Answer:
<em>EX=2.68 ≈ 3</em>
<em>Lakeside Olds should expect 3 automobiles lined up at opening</em>
Step-by-step explanation:
<u>Expected Value of a Discrete Probability Distribution</u>
Given a discrete probability distribution of values
x={x1,x2,x3...,xn}
And probabilities
p={p1,p2,p3,...,pn}
Provided the sum of all probabilities is 1, then the expected value of the distribution is
EX =
The data refers to the number of automobiles lined up at Lakeside Olds at opening time for service:
x={1,2,3,4}
And probabilities
p={ 0.40 , 0.03 , 0.06 , 0.51 }
Checking the sum: 0.40 + 0.03 + 0.06 + 0.51 = 1
Now compute the expected value
EX= 1*0.40+2*0.03+3*0.06+4*0.51
EX=2.68 ≈ 3
Lakeside Olds should expect 3 automobiles lined up at opening
Answer:
the third one
Step-by-step explanation:
y represents the number plus 5 and the x represents the other number
Answer: x=8
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
3x−x−1=15
3x+−x+−1=15
(3x+−x)+(−1)=15(Combine Like Terms)
2x+−1=15
2x−1=15
Step 2: Add 1 to both sides.
2x−1+1=15+1
2x=16
Step 3: Divide both sides by 2.
2x
/2
=
16
/2
x=8