The projected population of a certain ethnic group(in millions) can be approximated by p(t)equals=39.07 left parenthesis 1.021
right parenthesis Superscript t39.07(1.021)t where tequals=0 corresponds to 2000 and 0 less than or equals t less than or equals 500≤t≤50. a. Estimate the population of this group for the year 2010. b. What is the instantaneous rate of change of the population when tequals=10?
1 answer:
Answer:
- a) p(10) ≈ 48.10
- b) about 1.00 million/year
Step-by-step explanation:
a) Use t=10 in the given formula and do the arithmetic.
39.07×1.021^10 ≈ 48.0951 ≈ 48.10 . . . millions
__
b) The derivative of p(t) is ...
p'(t) = p(t)·ln(1.021) ≈ 0.8120·1.021^t
p'(10) ≈ 0.99954 ≈ 1.00 . . . . millions per year
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Answer:
P(even or odd)
Step-by-step explanation:
Given : One card is drawn at random from the deck of cards (refer given figure)
Solution:
<u>To find probability of getting even numbers </u>
favorable events of even no. = {2,4,6,8,10} = 5
total events = {1,2,3,4,5,6,7,8,9,10} =10
Thus , probability of getting even numbers =
favorable events of even no. / total events
= 5/10 = 1/2
<u>To find probability of getting odd numbers </u>
favorable events of even no. = {1,3,5,7,9} = 5
total events = {1,2,3,4,5,6,7,8,9,10} =10
Thus , probability of getting odd numbers =
favorable events of odd no. / total events
= 5/10 = 1/2
Option A = P(even or odd) =
using formula : [tex]P(A\cup B)= P(A)+P(B)-P(A\cap B)[/tex]
--(1)
So, we have already calculated the value of
To calculate
= P(even and odd)
But no card can be odd and even both
Thus, =0
putting all these value in (1)
⇒
⇒
Thus , P(even or odd) = 1
Now, consider Option 2 : P(even and odd)
Since no card can be odd and even both so P(even and odd) =0
Now, consider Option 3 : P(not yellow)
favorable event of not yellow = {1,2,3,4,5,7,8,9,10}=9
total events = {1,2,3,4,5,6,7,8,9,10} =10
Thus, P(not yellow) = 9/10
Hence OPTION A has probability equal to 1