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posledela
3 years ago
13

Why would there be a difference between percentages with the Empirical Rule and the actual percentages?

Mathematics
1 answer:
IgorC [24]3 years ago
5 0

Answer:

Given that \int\limits^a_b {\frac{1}{2\pi } e^{-1/2} x} \, dx   is not the easiest equation to solve, and that

the values at the convenient ± 1 sigma,± 2 sigma,± 3 sigma are irrational

using the "rule-of-thumb" "empirical rule" is typically "close enough"

given that the whole probability analysis by definition has some uncertainty in it

Step-by-step explanation:

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what grade u in ?! i will help u

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Marcus has 1,525 marbles and divides the marbles into 15 bags to sell each bag. How many marbles are in each bag and how many ma
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In 6th grade, the ratio of boys to girls is 4 to 5. If there are 145 girls,how many boys are there
swat32

Answer: 181.25

Step-by-step explanation:

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4 0
4 years ago
Need help with problem, been stuck on it for a while
Alinara [238K]
Well, if you look at the picture, the "radius" is 1/4, so r = 1/4.

\bf \textit{diameter of a circle}\\\\
d= 2r\qquad \boxed{r=\frac{1}{4}}\implies d=2\cdot \cfrac{1}{4}\implies d=\cfrac{1}{2}\\\\
-------------------------------\\\\
\textit{circumference of a circle}\\\\
C=2\pi r\qquad \boxed{r=\frac{1}{4}~,~\pi =\cfrac{22}{7}}\implies C=2\left( \frac{22}{7} \right)\left( \frac{1}{4} \right)
\\\\\\
C=\cfrac{2\cdot 22\cdot 1}{7\cdot 4}\implies C=\cfrac{11}{7}\\\\
-------------------------------\\\\

\bf \textit{area of a circle}\\\\
A=\pi r^2\qquad \boxed{r=\frac{1}{4}~,~\pi =\cfrac{22}{7}}\implies A=\left( \frac{22}{7} \right)\left( \frac{1}{4} \right)^2
\\\\\\
A=\cfrac{22}{7}\cdot \cfrac{1^2}{4^2}\implies A=\cfrac{22}{7}\cdot \cfrac{1}{16}\implies A=\cfrac{11}{56}
8 0
4 years ago
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