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4 years ago

13

Use the order of operations to solve. Please write the problem number on your paper and show your work for this problem. Work wi

ll be submitted at the end of the quiz.
(6 + 7) × (43 − 2) =

2 answers:

umka21 [38]4 years ago

5 0

To this question, 533 is the answer

Sedaia [141]4 years ago

4 0

Answer:

533

Step-by-step explanation:

(6+7)·(43-2)

do the parentheses ()

13·41

533

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X4 - 4k3 + 4k Factor out the GCF of all three terms<br>

Tatiana [17]

Answer:

Step-by-step explanation:

There isn't one. The first term is x^4.

The other two are stated in terms of k. There is nothing common here.

You can do 2 of the three. Terms 2 and 3 have a common factor of 4k, so what you can get is

x^4 - 4k(k^2 - 1)

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3 years ago

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

Anit [1.1K]

Answer:

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:

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

c) $23.818 \leq \sigma \leq 41.112$

And rounded :

$23.8 \leq \sigma \leq 41.1$

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:

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

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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

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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3 years ago

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tankabanditka [31]

Answer:

0.62

Step-by-step explanation:

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JulsSmile [24]

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Step-by-step explanation:

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3 years ago