Solve the equation as shown 2(x+6) = 22

Please help me with this please and thank you

Ierofanga [76]

Step-by-step explanation:

a geometric sequence means that every term is created by multiplying the previous term by a certain constant factor. this factor is called the "common ratio".

so,

6×r = 18

r = 18/6 = 3

and a quick check tells us this works also for the next terms (18×3 = 54, 54×3 = 162), so it is indeed a geometric sequence with common ratio 3.

a0 = 6

a1 = a0 × 3 = 6×3 = 18

a2 = a1 × 3 = a0 × 3×3 = 6×3² = 54

an = a0 × 3^n = 6 × 3^n

so,

f(x) = 6×3^x, x is integer, x >=0

4 0

2 years ago

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa

xxMikexx [17]

Answer:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

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:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

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:

$E_{1} =\frac{50*189}{225}=42$

$E_{2} =\frac{50*36}{225}=8$

$E_{3} =\frac{75*189}{225}=63$

$E_{4} =\frac{75*36}{225}=12$

$E_{5} =\frac{100*189}{225}=84$

$E_{6} =\frac{100*36}{225}=16$

And the expected values are given by:

Size Company/ Heal. Ins. Yes No Total

Small 42 8 50

Medium 63 12 75

Large 84 16 100

_____________________________________

Total 189 36 225

And now we can calculate the statistic:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

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:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0

3 years ago

3 letters without replacement 4 letters A B C D how many ways can this be done if the order of the choices matters

Veseljchak [2.6K]

Answer:

Since the order of choice matters, we will permute the values. a bit more explanation for this:

If the order of choice did NOT matter, ABC and BCA will be counted as one since order of choice does NOT matter

Since order of choice does matter, ABC , BCA and CAB are all different possibilities for the arrangement of the same 3 letters

Since we have 3 slots:

___ ___ ___

Now, for the first slot. You can out either one if the 4 alphabets in the first slot since no slot has been used as of now

So:

_4_ ___ ___

**Keep in mind that the 4 is the possible number of values this slot can have**

Now that one slot has been used, one of the 4 alphabets has been used and since we are not allowed to repeat the same alphabets, we are left with 3 more alphabets

we can put any one of the 3 alphabets in this second slot, Hence:

_4_ _3_ ___

Now that 2 of the 4 alphabets have been used, we are left with only 2 alphabets, so there are only 2 possible alphabets for slot 3

Therefore:

_4_ _3_ _2_

Now that we know the possible alphabets for all 3 slots, we will multiply them with each other to get the total possible number of 3 - alphabet words we can make with 4 alphabets

Total possible words = 4 * 3 * 2

Total possible words = 24

We could've used the formula for Permutation as well

8 0

3 years ago

Leyla drops a penny from a height of 150 m. How long will it take the penny to hit the ground? Use the formula h(t)=−4.9t^2+vt+h

torisob [31]

To determine the time it takes for the penny to reach the ground, we use the relation of height and time given above. We do as follows:

0 =−4.9t^2+(0)t+150
t = sqrt( 150/4.9)
t = 5.53 s

Hope this answers the question. Have a nice day.

4 0

3 years ago

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A wine store conducted a study. It showed that a customer does not tend to buy more or fewer bottles when more samples are offer

aniked [119]

We can conclude that samples won’t affect how many bottles of wine a customer will buy.

4 0

3 years ago

Read 2 more answers