Answer:
Step-by-step explanation:
13 - [4 *(8 - 5) + 2 - 9] * 2 =
13 - [ 4 (3) + 2 - 9] * 2 =
13 - [ 12 + 2 - 9] * 2 =
13 - 5 * 2 =
13 - 10 =
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
Answer:
95% confidence interval for p, the population proportion of students whose parents bought a car for them when they left for college
(0.4958 , 0.7041)
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the sample size 'n' = 85
The sample proportion
95% confidence interval for p, the population proportion of students whose parents bought a car for them when they left for college
<u><em>Step(ii):-</em></u>
Given that the level of significance
α = 0.05
Z₀.₀₅ = 1.96
(0.6 - 0.104138 , 0.6 +0.104138)
(0.4958 , 0.7041)
<u><em>Final answer:-</em></u>
95% confidence interval for p, the population proportion of students whose parents bought a car for them when they left for college
(0.4958 , 0.7041)
Step-by-step explanation:
Area =144m
2
Length =16m
Breadth=
16
144
=9m
Perimeter =2(l+b)
=2(16+9)
=2(25)=50m
Cost =6×50=300Rs.
