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stira [4]
3 years ago
12

Find the seventeenth term of the arithmetic sequence -4, -9, -15

Mathematics
1 answer:
Rufina [12.5K]3 years ago
6 0
When facing problems where you have to find a certain term of a sequence, look really closely for patterns. Some sequences have a really easy pattern like adding or subtracting numbers. Other sequences have more complicated patterns.

In this case, based solely on the three numbers you are given you can observe that the numbers are decreasing.

Here's what we can see so far:

To get from the 1st term to the 2nd term:
subtract 5 from the starting number
To get from the 2nd term to the 3rd term:
subtract 6 from the above result

Even though we only have a small piece of the pattern, let's try extending it! We're going to continue subtracting, but we're going to increase the amount that we take away every time.

Try solving it yourself before you look ahead! :)

So continuing the pattern:

Start at -4

To get from the 1st term to the 2nd term:
subtract 5
(-4 - 5) = -9

To get from the 2nd term to the 3rd term:
subtract 6
(-9 - 6) = -15

To get from the 3rd term to the 4th term:
subtract 7
(-15 - 7) = -22

To get from the 4th term to the 5th term:
subtract 8
(-22 - 8) = -30

To get from the 5th term to the 6th term:
subtract 9
(-30 - 9) = -39

[...continue this pattern until you reach the 17th term...]

Before you just write down the answer, make sure you do your own work! I calculated that the 17th term was -204.

Try practicing with some other sequences of numbers! You've got this!
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A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,
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Answer:

a) The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =85.82-81.90=3.92  

b) SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975  

And the Margin of error is given by:

Me= z_{\alpha/2} * SE=1.96*0.975=1.910

c) The 95% confidence interval would be given by 2.009 \leq \mu_1 -\mu_2 \leq 5.83.

Step-by-step explanation:

Notation and previous concepts

n_1 =35 represent the sample of ships that carry fewer than 500 passengers

n_2 =44 represent the sample of ships that carry 500 or more passengers

\bar x_1 =85.82 represent the mean sample of of ships that carry fewer than 500 passengers

\bar x_2 =81.90 represent the mean sample of of ships that carry 500 or more passengers

\sigma_1 =4.55 represent the population deviation of ships that carry fewer than 500 passengers

\sigma_2 =3.97 represent the sample deviation of ships that carry 500 or more passengers

\alpha=0.05 represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})} (1)  

Part a

The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =85.82-81.90=3.92  

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96  

The standard error is given by the following formula:  

SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}  

And replacing we have:  

SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975  

And the Margin of error is given by:

Me= z_{\alpha/2} * SE=1.96*0.975=1.910

Part c: What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

Confidence interval  

Now we have everything in order to replace into formula (1):  

3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009  

3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830  

So on this case the 95% confidence interval would be given by 2.009 \leq \mu_1 -\mu_2 \leq 5.83.

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No solution

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No because if that happened they wouldn’t be able to do it again so yeah
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