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aivan3 [116]
3 years ago
10

Which value is the 10th term in the sequence: -62, -48, -34, -20, -6....?

Mathematics
2 answers:
Karolina [17]3 years ago
8 0
B) 64, because all these numbers have a difference of 14, and it you keep going, the 10th term will be 64.
notka56 [123]3 years ago
7 0
Using the sequence, you can tell that this is an arithmetic sequence. Therefore, the common difference is 14. The formula is 14n - 76. Using this, plug in n=10 and the answer comes out to be 64, or B.
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PLS HELP ASAP THANKS ILL GIVE BRAINLKEST
Elis [28]

Answer:

5

Step-by-step explanation:

<u>move constant</u>

x+7 ≥ 12

<u>subtract</u>

x≥ 12-7

<u>solution</u>

x ≥ 5

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
Use the distance formula to determine whether ABCD below is a parallelogram. A(-3,2) B(-3,3) C (5,-3) D (-1.-5)
Pani-rosa [81]

Answer:

ABCD is not a parallelogram

Step-by-step explanation:

Use the distance formula to determine whether ABCD below is a parallelogram. A(-3,2) B(-3,3) C (5,-3) D (-1.-5)

We have to find the length of the sides of the parallelogram using the formula below

= √(x2 - x1)² + (y2 - y1)² when given vertices (x1, y1) and (x2, y2)

For side AB

A(-3,2) B(-3,3)

= √(-3 -(-3))² + (3 -2)²

= √0² + 1²

= √1

= 1 unit

For side BC

B(-3,3) C (5,-3)

= √(5 -(-3))² + (-3 -3)²

= √8² + -6²

= √64 + 36

= √100

= 10 units

For side CD

C (5,-3) D (-1.-5)

= √(-1 - 5)² + (-5 - (-3))²

= √-6² + -2²

= √36 + 4

= √40 units

For sides AD

A(-3,2) D (-1.-5)

= √(-1 - (-3))² + (-5 -2)²

= √(2² + -7²)

= √(4 + 49)

= √53 units

A parallelogram is a quadrilateral with it's opposite sides equal

From the above calculation

Side AB ≠ CD

BC ≠ AD

Therefore, ABCD is not a parallelogram

3 0
3 years ago
Need help with this stuck on another problem
Cerrena [4.2K]

just remove perenthesis

5 0
3 years ago
21. Which number below is equivalent to 3.82 x 10-5?
IceJOKER [234]

Answer:

19.10

Step-by-step explanation:

If im correct,

You subtract 10 and 5. Wich is 5. Then multiply that by 3.82.

3 0
3 years ago
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