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liq [111]
3 years ago
6

The weight of male american between the ages 18 to 25 is normally distributed, with mean of 162.7 lb and a standard deviation of

28 lb. One definition of obesity states that a person is obese if the z-score of his or her weight is greater than 2. What is the obesity threshold weight for a young adult american male? Round your answer to the nearest tenth.
Mathematics
2 answers:
MArishka [77]3 years ago
6 0

Answer:

218.7 lb.

Step-by-step explanation:

We have been given that the  weight of male american between the ages 18 to 25 is normally distributed, with mean of 162.7 lb and a standard deviation of 28 lb.

One definition of obesity states that a person is obese if the z-score of his or her weight is greater than 2.

We will use z-score formula to solve our given problem.

z=\frac{x-\mu}{\sigma}, where,

z=\text{ z-score},

x=\text{ Random sample-score},

\mu=\text{ Mean},

\sigma=\text{ Standard deviation}.

Upon substituting our given values in z-score formula we will get,

2=\frac{x-162.7}{28}

Now let us solve for x.

Upon multiplying both sides of our equation by 28 we will get,

2*28=\frac{x-162.7}{28}*28

56=x-162.7

Let us add 162.7 to both sides of our equation.

56+162.7=x-162.7+162.7

218.7=x

Therefore, the obesity threshold weight for a young adult American male is 218.7 pounds.

marusya05 [52]3 years ago
3 0

Answer: 218.7 lbs

Step-by-step explanation: apex

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Strike441 [17]

Answer:

25 one-dollar coins, 16 half-dollar coins, and 164 quarters

Step-by-step explanation:

First, set up equations based on the information given:

0.25q+0.50h+1.00d=74

\displaystyle{h=\frac{3}{5}d+1}

q=4(d+h)

Then, substitute <em>q</em> in the first equation with the expression from the third equation:

0.25[4(d+h)]+0.50h+1.00d=74\\1d+1h+0.50h+1.00d=74\\2d+1.5h=74

Next, substitute <em>h</em> in that equation with the expression from the second equation:

\displaystyle{2d+1.5(\frac{3}{5}d+1)=74}

2d+0.9d+1.5=74\\2.9d+1.5=74

Solve for <em>d</em>, the number of one-dollar coins:

2.9d+1.5=74\\2.9d=72.5\\d=25

Substitute 25 for <em>d</em> in the second equation to find <em>h</em>, the number of half-dollar coins:

\displaystyle{h=\frac{3}{5}d+1}

\displaystyle{h=\frac{3}{5}(25)+1}

h=15+1\\h=16

Substitute 25 for <em>d</em> and 16 for <em>h</em> in the third equation to find <em>q</em>, the number of quarters:

q=4(d+h)\\q=4(25+16)\\q=4(41)\\q=164

Then, verify that the coins total $74:

0.25(164)+0.50(16)+1.00(25)=74\\41+8+25=74\\74=74\\\text{Check.}

Next, verify that the number of half-dollar coins is one more than three-fifths of the number of one-dollar coins:

\displaystyle{h=\frac{3}{5}d+1}

\displaystyle{16=\frac{3}{5}(25)+1}

16 = 15 + 1\\16 = 16\\\text{Check.}

Finally, verify that the number of quarters is four times the number one-dollar and half-dollar coins together:

q=4(d+h)\\164=4(25+16)\\164=4(41)\\164=164\\\text{Check.}

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