Solve the nonlinear system of equations
1 answer:
Answer:
There are no real solutions. The complex solutions are ...
(x, y) = (2.5-i√1.75, -2.5+i√1.75) or (2.5+i1.75, -2.5-i1.75)
Step-by-step explanation:
Subtract the first equation from the second.
y -y = (x^2 -6x +8) -(-x)
0 = x^2 -5x +8
Rewrite to facilitate completing the square:
x^2 -5x = -8
x^2 -5x +6.25 = -1.75 . . . . . add 6.25 to complete the square
(x -2.5)^2 = -1.75 . . . . . write as a square
x -2.5 = ±i√1.75 . . . . . . take the square root
x = 2.5 ± i√1.75 . . . . . . add 2.5; y is the opposite of this
Solutions are ...
- x = 2.5 +i√1.75, y = -2.5 -i√1.75
- x = 2.5 -i√1.75, y = -2.5 +i√1.75
_____
The graph shows the equations have no point of intersection, meaning there are no real solutions. The complex solutions are shown above.
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Answer:
The experimental probability of throwing a five is 0.087.
Step-by-step explanation:
Given:
Number of trials (n) = 100
Number of times 5 appears (x) = 14
Let the event of occurrence of 5 be success and the probability represented by 'p'. So, all the other numbers occurrence is failure and its probability is represented as 'q'.
Probability of success is given as:
Favorable event is occurrence of 5. So, its number is 1 as there is only one 5 in the die. Total outcomes are 6 as there are six numbers. So,
Now, probability of failure is given by the formula:
Now in order to find the experimental probability of 14 successes out of 100 trials, we apply Bernoulli's theorem which is given as:
Plug in all the given values and find the probability of 14 successes. This gives,
Therefore, the experimental probability of throwing a five is 0.087.
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p =
Therefore, the probability is given by:
P
P
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.