Two lines or equations are described in each part. Decide whether each system has one solution, no solution, or infinitely many
solutions. Show your work AND explain your reasoning for each of your answers.
(a) Line a has a rate of change of 7.
Line b has a rate of change of -7
(b) 2x + 3y = 5.5
4x + 6y = 11
(c) These two graphed lines:
(d) Line c and line d are parallel.
(a) Line a has a rate of change of 7.
Line b has a rate of change of -7
(b) 2x + 3y = 5.5
4x + 6y = 11
(c) These two graphed lines:
(d) Line c and line d are parallel.
1 answer:
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Answer:
(a) 50%
(b) 47.5%
(c) 2.5%
Step-by-step explanation:
According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, and standard deviation,
.
Here the number of tosses is, <em>n</em> = 2500.
Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.
The mean and standard deviation are:
(a)
To not lose any money the even rolls has to be 1250 or more.
Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.
Thus, the probability that by the end of the evening you will not have lost any money is 50%.
(b)
If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.
Then for number of "even rolls" as 1300,
1300 = 1250 + 2 × 25
= μ + 2σ
Then P (μ + 2σ) for a normally distributed data is 0.975.
⇒ 1300 is at the 97.5th percentile.
Then the area between 1250 and 1300 is:
Area = 97.5% - 50%
= 47.5%
Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.
(c)
To win $100 or more the number of even rolls has to at least 1300.
From part (b) we now 1300 is the 97.5th percentile.
Then the probability that you will win $100 or more is:
P (Win $100 or more) = 100% - 97.5%
= 2.5%.
Thus, the probability that you will win $100 or more is 2.5%.