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Advocard [28]
2 years ago
13

PLEASE HELP! The AHS football team did a weigh-in at the start of training camp. The weights of the players were distributed nor

mally with a mean of 98kg and a standard deviation of 6kg. What percentage of players are under 86kg?

Mathematics
1 answer:
Viktor [21]2 years ago
8 0
<h3>Answer:  2.5%</h3>

=============================================================

Explanation:

Notice how 98-2*6 = 98-12 = 86. So we're 2 standard deviations below the mean.

Start at the center, which is where the highest point is located. The value 98 is at the center. Then move 2 spots to the left, which has us move 2 standard deviations lower than the mean, and we arrive at 86. Refer to the diagram below.

From here, we add up the percentages in that diagram that are below the 86 mark. Those percentages are the 0.15% and 2.35%

So we get (0.15%) + (2.35%) = 2.5% which is the answer

Side note: This diagram is using the Empirical Rule. Some textbooks call it the "68-95-99.7 rule", but that's more clunky of a name in my opinion. The three values 68, 95 and 99.7 refer to the approximate percentage values within 1, 2 and 3 standard deviations of the mean.

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-1.2= y/2.4 -1.7<br> Solve for y
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{\huge{\boxed{\sf{Question}}

-1.2= y\div2.4 -1.7

Solve for y

{\huge{\boxed{\sf{Answer\:with\:explanation }}}

let's put our equation first

-1.2= y\div2.4 -1.7

Flip the equation

y\div2.4-1.7=-1.2

Now follow the rule of BOMDAS/PEMDAS

{\huge{\boxed{\sf{BOMDAS }}}

  • Brackets
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{\huge{\boxed{\sf{PEMDAS}}}

  • Parenthesis
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_________________________

{\huge{\boxed{\sf{Solve\:the\:equation}}}

_________________________

y\div2.4-1.7=-1.2

{\underline{simplify}}

0.416667y-1.7=-1.2

add 1.7 to both sides

0.416667y=0.5

Divide both sides by 0.416667

y=1.2

{\huge{\boxed{\sf{Answer}}}

y would equal 1.2

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