a. Let be a random variable representing the weight of a ball bearing selected at random. We're told that , so
where . This probability is approximately
b. Let be a random variable representing the weight of the -th ball that is selected, and let be the mean of these 4 weights,
The sum of normally distributed random variables is a random variable that also follows a normal distribution,
so that
Then
c. Same as (b).
Answer and Step-by-step explanation:
The parent function is just f(x) = x.
See, the transformations equation of a parent function is as follows:
f(x) = a(x - h) + k
Where:
<u>a</u> is the scaling factor,
<u>h</u> is the horizontal (left and right) changing factor, and
<u>k</u> is the vertical (up and down) changing factor.
So, now we apply the transformations to this parent function by writing it in function g(x).
Vertically stretch by factor of 2: g(x) = 2x
Shifted up 5 units: g(x) = 2x + 5
Shifted left 4 units: g(x) = 2(x - 4) + 5
<u>g(x) = 2(x - 4) + 5 is the function.</u>
This function accomplishes these transformation by having 2 be the scale factor, 4 be the horizontal shift, and 5 be the the vertical shift.
<em><u>#teamtrees #PAW (Plant And Water)</u></em>
Joe will have eaten 24/30
Jane will have eaten 5/30
24 / 5 = 4.8
Joe has eaten 4.8 times as much pizza as Jane.
so ar^ = 2000 and you told me r = 10
a(10^5) = 2000
a = 2000/100000 = 1/50