What is 12 2/3 X 2 1/2

What is the sum of the first five terms of a geometric series with a1 = 6 and r = 1/3?

tresset_1 [31]

The sum of the terms of a geometric sequence with common ratio lesser than 1 is calculated through the equation,

Sn = (a1) x (1 - r^n) / (1 - r)
Substituting the known values,
S5 = (6) x (1 - (1/3)^5) / (1 - 1/3) = 242/27
Thus, the sum of the first five terms is approximately equal to 8.96.

6 0

3 years ago

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A roller coaster has four cars each with the same number of seats. these are 21 passengers seated in the roller coaster, but 3 s

belka [17]

Given:

Number of passengers seated in the roller coaster = 21

Empty seats = 3

Number of cars in roller coaster = 4 (each with the same number of seats)

To find:

An equation that can be used to determine the number of seats in each car.

Solution:

Let s be the number of seats in each car.

Total number of seats in 4 cars = 4s

Using the given information,

Total number of seats = Occupied seated + Empty seats

= 21 + 3

= 24

Now, the required equation is

$4s=24$

Therefore, the required equation is $4s=24$ .

Divide both sides by 4.

$s=\dfrac{24}{4}$

$s=6$

Therefore, the number of seats in each car is 6.

6 0

3 years ago

Help with question pls I don't understand

Mazyrski [523]

Answer:

15x^3y+6x^2y^2+9xy^3

Step-by-step explanation:

3xy ( 5x^2 +2xy +3y^2)

Distribute the 3xy to all terms in the parentheses

3xy*5x^2+3xy*2xy+3xy*3y^2

15x^3y+6x^2y^2+9xy^3

6 0

2 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

3 years ago

Jim says that the marked yellow paths show the shortest path to the tent. Is he correct ? Explain

aev [14]

Is there a picture so I can help

6 0

3 years ago