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shusha [124]
3 years ago
8

Help with this question, please!

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
6 0

The diet will include <u>123.75 </u>daily grams of protein

<h3>Further explanation</h3>

One variable linear equation is an equation that has a variable and the exponent number is one.  

Can be stated in the form:  

\large{\boxed{\bold{ax=b}}

or  

ax + b = c, where a, b, and c are constants, x is a variable  

We complete the task :

A nutritionist planning a diet for a football player wants him to consume 3,500 Calories and 725 grams of food daily. Calories from fat and protein will be 45% of the total Calories. There are 4, 4, and 9 Calories per gram for protein, carbohydrates, and fat, respectively. How many daily grams of protein will the diet include?

  • 243.75
  • 481.25
  • 123.75
  • 120

Calories from fat and protein will be 45% (0.45) of the total Calories

Calories from fat and protein = 0.45 x 3500 = 1575 calories

Calories from carbohydrates = 3500 - 1575 = 1925 calories

So mass for carbohydrates = 1925 : 4(4 calories/gram) = 481.25 grams

Then mass for fat and protein = 725 - 481.25 = 243.75 grams

We can make equation for fat and protein

1. 4 calories. mass of protein(p) + 9 calories . mass of fat(f) = 1575 calories

2. mass of protein(p) + mass of fat(f) = 243.75

we can simplify :

1. 4p+9f = 1575

2. p + f = 243.75 ⇒ f = 243.75 - p

we substitute equation 2 into equation 1 :

4p + 9(243.75-p) = 1575

4p + 2193.75-9p=1575

-5p = -618.75

p = 123.75

<h3>Learn more</h3>

linear equation

brainly.com/question/99841

Keywords : one variable, linear equation ,diet, nutritionist,calories, fat, proteins, carbohydrates, daily grams

#LearnWithBrainly

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An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do
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Answer:

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Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

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P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

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