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sladkih [1.3K]
2 years ago
7

Original price of a cd is 12.50. what is the price with a discount of 39%

Mathematics
1 answer:
vredina [299]2 years ago
6 0

Answer:

original price of a cd is 12.50. what is the price with a discount of 39%

Step-by-step explanation:

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Vehicles entering an intersection from the east are equally likely to turn left, turn right, or proceed straight ahead. If 50 ve
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Answer:

The probability that at least two-third of vehicles in the sample turn is 0.4207.

Step-by-step explanation:

Let <em>X</em> = number of vehicles that turn left or right.

The proportion of the vehicles that turn is, <em>p</em> = 2/3.

The nest <em>n</em> = 50 vehicles entering this intersection from the east, is observed.

Any vehicle taking a turn is independent of others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 50 and <em>p</em> = 2/3.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>X</em> if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

np=50\times \frac{2}{3}=33.333>10\\\\n(1-p)=50\times \frac{1}{3}= = 16.667>10

Thus, a Normal approximation to binomial can be applied.

So, X\sim N(np, np(1-p))

Compute the probability that at least two-third of vehicles in the sample turn as follows:

P(X\geq \frac{2}{3}\times 50)=P(X\geq 33.333)=P(X\geq 34)

                        =P(\frac{X-\mu}{\sigma}>\frac{34-33.333}{\sqrt{50\times \frac{2}{3}\times\frac {1}{3}}})

                        =P(Z>0.20)\\=1-P(Z

Thus, the probability that at least two-third of vehicles in the sample turn is 0.4207.

6 0
3 years ago
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