Complete the table of values for the function f(x) = (1/3)x for a, b, and c. x : -2, -1, 0, 1, 2. f(x) : a, b, c, 1/3, 1/9
balu736 [363]
Given:
The function is
To find:
The missing values for a, b, and c in the given table.
Solution:
We have,
At x=-2,
![\left[\because \left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^{n}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20%5Cleft%28%5Cdfrac%7Ba%7D%7Bb%7D%5Cright%29%5E%7B-n%7D%3D%5Cleft%28%5Cdfrac%7Bb%7D%7Ba%7D%5Cright%29%5E%7Bn%7D%5Cright%5D)
At x=-1,
At x=0,
![\left[\because a^0=1, a\neq 0\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbecause%20a%5E0%3D1%2C%20a%5Cneq%200%5Cright%5D)
Therefore, the missing values of a,b and c are 9, 3, and 1 respectively.
Answer:
The range of the 95% data (X) = 238.3 days < X < 289.9 days
Step-by-step explanation:
Given;
mean of the normal distribution, m = 264.1 days
standard deviation, d = 12.9 days
between two standard deviation below and above the mean is 96% of all the data.
two standard deviation below the mean = m - 2d
= 264.1 - 2(12.9)
= 238.3 days
two standard deviation above the mean = m + 2d
= 264.1 + 2(12.9)
= 289.9 days
The middle of the 95% of most pregnancies would be found in the following range;
238.3 days < X < 289.9 days
