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S_A_V [24]
4 years ago
14

Please help worth 20 points will mark brainliest

Mathematics
1 answer:
ddd [48]4 years ago
3 0

Answer:

P(t)= 100t + 150

Step-by-step explanation:

If the price for 6 hours of studio time is 600 dollars, that means every hour the studio is used it costs 100 dollars. Since the fixed fee is 150 dollars, that is added to the cost of how many hours the studio is used.

I hope this helped you. If it did Brainilest is appreciated.

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A certain bowler can bowl a strike 85 % of the time. What is the probability that she ​a) goes three consecutive frames without
Artist 52 [7]

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.

3 0
4 years ago
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