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stepladder [879]
2 years ago
6

Please help! im doing a test right now

Mathematics
1 answer:
GaryK [48]2 years ago
4 0

Answer:

a.) Time = 6 hours

b.) Average Speed = 410 miles per hours

c.) Distance = 196 miles

Step-by-step explanation:

<h3><u>a.</u>)</h3>

A coach drives 360 miles at a speed of 60 mph.

So,

Time = Distance ÷ Speed

Time = 360 ÷ 60

Time = 6 hours

<h3><u>b.</u>)</h3>

A plan flies 1640 miles in 4 hours.

So,

Average Speed = Distance ÷ Time

Average Speed = 1640 ÷ 4

Average Speed = 410 miles per hours

<h3><u>c.</u>)</h3>

A bus drives 3½ hour at an average Speed of 56 mph.

So,

3½ hour = 3.5 hour

Distance = Average Speed × Time

Distance = 56 × 3.5

Distance = 196 miles

<u>-TheUnknownScientist 72</u>

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Answer:

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

Further Explanation:

Let's label the three scenarios like so

  • scenario A: selecting a black card
  • scenario B: selecting a red card that is less than 5
  • scenario C: selecting anything that doesn't fit with the previous scenarios

The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.

-----------------

Now onto scenario B.

The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.

So far the fractions we found for scenarios A and B were: 1/2 and 2/13

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  • 2/13 = 4/26

Then add them up

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Next, subtract the value from 1

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The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.

-----------------------------------

Here's a table to organize everything so far

\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}

What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below

\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}

Then we add up the results of that fourth column to compute the expected value.

(1/2) + (18/13) + (-9/26)

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