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3 years ago

Can anyone help me fast?

Mathematics

2 answers:

Yanka [14]3 years ago

7 0

Answer:

b=8

Step-by-step explanation:

b²+15²=17²

b²+225=289

b²=289-225

b²=64

b= 8

I hope I helped you^_^

andrew-mc [135]3 years ago

3 0

Answer:

b = 8

Step-by-step explanation:

c^2 = a^2 + b^2

17^2 = 15^2 + b^2

289 = 225 + b^2

b^2 = 289 - 225

b^2 = 64

b = 8

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Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

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

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Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

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

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

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

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

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

And replacing into the formula for the interval we got:

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

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

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

$23.818 \leq \sigma \leq 41.112$

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3 years ago

What are the zeros of the polynomial function f(x) = x^3-10x^2+24x

Ludmilka [50]

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