(1/(√(2π)* 2.67))*e^((-(10.59-4.11)^2)/(2*7.15))

asked by guest
on Nov 17, 2024 at 8:16 am



You asked:

Evaluate the expression: 12π⋅2.67e−(10.59−4.11)22⋅7.15\frac{1}{\sqrt{2 \pi} \cdot 2.67} {e}^{\frac{-{\left( 10.59 - 4.11 \right)}^{2}}{2 \cdot 7.15}}

MathBot Answer:

12π⋅2.67e−(10.59−4.11)22⋅7.15=502267πe5248817875≈0.00792757085269139177627395012868\frac{1}{\sqrt{2 \pi} \cdot 2.67} {e}^{\frac{-{\left( 10.59 - 4.11 \right)}^{2}}{2 \cdot 7.15}} = \frac{50 \sqrt{2}}{267 \sqrt{\pi} e^{\frac{52488}{17875}}} \approx 0.00792757085269139177627395012868

12π⋅2.67e−(10.59−4.11)22⋅7.15=12⋅π⋅(2+67100)⋅e−⋅(10+59100−(4+11100))2⋅12⋅(7+15100)=12⋅π⋅(2+67100)⋅e−⋅(10+59100−(4+11100))2⋅12⋅(7+15100)=12⋅π⋅267100⋅e−⋅(10+59100−(4+11100))2⋅12⋅(7+15100)=1267100⋅2⋅π⋅e−⋅(10+59100−(4+11100))2⋅12⋅(7+15100)=50267⋅2⋅1π⋅e−⋅(10+59100−(4+11100))2⋅12⋅(7+15100)=50267⋅2⋅1π⋅e−⋅(10+59100−411100)2⋅12⋅(7+15100)=50267⋅2⋅1π⋅e−⋅(1059100−411100)2⋅12⋅(7+15100)=50267⋅2⋅1π⋅e−⋅(16225)2⋅12⋅(7+15100)=50267⋅2⋅1π⋅e−⋅26244625⋅12⋅(7+15100)=50267⋅2⋅1π⋅e−⋅26244625⋅12⋅(7+320)=50267⋅2⋅1π⋅e−⋅26244625⋅12⋅14320=50267⋅2⋅1π⋅e−⋅26244625⋅114310=50267⋅2⋅1π⋅e−5248817875=50267⋅2⋅1π⋅1e5248817875=50267⋅2⋅1π⋅e−5248817875\begin{aligned}\frac{1}{\sqrt{2 \pi} \cdot 2.67} {e}^{\frac{-{\left( 10.59 - 4.11 \right)}^{2}}{2 \cdot 7.15}}&=\frac{1}{\sqrt{2 \cdot \pi} \cdot \left(2 + \frac{67}{100}\right)} \cdot e^{- \cdot \left(10 + \frac{59}{100} - \left(4 + \frac{11}{100}\right)\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{1}{\sqrt{2} \cdot \sqrt{\pi} \cdot \left(2 + \frac{67}{100}\right)} \cdot e^{- \cdot \left(10 + \frac{59}{100} - \left(4 + \frac{11}{100}\right)\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{1}{\sqrt{2} \cdot \sqrt{\pi} \cdot \frac{267}{100}} \cdot e^{- \cdot \left(10 + \frac{59}{100} - \left(4 + \frac{11}{100}\right)\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{1}{\frac{267}{100} \cdot \sqrt{2} \cdot \sqrt{\pi}} \cdot e^{- \cdot \left(10 + \frac{59}{100} - \left(4 + \frac{11}{100}\right)\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \left(10 + \frac{59}{100} - \left(4 + \frac{11}{100}\right)\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \left(10 + \frac{59}{100} - \frac{411}{100}\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \left(\frac{1059}{100} - \frac{411}{100}\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \left(\frac{162}{25}\right)^{2} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \frac{26244}{625} \cdot \frac{1}{2 \cdot \left(7 + \frac{15}{100}\right)}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \frac{26244}{625} \cdot \frac{1}{2 \cdot \left(7 + \frac{3}{20}\right)}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \frac{26244}{625} \cdot \frac{1}{2 \cdot \frac{143}{20}}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \cdot \frac{26244}{625} \cdot \frac{1}{\frac{143}{10}}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \frac{52488}{17875}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot \frac{1}{e^{\frac{52488}{17875}}}\\&=\frac{50}{267} \cdot \sqrt{2} \cdot \frac{1}{\sqrt{\pi}} \cdot e^{- \frac{52488}{17875}}\end{aligned}

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

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