The answer is 9.5 ( rounded, answer is 9.486833)
Let f(x)= y = x3 - 6x2 + 11x - 6
X intercept means a point on the graph which intercepts the x axis
And Y-intercept mean a point on the graph which intercepts the y axis
Meaning at the x intercept the component of y is zero because there is no y on the x axis
And at the y intercept the x component of the coordinate is zero because on the y axis there is no x value
So in finding the y intercept of the graph wherever u see x place zero there (ie x= 0 on the y axis)
So with x = 0
y = (0)3- 6(0)2 + 11(0) - 6
y = -6
Therefore the y intercept is (0,-6) in coordinate form or -6
X-intercept
y = x3 - 6x2 + 11x - 6
After factorization and simplification
y = (x-1)(x-2)(x-3)
On the x axis y= 0
0= (x-1)(x-2)(x-3)
Either x-1=0
x=1
(1,0)
Or x-2=0
x=2
(2,0)
Or x-3=0
x=3 (3,0)
Therefore the x intercepts are 1,2 and 3
And they are expressed in coordinate form as (1,0) ,(2,0) and
(3,0)
Answer:
the probability that a randomly selected U.S. adult weighs less than the overweight(but not obese) is 0.394
Step-by-step explanation:
Given the data in the question;
Underweight Healthy Weight Overweight (not Obese) Obese
Probability 0.017 0.377 0.343 0.263
so
P( underweight) = 0.017
P( Healthy Weight) = 0.377
P( Overweight (not Obese) ) = 0.343
P( Obese ) = 0.263
now, the probability that a randomly selected U.S. adult weighs less than the overweight(but not obese) range will be;
P( weigh less than overweight(but not obese) = P( underweight) + P( Healthy Weight)
P( weigh less than overweight(but not obese) = 0.017 + 0.377
P( weigh less than overweight(but not obese) = 0.394
Therefore, the probability that a randomly selected U.S. adult weighs less than the overweight(but not obese) is 0.394
Given:
x, y and z are integers.
To prove:
If 
is even, then at least one of x, y or z is even.
Solution:
We know that,
Product of two odd integers is always odd. ...(i)
Difference of two odd integers is always even. ...(ii)
Sum of an even integer and an odd integer is odd. ...(iii)
Let as assume x, y and z all are odd, then is even.
is always odd. [Using (i)]
is always odd. [Using (i)]
is always even. [Using (ii)]
is always odd. [Using (iii)]
is always odd.
So, out assumption is incorrect.
Thus, at least one of x, y or z is even.
Hence proved.
Not completely sure but i think y is 3
