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ryzh [129]
2 years ago
7

The function f gives the number of copies a book has sold w weeks after it was published. The equation f(w) = 500 * 2^w defines

this function.
Select all domains for which the average rate of change could be a good measure for the number of books sold.

A. 0 ≤ w ≤ 2
B. 0 ≤ w ≤ 7
C. 5 ≤ w ≤ 7
D. 5 ≤ w ≤ 10
E. 0 ≤ w ≤ 10
Mathematics
1 answer:
slavikrds [6]2 years ago
3 0

Good models are expected to reflect the number of books sold between zero and w numbers of weeks where w is defined.

<h3>What is a model?</h3>

A mathematical model is an equation that serves as a representation of a  real life situation. It could be used for investigative or predictive purposes as required.

The following are all domains for which the average rate of change could be a good measure for the number of books sold:

  • 0 ≤ w ≤ 2
  • 0 ≤ w ≤ 7
  • 0 ≤ w ≤ 10

Hence, good models are expected to reflect the number of books sold between zero and w numbers of weeks where w is defined.

Learn more about mathematical models: brainly.com/question/731147

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The following equation is being multiplied by the LCD. Complete the multiplication to eliminate the denominators
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Answer:

(x+2)(x-2) -1(3x)=(x-3)(x-2)

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When two six-sided dice are rolled, there are 36 possible outcomes. Find the probability that the sum is not 5.
zaharov [31]

Answer:

p(Not\ 5) = \frac{8}{9}

Step-by-step explanation:

Given:

Txo six-sided dice are rolled.

Total number of outcomes n(S) = 36

We need to find the probability that the sum is not equal to 5 p(Not 5).

Solution:

Using probability formula.

P(E)=\frac{n(E)}{n(S)}  ----------------(1)

Where:

n(E) is the number of outcomes favourable to E.

n(S) is the total number of equally likely outcomes.

The sum of two six-sided dice roll outcome is equal to 5 as.

Outcome as 5: {(1,4), (2,3), (3,2), (4,1)}

So, the total favourable events n(E) = 4

Now, we substitute n(E) and n(s) in equation 1.

P(5)=\frac{4}{36}

p(5) = \frac{1}{9}

Using formula.

p(Not\ E) + p(E) = 1

p(Not\ 5) + p(5) = 1

Now we substitute p(5) in above equation.

p(Not\ 5) + \frac{1}{9} = 1

p(Not\ 5) = 1-\frac{1}{9}

p(Not\ 5) = \frac{9-1}{9}

p(Not\ 5) = \frac{8}{9}

Therefore, the sum of two six-sided dice roll outcome is not equal to 5.

p(Not\ 5) = \frac{8}{9}

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