Answer: 0.21
Step-by-step explanation:
Let A denotes the event that employees are college graduates.
B denotes that the event that employees have more than ten years of experience.
As per given , we have
P(A)=0.51
P(B)=0.47
P(A∪B)=0.77
We know that , 
Hence, the probability that a randomly selected employee will have more than ten years of experience and be a college graduate = 0.21
Step-by-step explanation:
x approches negative infinity
so : |x|= -x
replace it in the expression
so :
= limx⇒-∞( -x/x) you take the terms with the highest degree
so limx⇒-∞ (-x/x) = -1
Idk the answer man u have to do it by urself
Answer:
a) x = 1225.68
b) x = 1081.76
c) 1109.28 < x < 1198.72
Step-by-step explanation:
Given:
- Th random variable X for steer weight follows a normal distribution:
X~ N( 1154 , 86 )
Find:
a) the highest 10% of the weights?
b) the lowest 20% of the weights?
c) the middle 40% of the weights?
Solution:
a)
We will compute the corresponding Z-value for highest cut off 10%:
Z @ 0.10 = 1.28
Z = (x-u) / sd
Where,
u: Mean of the distribution.
s.d: Standard deviation of the distribution.
1.28 = (x - 1154) / 86
x = 1.28*86 + 1154
x = 1225.68
b)
We will compute the corresponding Z-value for lowest cut off 20%:
-Z @ 0.20 = -0.84
Z = (x-u) / sd
-0.84 = (x - 1154) / 86
x = -0.84*86 + 1154
x = 1081.76
c)
We will compute the corresponding Z-value for middle cut off 40%:
Z @ 0.3 = -0.52
Z @ 0.7 = 0.52
[email protected] < x < [email protected]
-.52*86 + 1154 < x < 0.52*86 + 1154
1109.28 < x < 1198.72
