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Zarrin [17]
3 years ago
7

Adam is going to cook a turkey for 14 people and wants to allow 3/4 lb of turkey for each person. 1lb = 450g

Mathematics
2 answers:
Zarrin [17]3 years ago
7 0
(You forgot to add the question, but I guess you want to know the mass of the turkey for 14 people :p )
By hypothesis, 1 lb = 450g. Adam wants that each of the 14 person, get 3/4 lb. Let x be the mass of 3/4 lb in grams, using the rule of three:

1 lb --> 450g
3/4 lb --> x

x = 3/4 * 450 = 337.5
So 3/4 lb = 337.5 g

Adam wants to cook for 14 people and wants to allow 3/4 lb of turkey for each person, so 337.5 for each.
So to know the total mass of the turkey, you multiply the mass for one person by the total  number of people:
337.5 * 14 = 4725
So the mass of the turkey for 14 people is 4725g, or 10.5 lb (3/4 * 14)

Hope this Helps! :)
timofeeve [1]3 years ago
6 0
3/4 = .75

.75 x 14 = 10.5 = 10 1/2

convert to grams = 10.5 * 450 = 4,725 grams

so he needs to cook a 10 1/2 lbs or  4,725 grams turkey for 14 people to allow 3/4 lbs each person.
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There were 54 sporting events in the year 1896 hosted by the competition

What is the function to determine the number of events in 1896?

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1 year ago
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3 years ago
If 2a:3b=5:6 and 3b:2c=36:15,then find a:c​
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a:c=2.5:1.25

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The incubation time for hummingbird eggs is approximately normal and has a mean of 16 days and standard deviation of 2 days. Let
Veseljchak [2.6K]

Answer:

0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 16 days and standard deviation of 2 days.

This means that \mu = 16, \sigma = 2

Probability that the length of hatching times is between 15 and 18 days.

This is the p-value of Z when X = 18 subtracted by the p-value of Z when X = 15. So

X = 18

Z = \frac{X - \mu}{\sigma}

Z = \frac{18 - 16}{2}

Z = 1

Z = 1 has a p-value of 0.8413

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Z = \frac{X - \mu}{\sigma}

Z = \frac{15 - 16}{2}

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Z = -0.5 has a p-value of 0.3085

0.8413 - 0.3085 = 0.5328

0.5328 = 53.28% probability that the length of hatching times is between 15 and 18 days.

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3 years ago
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