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2 years ago

If average of 3 quantities is 15 , sum of quantities is....?

Mathematics

1 answer:

Verizon [17]2 years ago

8 0

Answer:

Step-by-step explanation:

Let the three quantities be x, y and z.

$\because \: average = \frac{sum \: of \: quantities}{no \: of \: quantities} \\ \\ \therefore \: 15 = \frac{x + y + z}{3} \\ \\ 15 \times 3 = x + y + z \\ \\ \huge \purple{ 45 = x + y + z}$

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Mazyrski [523]

Answer:

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Step-by-step explanation:

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3 years ago

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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

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

It is provided that:

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Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

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

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

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3 years ago

Keith_Richards [23]

Answer:

Step-by-step explanation:

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3 years ago

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so, just substitute with the givens in the equation to calculate the parameter:
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