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WARRIOR [948]
3 years ago
10

This expression represents the average cost per game, in dollars, at a bowling alley, where n represents the number of games: .

What is the average cost per game if James bowls 4 games?
Mathematics
1 answer:
koban [17]3 years ago
4 0

Answer:

$4.75

Step-by-step explanation:

The expression given for then average cost is :

(3n + 7 / n)

If the number of games, n = 4

The average cost becomes :

(3n + 7) / n

Put n = 4 in the equation

(3(4) + 7) / 4

(12 + 7) / 4

19 / 4

= 4 3/4

= $4.75

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Answer:

Probability that the first vehicle selected is a motorcycle and the second vehicle is a van is (24/187) or 0.1283.

Step-by-step explanation:

We are given that an automobile manufacturing plant produced 34 vehicles today: 16 were motorcycles, 9 were trucks, and 9 were vans.

Plant managers are going to select two of these vehicles for a thorough inspection. The first vehicle will be selected at random, and then the second vehicle will be selected at random from the remaining vehicles.

As we know that, <u>Probability of any event</u>  =  \frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}

<u>Now, Probability that the first vehicle selected is a motorcycle is given by;</u>

                   =  \frac{\text{Number of motorcycles}}{\text{Total number of vehicles}}

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Total number of vehicles = 16 + 9 + 9 = 34

So, <em>Probability that the first vehicle selected is a motorcycle</em> =  \frac{16}{34}

<u>Similarly, Probability that the second vehicle is a van is given by;</u>

              =   \frac{\text{Number of vans}}{\text{Total number of remaining vehicles}}

Here, Number of vans = 9

And Total number of remaining vehicles after selecting one motorcycle = 34 - 1 = 33

So,<em> Probability that the second vehicle selected is a van</em> =  \frac{9}{33}

Therefore, the probability that the first vehicle selected is a motorcycle and the second vehicle is a van  =  \frac{16}{34}\times \frac{9}{33}

                                               =  \frac{24}{187}  =  <u>0.1283</u>

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