I say it's either A. or C. My main guess goes to C.
A population of size P increasing at the rate of 2% may be modelled as follows
P = P0<span> e</span>0.02 t<span> , where t is the number of years after t = 0 and P</span>0<span> is the population at t = 0. </span>
Let t = 0 corresponds to January 1, 2000 and therefore t = 4 corresponds to December 31. But P is 2,000,000 when t = 4. Hence
2,000,000 = P0<span> e</span>0.02*4
<span>Solve the above for P0</span>
P0<span> = 2,000,000 / e</span>0.02*4<span> = 1 846 000 (rounded to the nearest thousand) </span>
<span>P0 is the population at t = 0 or on January 1, 2000.
A. is the correct answer</span>
Answer:
1/6 is the probability for each event, so P of all three = 1/(6^3)=1/216
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
Given that t and w vary inversely then the equation relating them is
t = 
← k is the constant of variation
To find k use the condition t = 
when w = 4
k = tw = × 4 = 
t = 
← equation of variation
When w = 9, then
t = 
= 
4 plus negative 3 is 1.
When adding negative numbers to positive ones, you should treat it as a normal subtraction, so 4-3.
