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umka2103 [35]
2 years ago
14

The number of potholes in any given 1 mile stretch of freeway pavement in Pennsylvania has a bell-shaped distribution. This dist

ribution has a mean of 47 and a standard deviation of 7. Using the empirical rule, what is the approximate percentage of 1-mile long roadways with potholes numbering between 40 and 68?
Mathematics
1 answer:
alexira [117]2 years ago
5 0
<h3>Answer:   83.85%</h3>

This value is approximate.

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

Explanation:

Let's compute the z score for x = 40

z = (x-mu)/sigma

z = (40-47)/7

z = -1

We're exactly one standard deviation below the mean.

Repeat these steps for x = 68

z = (x-mu)/sigma

z = (68-47)/7

z = 3

This score is exactly three standard deviations above the mean.

Now refer to the Empirical Rule chart below. We'll add up the percentages that are between z = -1 and z = 3. This consists of the two pink regions (each 34%), the right blue region (13.5%) and the right green region (2.35%). These percentages are approximate.

34+34+13.5+2.35 = 83.85

<u>Roughly 83.85%</u> of the one-mile roadways have between 40 and 68 potholes.

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3 years ago
How do you simplify this?<br>x²y+xy² / y²+2/5 × xy​​​
polet [3.4K]

\huge \boxed{\mathbb{QUESTION} \downarrow}

  • How do you simplify this?
  • x²y+xy² / y²+2/5 × xy

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\sf\frac{  { x  }^{ 2  }  y+x { y  }^{ 2  }    }{  { y  }^{ 2  }  + \frac{ 2  }{ 5  }   \times  xy  } \\

Factor the expressions that are not already factored.

_____

<u>How </u><u>to</u><u> factorise</u><u> </u><u>:</u><u>-</u>

<u>NUMERATOR</u> \downarrow

\sf \: x ^ { 2 } y + x y ^ { 2 }

Factor out xy.

\sf \: xy\left(x+y\right)

<u>DENOMINATOR</u> \downarrow

\sf{ y  }^{ 2  }  + \frac{ 2  }{ 5  }   \times  xy \\

Factor out 1/5.

\sf \: {\frac{1}{5}y\left(2x+5y\right)}  \\

_____

Continuing...

\sf\frac{xy\left(x+y\right)}{\frac{1}{5}y\left(2x+5y\right)}  \\

Cancel out y in both the numerator and denominator.

\sf\frac{x\left(x+y\right)}{\frac{1}{5}\left(2x+5y\right)}  \\

Expand the expression.

\sf\frac{x^{2}+xy}{\frac{2}{5}x+y}  \\

This can further simplified to as \downarrow

=    \boxed{\boxed{\bf\frac{5x\left(x  +y\right)}{2x+5y}}}

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sorry idk if that made sense or not

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