The correct format of the question is
At the end of 2006, the population of Riverside was 400 people. The population for this small town can be modeled by the equation below, where t represents the number of years since the end of 2006 and P represents the number of people. 
Based on this model, approximately what was the increase in the population of Riverside at the end of 2009 compared to the end of 2006?
(A) 291
(B) 691
(C) 1040
(D) 1440
Answer:
The increase in the population at the end of 2009 is 291 people
Step-by-step explanation:
We are given the equation as
where
P = No of People
t= No of Years
it is given that in the year 2006 the population is 400
this will only happen when we take t= 0
so for
Year value of t
2006- t = 0
2007- t = 1
2008- t = 2
2009 t = 3
No of people in 2009 will be
= 400*1.728
P = 691.2
Since the equation represents no of people so it can't be in decimals, Therefore the population will be 691
Increase = P(2009) - P(2006)
= 691 - 400
= 291
The increase in the population at the end of 2009 is 291 people.
(-3x+4) + (5x + 12) = 2x +16
The answer is D.
Hope it helps!
Answer:
150
Step-by-step explanation:
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In order to answer the above question, you should know the general rule to solve these questions.
The general rule states that there are 2ⁿ subsets of a set with n number of elements and we can use the logarithmic function to get the required number of bits.
That is:
log₂(2ⁿ) = n number of <span>bits
</span>
a). <span>What is the minimum number of bits required to store each binary string of length 50?
</span>
Answer: In this situation, we have n = 50. Therefore, 2⁵⁰ binary strings of length 50 are there and so it would require:
log₂(2⁵⁰) <span>= 50 bits.
b). </span><span>what is the minimum number of bits required to store each number with 9 base of ten digits?
</span>
Answer: In this situation, we have n = 50. Therefore, 10⁹ numbers with 9 base ten digits are there and so it would require:
log2(109)= 29.89
<span> = 30 bits. (rounded to the nearest whole #)
c). </span><span>what is the minimum number of bits required to store each length 10 fixed-density binary string with 4 ones?
</span>
Answer: There is (10,4) length 10 fixed density binary strings with 4 ones and
so it would require:
log₂(10,4)=log₂(210) = 7.7
= 8 bits. (rounded to the nearest whole #)
Answer:
-1.25737973739
Step-by-step explanation:
plz make it brillent ans
