The3Mathketeers

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 #1
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We should consider the information from both flights, put that information together, and then solve for the distance of the trip. We will take advantage of the relationship of distance, rate, and time with the formula d = rt.

 

Let d be the distance of the one-way trip from Penthaven to Jackson, in miles

Let r be the rate of speed of the airplane during no wind, in miles per hour

let t be the time of the one-way trip from Penthaven to Jackson, in hours

 

Case 1) No wind

Here, we know the total time of the round trip without wind is 3 hours and 20 minutes. Converting into hours, \(t_1 = 3 \text{ hr} + 20 \text{ min} * \frac{1 \text{ hr}}{60 \text{min}} = 3\frac{1}{3} \text{ hr} = \frac{10}{3} \text{ hr}\). Both the distance d and the rate of speed r of the trip is unknown. However, the given information concerns a round trip, which means \(2d = \frac{10}{3}r \text{ so } d = \frac{5}{3}r\).

 

Case 2) 70 miles per hour of wind

Here, we know the total time of the round trip with wind is 3 hours and 50 minutes. Converting into hours, \(t_2 = \text{ hr} + 50 \text{ min} * \frac{1 \text{ hr}}{60 \text{ min}} = 3\frac{5}{6} \text{ hr} = \frac{23}{6} \text{ hr}\). Both the distance d and the rate of speed r of the trip is unknown again. However, there is some relationship between rates of speeds. For example, while the plan is flying with the aide of the wind, then plane is flying at r + 70 and at r - 70 on the return flight. If d = rt, then t = d/r, so we can sum the individual times.

\(\frac{d}{r + 70} + \frac{d}{r - 70} = \frac{23}{6}\).

 

Now, solve for d.

\(\begin{cases} d = \frac{5}{3}r \\ \frac{d}{r + 70} + \frac{d}{r - 70} = \frac{23}{6} \end{cases} \\ r = \frac{3}{5}d \text{ so } r^2 = \frac{9}{25}d^2\\ d(r - 70)+d(r+70) = \frac{23}{6}\left(r^2 - 4900\right) \\ 6dr - 420d + 6dr + 420d = 23\left(r^2 - 4900\right) \\ 12dr = 23r^2 - 112700 \\ 12d * \frac{3}{5}d = 23*\frac{9}{25}d^2 - 112700 \\ 180d^2 = 207d^2 - 2817500\\ 27d^2 = 2817500 \\ d = \sqrt{\frac{2817500}{27}} \approx 323.035 \text{ mi}\)

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12.10.2024
 #1
avatar+189 
+1

Since a, b, c, d, and e are all variations of a similar question, I think it is fair just to do problems a and d and leave the rest.

 

Expected value is the average outcome of running a long-term experiment, which is calculated with the formula \(E(X) = \sum_{i = 1}^n x_iP(x_i)\) where E(X) is the expected value, xi is a specific outcome, and P(xi) is the probability of xi occurring. Let's apply this to problem a.

 

Upon drawing a slip of paper from the bag, 2 possible outcomes can occur: 3 or 8 with their respective probabilities. With the unmodified bag, 3 occurs with a probability of 8/10 because there are eight 3's and ten slips of paper. Likewise, 8 occurs with a probability of 2/10 because there are two 8's and ten slips of paper. Substitute this information into the formula and the answer follows naturally.

 

\(\begin{align} E(X) &= 3 * \frac{8}{10} + 8 * \frac{2}{10} \\ &= \frac{24}{10} + \frac{16}{10} \\ &= 4 \end{align}\)

 

Therefore, E(X) = 4 for problem a.

 

Next, let's consider problem d. Here, the expected value is given, so E(X) = 6. Adding 8's to the original contents of the bag increases both the number of 8's and the total number of 8's in the bag. Let a represent the number of 8's to add to the bag such that E(X) = 6. This would result in the following equation.

 

\(\begin{align*} E(X) &= \sum_{i = 1}^n x_iP(x_i) \\ 6 &= 3 * \frac{8}{10 + a} + 8 * \frac{2 + a}{10 + a} \\ 60 + 6a &= 24 + 16 + 8a \\ 20 &= 2a \\ a &= 10 \end{align*}\)

 

Therefore, add ten 8's to the bag. To be a proper mathematics steward, note that 10 + a appears in the denominator. This means a ≠ -10 under any circumstance. However, this is a non-issue regardless because the variable a represents the number of 8's added, so a negative value for a is nonsensical anyway.

04.10.2024
 #1
avatar+189 
0

This question appears to be a duplicate. Go to https://web2.0calc.com/questions/domain_60627#r1 to see the explanation for how to determine the domain of this function.

07.09.2023