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3 years ago

What is the product? -1/4 (-6/11)

1 answer:

8 0

<span> -1/4 (-6/11)
To find the product, all you have to do is multiply the numerator by the other numerator and the denominator by the other denominator
6/44
Simplify
Final Answer: 3/22

*Note: Whenever you multiply or divide a negative number by another negative number, it automatically becomes a positive number.</span>

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Answer the following 4 questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Shkiper50 [21]

1. A) 5 (It would be B. because it does equal 4.09 but it has to be 5 because it is the greatest integer in all of them)

2. D) 10 (Just like 1. it would be B. also BUT 10 is the greatest integer you can have out of all of them)

3. A) 6 (Just like the other 2 problems at the top^ it would be C. if not for the fact it has a higher number.)

4. A) A value we can put in place of a variable (such as x) that makes the equation true.

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3 years ago

Read 2 more answers

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

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

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

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

You begin solving the equation 3+4x=51 by subtracting 3 from both sides. Which is the best choice for Step 2? Subtract 48 from b

Angelina_Jolie [31]

You will divide both sides by 4.

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3 years ago

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Suppose a new car is purchased for $46,357 and depreciates by 22% over the first year of ownership. If the car is driven 12,170

miss Akunina [59]

Answer:check the pic

Step-by-step explanation:

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3 years ago

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Write down the 3rd term in the sequence given by: T(n) = n2 + n

Sergio [31]

The solution for the problem is:
I will first get the first five terms so that I could easily locate the third term of this problem:So, substituting the values:
T(1) = 1^2 = 1T(2) = 2^2 = 4T(3) = 3^2 = 9T(4) = 4^2 = 16T(5) = 5^2 =25
So the third terms is T(3) = 3^2 = 9

8 0

3 years ago