The answer to 10+10 is 20. I hope I made it for the dog.
V = lwh
2x³ + 17x² + 46x + 40 = l(x + 4)(x + 2)
2x³ + 12x² + 16x + 5x² + 30x + 40 = l(x + 4)(x + 2)
2x(x²) + 2x(6x) + 2x(8) + 5(x²) + 5(6x) + 5(8) = l(x + 4)(x + 2)
2x(x² + 6x + 8) + 5(x² + 6x + 8) = l(x + 4)(x + 2)
(2x + 5)(x² + 6x + 8) = l(x + 2)(x + 4)
(2x + 5)(x² + 2x + 4x + 8) = l(x + 4)(x + 2)
(2x + 5)(x(x) + x(2) + 4(x) + 4(2)) = l(x + 4)(x + 2)
(2x + 5)(x(x + 2) + 4(x + 2)) = l(x + 4)(x + 2)
(2x + 5)(x + 4)(x + 2) = l(x + 4)(x + 2)
(x + 4)(x + 2) (x + 4)(x + 2)
2x + 5 = l
Since we already have X, I'm guessing all the question asks you to do is plug it in and solve it to get a Y= equation
If I'm right, then the process is
Y=2(3)+1 (putting 3 in place of x)
Y=6+1
Y=7
Which makes 7 your answer
Only answer that can give u zero is zero
Answer:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.
This means that 
80% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).
