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3 years ago

Evaluate the geometric series . 254 255 508 510

2 answers:

5 0

A1 is the first term in the sequence. To find the explicit formula, you will need to be given the first term and use the value in the formula. r is the common ratio for the geometric sequence. You will be given this value or be given information to compute it.

mihalych1998 [28]3 years ago

3 0

Answer with explanation:

The given geometric series is : 254, 255, 508, 510,....

→If you will look at the first term and third term , it is

254, 254 × 2,

So,First Half of the series will be as follows :

254×1, 254×2, 254×3,254×4,254×5........

→And if you will look at the second term and fourth term, it is as follows:

255, 255×2,

So,Second Half of the series will proceed as follows:

255×1, 255×2, 255×3, 255×4,.........

⇒So, the whole series will be as follows:

254, 255, 254×2,255×2,254×3,255×3,254×4,255×4,....

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3 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

5 0

3 years ago