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zalisa [80]
3 years ago
7

A nutritionist planning a diet for a rugby player wants him to consume 3,650 Calories and 650 grams of food daily. Calories from

carbohydrates and fat will be 70% of the total Calories. There are 4, 4, and 9 Calories per gram for protein, carbohydrates, and fat, respectively. How many daily grams of fat will the diet include?
210
300
350
140
Mathematics
2 answers:
Tcecarenko [31]3 years ago
8 0

Answer:

300

Step-by-step explanation:

because when multiply then divide all of them then find a percent 300 is your answer

9966 [12]3 years ago
5 0

Answer:

210 is the correct answer!

Hope this helps :)

Step-by-step explanation:

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The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds
Feliz [49]

Answer:

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 200, \sigma = 50

Find the probability that he weighs between 170 and 220 pounds.

This is the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 170.

X = 220

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

Z = \frac{220 - 200}{50}

Z = 0.4

Z = 0.4 has a pvalue of 0.6554

X = 170

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

Z = \frac{170 - 200}{50}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.6554 - 0.2743 = 0.3811

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.

6 0
3 years ago
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