First we use product rule
y=x^2lnx
dy/dx = x^2 d/dx (lnx) + lnx d/dx (x^2)
dy/dx = x^2 (1/x) + lnx (2x)
dy/dx = x + 2xlnx
now taking second derivative:
d2y/dx2 = 1 + 2[x (1/x) + lnx (1)]
d2y/dx2 = 1 + 2[1+lnx]
1+2+2lnx
3+2lnx is the answer
The required probability is 
<u>Solution:</u>
Given, a shipment of 11 printers contains 2 that are defective.
We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.
Now, we know that, 
Probability for first draw to be non-defective 
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective 
(printers after first slot = 10; total defective printers = 2)
Then, total probability 
The answer is 0.6666666666 but to simplify it you can put 0.6
Answer:
he should choose the $800 for 4 days job.
Step-by-step explanation:
Answer:
using the given points substitute it into the equation y=mx+c
therefore
y=4
m=-2/3
X=6
c=?
so we should c=intercept
Step-by-step explanation:
y=mx+c
4=-2/3(6)+c
4=-4+c
c=4+4=8
therefore equation of the graph is y=-2/3X+8