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Sav [38]
3 years ago
12

Find the sticker price and dealer's cost for the Tarop Series car. The base price given is $24,990.00. The options for the Tarop

are leather interior for $868.00, custom stereo for $452.00, and computer package for $1,260.00. Destination charges total $630.00. The dealer pays 76% of the base price and 88% of the options. what is the sticker price?
Mathematics
2 answers:
Viefleur [7K]3 years ago
5 0
24,990.00 × 76% = 18992.40
868.00 + 452.00 + 1,260.00 = 2580
2580 × 88% = 2270.40
18992.40 + 2270.40 + 630.00 = 21892.80
podryga [215]3 years ago
4 0

Alright, lets get started.

The base price is given : $ 24990

The dealer pays 76% of the base price means

he pays for base price : 24990 * \frac{76}{100}

he pays for base price : $ 18992.4

For options, total : 868 + 452 + 1260 = 2580

He pays 88% for options, means

he pays for options : 2580 * \frac{88}{100}

he pays for options : $ 2270.4

So, total dealer pays : 18992.4 + 2270.4 + destination charges

So, total dealer pays : 18992.4 + 2270.4 + 630 = $ 21892.8

Hence dealer's cost : $ 21892.8

Price of sticker will be : 24990 + 868 + 452 + 1260 + 630

Price of sticker will be : $ 28200

dealer's cost : $ 21892.8

Hope it will help :)

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

Further Explanation:

Let's label the three scenarios like so

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  • 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.

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

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.

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Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.

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