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2 years ago

Marni has 11 ounces of walnuts. She needs 6 ounces for one recipe and pound for another. There are 16 ounces in 1 pound.

2 answers:

4 0

11-6=5 ounces. She doesn’t have enough for both of the recipes. She will need 11 more ounces to be able to make the recipes. 5+11=16 and 16 ounces are in a pound.

Liula [17]2 years ago

3 0

Answer:

No, she doesn't have enough.

She needs 22 ounces in total, so needs 11 more ounces.

Step-by-step explanation:

1. Marni has 11 ounces of walnuts. She needs 6 ounces for one recipe and 16 ounces for another.

1 pound = 16 ounces, so convert any measurements using pounds to ounces.

2. Marni needs 6 + 16 ounces for both recipes. That's 22 ounces in total.

Add together the amount of walnuts needed for both recipes.

3. 6 + 16 > 11;

Marni does not have enough ounces of walnuts. She will need at least 22 ounces, which is 11 more than what she currently has.

Hope this helped!

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

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

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

Just need someone to check my answer

mixas84 [53]

The correct answer is 216x⁶y⁵.

Explanation:
The first thing we do is raise the last monomial to the third power.

(4xy)(2x²y)(3xy)³
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=4xy(2x²y)(27x³y³).

Now we can multiply the first two monomials. When we multiply powers with the same base, we add the exponents:
8x³y²(27x³y³).

We multiply these last two monomials, again adding the exponents:
216x⁶y⁵.

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