Step-by-step explanation:
1) B
2) A
3) C
4) C
yeahhh that's about right
Hi there!
Our interval is from 0 to 3, with 6 intervals. Thus:
3 ÷ 6 = 0.5, which is our width for each rectangle.
Since n = 6 and we are doing a right-riemann sum, the points we will be plugging in are:
0.5, 1, 1.5, 2, 2.5, 3
Evaluate:
(0.5 · f(0.5)) + (0.5 · f(1)) + (0.5 · f(1.5)) + (0.5 · f(2)) + (0.5 · f(2.5)) + (0.5 · f(3)) =
Simplify:
0.5( -2.75 + (-3) + (-.75) + 4 + 11.25 + 21) = 14.875
Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
875 × .64= 560
answer: 560 employees eat lunch at the cafeteria on Wednesday.
