3x+2=x+8
Slandered Form
2x-6=0
Factorizations
2(x-3)=0
Solutions
x=6/2=3 (6 divided by 2 = 3)
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
P + s = 200......p = 200 - s
20p + 15s = 3400
20(200 - s) + 15s = 3400
4000 - 20s + 15s = 3400
-20s + 15s = 3400 - 4000
-5s = - 600
s = -600/-5
s = 120 <=== there were 120 standard tickets sold
p + s = 200
p + 120 = 200
p = 200 - 120
p = 80 <=== there were 80 premium tickets sold
Answer:
HJ
Step-by-step explanation:
we know that
If two lines are parallel, then their slopes are the same
so
The slope of the line that is parallel to a line that has a slope of 3 is equal to 3
Verify the slope of the blue and red line , because their slopes are positive
<em>Blue line</em>
we have
C(-3,0),D(3,2)
The slope m is equal to
m=(2-0)/(3+3)
m=2/6
m=1/3
<em>Red line</em>
we have
H(-1,-4),J(1,2)
The slope m is equal to
m=(2+4)/(1+1)
m=6/2
m=3
therefore
The answer is the red line HJ
Both of these conditions must be true in order for the assumption that the binomial distribution is approximately normal. In other words, if 
and 
then we can use a normal distribution to get a good estimate of the binomial distribution. If either np or nq is smaller than 5, then a normal distribution wouldn't be a good model to use.
side note: q = 1-p is the complement of probability p
