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djyliett [7]
3 years ago
12

At 1 P. M. the family resumes their trip, and they reach Helena at 5 P. M. The total distance from the family’s home to Helena i

s 485 miles. Including the 30-minute lunch stop, what is the average speed over the entire trip?
Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer: 60.6 miles per hour

Step-by-step explanation: 5 hours + 3 hours + 30 mins = 8 hours 30 mins 485 divided by 8

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Suppose the standard deviation of X is 6 and the standard deviation of Y is 8. Answer the following two questions, rounding to t
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Recall that

Var[<em>aX</em> + <em>bY</em>] = <em>a</em> ² Var[<em>X</em>] + 2<em>ab</em> Cov[<em>X</em>, <em>Y</em>] + <em>b</em> ² Var[<em>Y</em>]

Then

Var[3<em>X</em> - 7<em>Y</em>] = 9 Var[<em>X</em>] - 42 Cov[<em>X</em>, <em>Y</em>] + 49 Var[<em>Y</em>]

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The general result is easy to prove: by definition,

Var[<em>X</em>] = E[(<em>X</em> - E[<em>X</em>])²] = E[<em>X </em>²] - E[<em>X</em>]²

Cov[<em>X</em>, <em>Y</em>] = E[(<em>X</em> - E[<em>X</em>]) (<em>Y</em> - E[<em>Y</em>])] = E[<em>XY</em>] - E[<em>X</em>] E[<em>Y</em>]

Then

Var[<em>aX</em> + <em>bY</em>] = E[((<em>aX</em> + <em>bY</em>) - E[<em>aX</em> + <em>bY</em>])²]

… = E[(<em>aX</em> + <em>bY</em>)²] - E[<em>aX</em> + <em>bY</em>]²

… = E[<em>a</em> ² <em>X</em> ² + 2<em>abXY</em> + <em>b</em> ² <em>Y</em> ²] - (<em>a</em> E[<em>X</em>] + <em>b</em> E[<em>Y</em>])²

… = E[<em>a</em> ² <em>X</em> ² + 2<em>abXY</em> + <em>b</em> ² <em>Y</em> ²] - (<em>a</em> ² E[<em>X</em>]² + 2 <em>ab</em> E[<em>X</em>] E[<em>Y</em>] + <em>b</em> ² E[<em>Y</em>]²)

… = <em>a</em> ² E[<em>X</em> ²] + 2<em>ab</em> E[<em>XY</em>] + <em>b</em> ² E[<em>Y</em> ²] - <em>a</em> ² E[<em>X</em>]² - 2 <em>ab</em> E[<em>X</em>] E[<em>Y</em>] - <em>b</em> ² E[<em>Y</em>]²

… = <em>a</em> ² (E[<em>X</em> ²] - E[<em>X</em>]²) + 2<em>ab</em> (E[<em>XY</em>] - E[<em>X</em>] E[<em>Y</em>]) + <em>b</em> ² (E[<em>Y</em> ²] - E[<em>Y</em>]²)

… = <em>a</em> ² Var[<em>X</em>] + 2<em>ab</em> Cov[<em>X</em>, <em>Y</em>] + <em>b</em> ² Var[<em>Y</em>]

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