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ratelena [41]
3 years ago
7

The formula P = 0.68x2 - 0.048x + 1 models the approximate population P, in thousands, for a species of fish in a local pond, x

years after 1997. During what year will the population reach 33.984 thousand fish?
Mathematics
1 answer:
marshall27 [118]3 years ago
8 0

Answer:

The answer to your question is 2184

Step-by-step explanation:

Data

Equation    P(x) = 0.68x² - 0.048x + 1

x = years

population = 33984

Process

1.- Substitute the population in the equation

                   33984 = 0.84x² - 0.048x + 1

2.- Equal to zero

                   0.84x² - 0.048x + 1 - 33984 = 0

3.- Simplify

                    0.84x² - 0.048x - 33983 = 0

4.- Solve for x

x = \frac{0.048 +- \sqrt{(0.048^{2}) - (4)(0.84)(-33982)}}{2(0.84)}

x = \frac{0.048 +- \sqrt{337.9}}{1.68}

x1 = 201.2

x2 = -201.2

Only x1 is correct because time can not be negative

5.- Calculate the year

Year = 1997 + 201

Year = 2184

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4 years ago
A certain bowler can bowl a strike 85 % of the time. What is the probability that she ​a) goes three consecutive frames without
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Answer:

a) 0.34% probability that she goes three consecutive frames without a​ strike.

b) 1.91% probability that she her first strike in the third ​frame

c) 99.66% probability that she has at least one strike in the first three ​frames.

d) 14.22% probability that she bowls a perfect game.

Step-by-step explanation:

For each frame, there are only two possible outcomes. Either there is a strike, or there is not. The probability of a strike happening in a frame is independent of other frames. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A certain bowler can bowl a strike 85 % of the time.

This means that p = 0.85

a) goes three consecutive frames without a​ strike?

This is P(X = 0) when n = 3. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.85)^{0}.(0.15)^{3} = 0.0034

0.34% probability that she goes three consecutive frames without a​ strike.

​b) makes her first strike in the third ​frame?

No strike during the first two(with a 15% probability)

Strike during the third(85% probability). So

P = 0.15*0.15*0.85 = 0.0191

1.91% probability that she her first strike in the third ​frame

c) has at least one strike in the first three ​frames? ​

Either there are no strikes, or there is at least one strike. The sum of the probabilities of these events is 100%.

From a), 0.34% probability that she goes three consecutive frames without a​ strike.

100 - 0.34 = 99.66

99.66% probability that she has at least one strike in the first three ​frames.

d) bowls a perfect game​ (12 consecutive​ strikes)?

This is P(X = 12) when n = 12. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 12) = C_{12,12}.(0.85)^{12}.(0.15)^{0} = 0.1422

14.22% probability that she bowls a perfect game.

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