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Naily [24]
3 years ago
7

Chef Andy needed to adapt his recipes for his European kitchen staff in Paris. He was​ 1-pound portions of​ entrecote, 2 ounces

of​ Béarnaise, and 4 ounces of asparagus. What are the metric conversions of each of these​ items?
Mathematics
1 answer:
Ronch [10]3 years ago
8 0

Answer:

1 pound = 0.454 kilograms

1 ounce = 0.0296 liters

1 ounce = 0.0283 kilograms

Step-by-step explanation:

You need to convert the quantities to their corresponding unit.

Convert pounds to kilograms (For the entrecote)

1 pound = 0.454 kilograms

Thus, 1 pound of entrecote = 0.454 kilograms

Ounces to liters (for the Béarnaise, since it is a liquid)

1 ounce = 0.0296 liters

2 ounces of Béarnaise = 0.0592 liters

Ounces to kilograms (for the asparagus)

1 ounce = 0.0283 kilograms

4 ounces of asparagus =0.1132 kilograms

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INT1 4.1.1 Line of Best Fit (classwork)
Lorico [155]

Answer:

a. 2401.06

b. 37.54%

c. 56.3%

Step-by-step explanation:

hopefully this is right

*note: I think you forgot to convert 4-2 back into yards.

a. Robbie's field of view to the North end includes parts not on the football field. so to find the area of the football field he can see, we need to find:

total area of what Robbie sees - area of non football field Robbie sees

what you shaded represents the total of what Robbie sees. it's a triangle. area of a triangle is 1/2(b)(h) where h is distance away and b is width of view.

total area = 1/2(170)(30.92) = 2628.56 yd

area of non football field (pipe to South end)

= 1/2(50)(9.1) = 227.5

so 2628.56 - 227.5 = 2401.06

b. area found in part a / total area of football field

2401.06 / (120*53.3) = .3754

.3754 * 100 = 37.54%

c. the chance of Robbie seeing the touchdown depends on how much of the (North) endzone he can see.

area of north endzone is

10 * 53.3 = 533.

area Robbie sees in endzone is

2628.56 - 2328.48 = 300.08

(found by total area Robbie sees - area of non endzone Robbie sees)

300.08 / 533 = 0.563

= 56.3%

7 0
3 years ago
12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that th
atroni [7]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

Confidence= 0.95 the confidence level desired

ME =0.03 represent the margin of error desired

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

The confidence level is 95% or 0.95, the significance is \alpha=0.05 and the critical value for this case using the normal standard distribution would be z_{\alpha/2}=1.96

Since we don't have prior information we can use \hat p= 0.5 as an unbiased estimator

Also we know that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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3 years ago
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3 years ago
What does 2+2 equal
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2+2=4

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