War between Japan and the United States.
Answer: possible values of Range will be values that are >=91 or <=998
Step-by-step explanation:
Given that :
Set Q contains 20 positive integer values. The smallest value in Set Q is a single digit value and the largest value in Set Q is a three digit value.
Therefore,
given that the smallest value in set Q is a one digit number :
Then lower unit = 1, upper unit = 9( this represents the lowest and highest one digit number)
Also, the largest value in Set Q is a three digit value:
Then lower unit = 100, upper unit = 999 ( this represents the lowest and highest 3 digit numbers).
Therefore, the possible values of the range in SET Q:
The maximum possible range of the values in set Q = (Highest possible three digit value - lowest possible one digit) = (999 - 1) = 998
The least possible range of values in set Q = (lowest possible three digit value - highest possible one digit value) = (100 - 9) = 91
Answer:
$21.17
Step-by-step explanation:
Simple Interest=P(1+r)^t
Compounded Interest=P(e^rt)
SI=668*(1.0925)^5
SI=1039.64
CI=668(e^(0.0925*5))
CI=1060.81
1060.81-1039.64=21.17
So the difference is $21.17, thus doing compounded interest is better.
Answer:
2x10^3
Step-by-step explanation:
Factor the numerator and denominator and cancel the common factors.
Answer:
a)
, b)
, c)
, d) 
Step-by-step explanation:
a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

b) The general solution is found after separating variables and integrating each sides:

Where
is the time constant and 
c) The time constant is:


The particular solution of the differential equation is:

d) The amount of radium after 300 years is:
