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2 years ago

Which is the sum of the first 6 terms of the geometric series? 7+21 +63

Mathematics

1 answer:

Gwar [14]2 years ago

5 0

Answer:

2,548

Step-by-step explanation:

First, we have to find out the rule. The rule is multiply the previous number by 3 to get the next number.

7x3=21

21x3=63

So, we would continue this pattern until we have the first 6 numbers which are 189, 567, and 1701.

The questions is asking for the SUM of the 6 numbers so you would add the those numbers and do this equation:

7+21+63+189+567+1701= 2548

Hope this helps! :)

You might be interested in

Options:<br> 14<br> 36<br> 16<br> 12

Lerok [7]

Answer:

12 million dollars.

Step-by-step explanation:

Let the sales in the first year be x million .

So the sales in the second year were x + 4 million, in the third year were 3(x + 4) and this is 48 million dollars.

So we have the equation 3(x + 4) = 48

3x + 12 = 48

3x = 36

x = 12 million dollars.

8 0

3 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .