Answer: 300
Step-by-step explanation:
Given , An electronics retailer receives a shipment of 6,000CDs to distribute to its stores.
i.e, Total CD's he has now = 6,000
Also, a quality control manager inspects a random sample(S) of 40CD's , the number of defective(D) CD's are 2.
Then , the proportion of defective( D) CD's = 
Now , the number of CDs in the shipment are likely to be defective = ( Total CD's) x (proportion of defective CD's)
= 6000 x 0.05
=300
Hence, the number of CDs in the shipment are likely to be defective =300
Answer:
I think the answer should be B
Answer:
$18 :)
Step-by-step explanation:
10% of 180 is 18
Answer:
x(t) = 2000 - e^(-k*t)
Step-by-step explanation:
Interpretation:
No . infected student = x
Total student = 2000
rate of infected students = dx / dt
not-infected student = 200 - x
The general rate at which student are infected can be expressed as below:
dx / dt = k * ( 2000 - x )
To develop an expression of x(t) we integrate the above expression by separating variables:
dx / (2000 - x ) = k * dt
Now integrate:
@ t = 0 , infected students x = 0
Hence,
C = - ln (2000)
Answer:
D
Step-by-step explanation:
