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xxMikexx [17]
3 years ago
7

According to a human modeling​ project, the distribution of foot lengths of​ 16- to​ 17-year-old boys is approximately Normal wi

th a mean of 25.2 centimeters and a standard deviation of 1.5 centimeters. In the United​ States, a​ man's size 11 shoe fits a foot that is 27.9 centimeters long. What percentage of boys of this age group will wear a size 11 shoe or​ larger? The percentage of boys​ 16- to​ 17-years-old who wear a size 11 shoe or larger is nothing​%.

Mathematics
1 answer:
anygoal [31]3 years ago
6 0

The percentage of boys​ 16- to​ 17-years-old who wear a size 11 shoe or larger is 3.59​%.

<u>Step-by-step explanation:</u>

Step 1: Sketch the curve.

The probability that X>27.9 is equal to the blue area under the curve.

Step 2:

Since μ=25.2 and σ=1.5 we have:

P ( X > 27.9 ) = P ( X−μ > 27.9−25.2 )

⇒ P ( X−μ/σ > 27.9−25.2/1.5)

Since Z=x−μσ and 27.9−25.21.5=1.8 we have:

P ( X>27.9 )=P ( Z>1.8 )

Step 3: Use the standard normal table to conclude that:

P (Z>1.8)=0.0359

Percentage = 0.0359(100) = 3.59%

Therefore , The percentage of boys​ 16- to​ 17-years-old who wear a size 11 shoe or larger is 3.59​%.

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A rectangle has an area of 102cm2. The length of the rectangle is 17 cm. What is the perimeter of the rectangle
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joe has a piggy bank with $8.90 split upon nickels, dimes, and quarters. The piggy bank contains 76 coins in all. The number of
kozerog [31]

Answer:

There are 38 dimes, 22 nickels and 16 quarters.

Step-by-step explanation:

Let n, d and q represent the # of nickels, dimes and quarters respectively.

Then n + d + q = 76

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Thus, the value of n nickels is $0.05n (and so on).

The total value of the coins is $0.05n + $0.10d + $0.25q = $8.90.

d = n + q allows us to eliminate d.

First, n + d + q = 76 becomes n + (n + q) + q = 76, and second:

$0.05n + $0.10(n + q) + $0.25q = $8.90.  Here we have succeeded in eliminating d from two different equations, and now we have these two different equations in two unknowns (n and q), which is solvable.

Simplifying both equations, we get:

2n + 2q = 76 and

5n + 10n + 10q + 25q = 890, or 15n + 35q = 890

Let's use the substitution method of solving linear equations:

Rewrite 2n  + 2q = 76 as n + q = 38, or n = 38 - q.  Substituting this result into the second equation, we get:

15(38 - q) + 35 q = 890, or

  570 - 15q + 35q = 890, or

   570 + 20 q = 890.  Then 20q = 890 - 570 = 320, and q = 320/20 = 16.

There are 16 quarters.  Thus, the number of nickels is n = 38 - 16 = 22.

Finally, since n + d + q = 76, 22 + d + 16 = 76, or:

22 + d = 60, or d = 60 - 22 = 38.

There are 38 dimes, 22 nickels and 16 quarters.

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