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3 years ago

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:

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

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

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Step-by-step explanation:

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vaieri [72.5K]

Answer:

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b) P=0.007

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d) P=0.366

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.

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

Then,

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c) In this case, the distribution is B(1200,0.01)

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

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