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Westkost [7]
3 years ago
11

The revenue, in dollars, of a company that makes toy cars can be modeled by the polynomial 3x2 + 4x – 60. The cost, in dollars,

of producing the toy cars can be modeled by 3x2 – x + 200. The number of toy cars sold is represented by x.
If the profit is the difference between the revenue and the cost, what expression represents the profit?

3x – 260
3x + 140
5x – 260
5x + 140
Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

5x – 260

i hope this helps you

Tems11 [23]3 years ago
4 0

For this case we have the following quadratic functions:

Revenue: 3x ^ 2 + 4x - 60

Cost: 3x ^ 2 - x + 200

Then, we observe that the profit is given by the following mathematical relationship:

Profit = Revenue - Cost

Substituting values we have:

Profit = (3x ^ 2 + 4x - 60) - (3x ^ 2 - x + 200)

Making the corresponding calculations we have:

Profit = x ^ 2 (3-3) + x (4 + 1) + (-60-200)\\Profit = 5x - 260

Answer:

An expression that represents the profit is:

Profit = 5x - 260

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Orly uses 2 cups of raisins for every 12 cups of trail mix she makes. How many cups of trail mix will she make if she uses 8 cup
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3 years ago
A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume
mixer [17]

Answer:

(a) 0.4096

(b) 0.64

(c) 0.7942

Step-by-step explanation:

The probability that the player wins is,

P(W)=0.80

Then the probability that the player losses is,

P(L)=1-P(W)=1-0.80=0.20

The player is playing the video game with 4 different opponents.

It is provided that when the player is defeated by an opponent the game ends.

All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)

(a)

The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.

The probability that the player defeats all four opponents in a game is,

P (Player defeats all 4 opponents) = P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096

Thus, the probability that the player defeats all four opponents in a game is 0.4096.

(b)

The probability that the player defeats at least two opponents in a game is,

P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) = 1-P(L)-P(WL)

                                    =1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64

Thus, the probability that the player defeats at least two opponents in a game is 0.64.

(c)

Let <em>X</em> = number of times the player defeats all 4 opponents.

The probability that the player defeats all four opponents in a game is,

P(WWWW) = 0.4096.

Then the random variable X\sim Bin(n=3, p=0.4096)

The probability distribution of binomial is:

P(X=x)={n\choose x}p^{x} (1-p)^{n-x}

The probability that the player defeats all the 4 opponents at least once is,

P (<em>X</em> ≥ 1) = 1 - P (<em>X</em> < 1)

             = 1 - P (<em>X</em> = 0)

             =1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942

Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.

3 0
3 years ago
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