The mean is 7.5 you add up all the numbers and then divide by the amount of numbers there is
Answer:
6 votes!
Step-by-step explanation:
Albert: 60% of 30= 18
Anthony: 40% of 30= 12
18 - 12= 6
Answer:
Interes= 162,58
Step-by-step explanation:
Dada la siguiente información:
Se deposita 4000 soles a una tasa de interés del 0,8% quincenas
<u>Primero, debemos calcular el valor final de la inversión:</u>
VF= [VP*( 1 + i)^n]
VF= 4.000*(1,008^5)
VF= 4.162,58
<u>Para calcular el interés ganado en 5 quincenas, debemos usar la siguiente formula:</u>
Interes= 4.162,58 - 4.000
Interes= 162,58
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
(x+2)(x+8)(x+k)=x^3+9x^2+6x-16
(x^2+10x+16)(x+k)=x^3+9x^2+6x-16
x^3+10x^2+16x+kx^2+10kx+16k=x^3+9x^2+6x-16
kx^2+10kx+16k=-x^2-10x-16
k(x^2+10x+16)=-x^2-10x-16
k=(-x^2-10x-16)/(x^2+10x+16)
k=-1
so the width is (x-1)
