.

Step-by-step explanation:

Just go to Desmos.com and plug it in.

7 0
3 years ago
Read 2 more answers
What is the same ration of 2/3
frez [133]

Answer:

4/6

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Other questions:
  • Find the area of the triangle
    12·1 answer
  • Where does each answer belong in the boxes?? Right answers only, help please!! :))
    5·1 answer
  • Jada is making a circular birthday invitation for her friends . The diameter of the circle is 12 cm. She bought 180 cm of ribbon
    13·1 answer
  • What is the additive inverse of -3 *<br>A 3<br>B 4<br>C -3<br>D 0
    9·2 answers
  • Urgent! Thankyou!! Extra points! <br> 1. Create a rational expression that will be undefined at x=1
    8·1 answer
  • Heeelpppppp Meeeee!!!1 will give brainliest<br> Stay Safe and at home
    9·1 answer
  • Which expression represents each mathematical phrase?
    5·1 answer
  • Which of the x-values are solutions to the following inequality? x&lt; 100 Help ASAP
    8·1 answer
  • Hey stars of the day i need help
    12·1 answer
  • Which equation is true for y?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!