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sweet [91]
2 years ago
5

Use the recursive formula f(n) = 0.4 . f(n-1) + 12 to determine the 2nd term if f(1) = 4.

Mathematics
2 answers:
ZanzabumX [31]2 years ago
7 0

Answer:

13.6 (Answer C)

Step-by-step explanation:

I think you meant  f(n) = 0.4 * f(n-1) + 12, where * represents multiplication.

Then f(2) = 0.4 * (4) + 12, or 1.6 + 12, or 13.6.

Irina18 [472]2 years ago
5 0

Answer:

Assuming you have f(n)=0.4f(n-1)+12 with f(1)=4, the answer is f(2)=13.6.

Step-by-step explanation:

I think that says f(n)=0.4f(n-1)+12 with f(1)=4.

Now we want to find f(2) so replace n with 2:

This gives you:

f(2)=0.4f(2-1)+12

f(2)=0.4f(1)+12

f(2)=0.4(4)+12

f(2)=1.6+12

f(2)=13.6

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Answer:

(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908  

(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308  

So then the confidence interval for the difference of means is given by:

-8.908 \leq \mu_M -\mu_F \leq 5.308

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X_{M}= 69.8 represent the mean for the age of grandmother

\bar X_{F}= 71.6 represent the mean for the age of grandfather

s_{M}= 8.38 represent the sample deviation for the age of grandmother

s_{F}= 6.65 represent the sample deviation for the age of grandfather

n_M = n_F= 10

Solution to the problem

For this case the confidence interval is given by:

(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}

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df=n_M +n_F-2=10+10-2=18

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: "=-T.INV(0.025,18)".And we see that t_{\alpha/2}=2.101

And replacing we got:

(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908  

(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308  

So then the confidence interval for the difference of means is given by:

-8.908 \leq \mu_M -\mu_F \leq 5.308

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