The future worth of a certain item or money deposited at a certain percentage is obtained by the equation,
F = P x (1 + i)^n
F is the future worth, P is present worth, i is the percentage, and n is the number of years.
n = 2013 - 1938
n = 75
Substituting the known values,
300,000 = 100 x (1 + i)^75
The value of i from the equation is 0.1126.
The answer to this item is 11.26%.
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Answer:
The answers to the questions are;
(a) P(At least 1 defective)
= 0.9883.
(b) P(At least 1 defective)
= 0.6409.
Step-by-step explanation:
There are 110 cards and 20 defectives.
a) The probability of at least one defective is given by
P(At least 1 defective) = 1 - P(0 defective)
P(0 defective) = 20C0 × (90C0)/(110C20) = 0.0116
1 - 0.0116 = 0.9883
b) For a set of 110 boards that has 5 defective and 105 non-defective
P(At least 1 defective) = 1 - P(0 defective)
P(0 defective) = (20C0)(90C5)/(110C5) = 0.35909
1-0.35909
= 0.6409
Answer:
Step-by-step explanation:
Initial Quantity of salt Q(0) = 100 kg
Capacity of tank = 1000 L
Inflow of liquid = 10 l /minute and outflow = same rate thus leaving the content of volume 1000 L at any time
Assume after entering constantly mixes and drains out.
Concentration flow = inflow - outflow
= 0.05(10) - 
i.e.
, Q(0) = 100

Use Q(0) = 5
-45 = A
So the equation would be

a) Initial concentratin = 
b) Use the solution fo rDE substitute t =1
Q(1) = 50-45e^(-0.01) = 5.4477 kg
c) As t becomes large Q = 50
So concentration limit = 50/1000 = 0.05
Answer:
is the number.
Step-by-step explanation:
First, rewrite the problem as an equation. 
Now, subtract 3 from both sides. 
Finally, divide both sides by 3. 
There’s the answer.