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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:
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
Keywords : one variable, linear equation ,diet, nutritionist,calories, fat, proteins, carbohydrates, daily grams
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Answer:
The calculated value of z = - 0.197 falls in the critical region therefore we reject the null hypothesis and conclude that at the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays
Step-by-step explanation:
We formulate the null and alternative hypotheses as
H0: p1= p2 there is no difference in population proportions of households that decorate their houses with lights for the holidays
against Ha : p1≠ p2 (claim) ( two sided)
The significance level is set at ∝= 0.05
The critical value for two tailed test at alpha=0.05 is ± 1.96
or Z∝= 0.05/2= ± 1.96
The test statistic is
Z = p1-p2/√pq(1/n1 +1/n2)
p1= proportions of households decorating in city 1 = 45/60=0.75
p2= proportions of households decorating in city 2 = 40/50= 0.8
p = the common proportion on the assumption that the two proportion are same.
p =
Calculating
p =60 (0.75) + 50 (0.8) / 110
p= 45+ 40/110= 85/110 = 0.772
so q = 1-p= 1- 0.772= 0.227
Putting the values in the test statistic and calculating
z= 0.75- 0.8/ √0.772*0.227( 1/60 + 1/50)
z= -0.05/√ 0.175244 ( 110/300)
z= -0.05/0.25348
z= -0.197
The calculated value of z = - 0.197 falls in the critical region therefore we reject the null hypothesis and conclude that at the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays