We know that
• The new charge was $0.50 to mail a letter weighing up to 1 ounce, and $0.21 for each additional ounce.
Based on the given information, the expression is
Where <em>x</em> represents additional ounces, and <em>y</em> represents a cost.
<h2>Hence, the function is</h2>
<em>u={1,2,3,4,5},A={2,4} and Beta {2,5,5}</em>
<em>now</em><em>,</em><em> </em><em>(AUB)</em><em>=</em><em>{</em><em>1</em><em>,</em><em>3</em><em>,</em><em>3</em><em>,</em><em>4</em><em>,</em><em>5</em><em>}</em>
<em>[</em><em>AUB</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>set</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>elements</em><em> </em><em>of</em><em> </em><em>set</em><em> </em><em>A</em><em> </em><em>and</em><em> </em><em>set</em><em> </em><em>B</em><em> </em><em>without </em><em>any</em><em> </em><em>repetition </em><em>]</em>
<em>n</em><em>(</em><em>AUB</em><em>)</em><em>=</em><em>5</em>
<em>n</em><em>(</em><em>AUB</em><em>)</em><em>is</em><em> </em><em>the</em><em> </em><em>total</em><em> </em><em>no</em><em> </em><em>of</em><em> </em><em>elements</em><em> </em><em>in</em><em> </em><em>set</em><em> </em><em>(</em><em>AUB</em><em>)</em>
Answer:
a) 35
b) 22
Step-by-step explanation:
a) The amount of numbers to the right of the vertical line is the amount of people. Count them up and you get 35.
b) The amount of numbers that are greater than 14.0 is the answer. There are 22 of those numbers.
Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
Let the 2016 revenue = x
Let the 2017 revenue = y
36/25 * 100 = the percent relationship
% relationship = 144%
y = 144% x
Answer 144% <<<<
