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3 years ago

5. Write down the nth term of each of the following G.Ps

Mathematics

1 answer:

yawa3891 [41]3 years ago

7 0

Answer:

1i) nth term = 12(-b/4)^(n-1)

ii) nth term = 3(-1/9)^(n-1)

2i) Sixth term= (-3b^5)/256

ii) Eight term = -1/1594323

Question:

The complete question as found on brainly (question-10746111):

Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.

i) 12,-3b,......sixth term

ii) 3,-1/3,..........;8th term?

Step-by-step explanation:

1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.

i) 12,-3b,......;sixth term

1st term = a= 12

r = common ratio = 2nd term /1st term

r = -3b/12 = -b/4

Then we would find the nth term

See attachment for details

ii) 3,-1/3,..........;8th term

1st term = a= 3

r = common ratio = 2nd term /1st term

r = (-⅓)/3 = (-⅓)(⅓) = -1/9

Then we would find the nth term

See attachment for details

2) we are to determine the sixth term in (i) and eight term in (ii)

See attachment for details

Sixth term= (-3b^5)/256

Eight term = -1/1594323

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A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

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Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

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Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

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