What is the value of 6x+7y−56x+7y−5 when x = 2 and y = 3?
1 answer:
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Answer:
The probability that there will be a total of 7 defects on four units is 0.14.
Step-by-step explanation:
A Poisson distribution describes the probability distribution of number of success in a specified time interval.
The probability distribution function for a Poisson distribution is:
Let <em>X</em> = number of defects in a unit produced.
It is provided that there are, on average, 2 defects per unit produced.
Then in 4 units the number of defects is, .
Compute the probability of exactly 7 defects in 4 units as follows:
Thus, the probability of exactly 7 defects in 4 units is 0.14.
Answer:
The zeros of the function are;
x = 0 and x = 1
Step-by-step explanation:
The zeroes of the function simply imply that we find the values of x for which the corresponding value of y is 0.
We let y be 0 in the given equation;
y = x^3 - 2x^2 + x
x^3 - 2x^2 + x = 0
We factor out x since x appears in each term on the Left Hand Side;
x ( x^2 - 2x + 1) = 0
This implies that either;
x = 0 or
x^2 - 2x + 1 = 0
We can factorize the equation on the Left Hand Side by determining two numbers whose product is 1 and whose sum is -2. The two numbers by trial and error are found to be -1 and -1. We then replace the middle term by these two numbers;
x^2 -x -x +1 = 0
x(x-1) -1(x-1) = 0
(x-1)(x-1) = 0
x-1 = 0
x = 1
Therefore, the zeros of the function are;
x = 0 and x = 1
The graph of the function is as shown in the attachment below;
