Solution :
Given
![$f(x)=x^3-x-1, x_1=1$](https://tex.z-dn.net/?f=%24f%28x%29%3Dx%5E3-x-1%2C%20x_1%3D1%24)
![$f'(x)=3x^2-1$](https://tex.z-dn.net/?f=%24f%27%28x%29%3D3x%5E2-1%24)
Let the initial approximation is ![$x_1 =1$](https://tex.z-dn.net/?f=%24x_1%20%3D1%24)
So by Newton's method, we get
![$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$](https://tex.z-dn.net/?f=%24x_%7Bn%2B1%7D%3Dx_n-%5Cfrac%7Bf%28x_n%29%7D%7Bf%27%28x_n%29%7D%2Cn%3D1%2C2%2C...%24)
![$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$](https://tex.z-dn.net/?f=%24x_2%3Dx_1-%5Cfrac%7Bf%28x_1%29%7D%7Bf%27%28x_1%29%7D%3D1-%5Cfrac%7B1%5E3-1-1%7D%7B3%281%29%5E2-1%7D%3D1.5%24)
![$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$](https://tex.z-dn.net/?f=%24x_3%3D1.5-%5Cfrac%7B%281.5%29%5E3-1.5-1%7D%7B3%281.5%29%5E2-1%7D%3D1.34782608%24)
![$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$](https://tex.z-dn.net/?f=%24x_4%3D1.34782608-%5Cfrac%7B%281.34782608%29%5E3-1.34782608-1%7D%7B3%281.34782608%29%5E2-1%7D%3D1.32520039%24)
![$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$](https://tex.z-dn.net/?f=%24x_5%3D1.32520039-%5Cfrac%7B%281.32520039%29%5E3-1.32520039-1%7D%7B3%281.32520039%29%5E2-1%7D%3D1.32471817%24)
![$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$](https://tex.z-dn.net/?f=%24x_6%3D1.32471817-%5Cfrac%7B%281.32471817%29%5E3-1.32471817-1%7D%7B3%281.32471817%29%5E2-1%7D%3D1.32471795%24)
are identical up to eight decimal places.
The approximate real root is x ≈ 1.32471795
∴ x = 1.32471795
Answer:
1/(x^3 + 6x^2 + 12x + 8)
Step-by-step explanation:
The first thing we do is rationalize this expression. (2+x)^-3 is written as
1/(2+x)^3
Then from there we can foil out the denominator. It is easiest to foil (2+x)(2+x) first and then multiply that product by (2+x).
(2+x)(2+x) = 4 + 4x + x^2
(4+4x+x^2)(2+x) = 8+8x+2x^2+4x+4x^2+x^3.
Then we combine like terms and put them in order to get:
x^3 + 6x^2 + 12x + 8
And of course we can't forget that this was raised to the negative third power, so our answer is 1/(x^3 + 6x^2 + 12x + 8)
Answer:
0.337 = 33.7% probability that one of the eight children has a food allergy.
Step-by-step explanation:
For each children, there are only two possible outcomes. Either they have a food allergy, or they do not. The probability of a child having food allergy is independent of any other child. This means that we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
7% of U.S. children 4 years of age or younger have a food allergy.
This means that ![p = 0.07](https://tex.z-dn.net/?f=p%20%3D%200.07)
A day care program has capacity for 8 children in that age range.
This means that ![n = 8](https://tex.z-dn.net/?f=n%20%3D%208)
What is the probability that one of the eight children has a food allergy?
This is P(X = 1).
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 1) = C_{8,1}.(0.07)^{1}.(0.93)^{7} = 0.337](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B8%2C1%7D.%280.07%29%5E%7B1%7D.%280.93%29%5E%7B7%7D%20%3D%200.337)
0.337 = 33.7% probability that one of the eight children has a food allergy.
Answer:
1.27 centimeters
Step-by-step explanation:
brainliest??