Question

Please help!

0.98 as a binomial: (1 + -0.2)
Enter the value for each variable.
a = __ b =___ n =___

Answer 1

a= 1 b= -0.02 c=5

Explanation:

took it on edge

Answer 2

For the answer to the question above,
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3

1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3

1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3

1 + 9x + 36x^2 + 84x^3

In my experience, up to the x^3 is often adequate to approximate a route.

(1+x) = 0.98

x = 0.98 - 1 = -0.02

Substituting:

1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3

approximation = 0.834

Checking the real value in your calculator:

(0.98)^9 = 0.834

So you have approximated correctly.

If you want to know how accurate your approximation is, write out the result of each in full:

1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728

(0.98)^9 = 0.8337477621

So it is correct to 4