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3 years ago

M + 5n; use m = 1, and n = 4

Mathematics

2 answers:

Nutka1998 [239]3 years ago

7 0

$m + 5n; \\ m = 1, \\ n = 4 \\ ======= \\ 1+5 \cdot4= \\ 1+20= \\ \boxed{21}
$

guapka [62]3 years ago

7 0

If "m" = 1 and If "n" = 4

Your problem becomes↓ $1 + 5(4)$

$5(4) = 20$

$1 + 20 = 21$

Therefore your answer is: $21$

Just do PEMDAS

$Parentheses$
$Exponents$
$Multiplication$
$Division$
$Addition$
$Subtraction$

From the looks of it we have used the "addition " and "multiplication" method of the PEMDAS strategy

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a) $\hat \sigma^2 =s^2 =30^2 = 900$

b) $567.277 \leq \sigma^2 \leq 1690.224$

Rounded to the nearest number would be:

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Step-by-step explanation:

Data given and notation

s=30 represent the sample standard deviation

$\bar x=290$ represent the sample mean

n=20 the sample size

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Part a

The best point of estimate for the population variance is the sample variance, so on this case:

$\hat \sigma^2 =s^2 =30^2 = 900$

Part b

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

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

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Since the Confidence 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(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma \leq \frac{(19)(30)^2}{10.117}$

$567.277 \leq \sigma^2 \leq 1690.224$

Rounded to the nearest number would be:

$567 \leq \sigma^2 \leq 1690$

Part c

In order to find the confidence interval for the deviation we just need to take the square root for the interval of the variance, and we got:

$23.818 \leq \sigma \leq 41.112$

And rounded :

$23.8 \leq \sigma \leq 41.1$

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Read 2 more answers

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