I believe the answer is 3 3/4.
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
Slope of the line is -3/2
Step-by-step explanation:
- Step 1: Given points are (1, -2) and (3, -5)
- Step 2: Equation for slope, m = (y2 - y1)/(x2 - x1)
⇒ m = (-5 - -2)/(3 - 1) = -3/2
C is the correct answer
Hope this helps :D
Answer:
The vector joining the ship to the rock is t= 7 i + 5 j
The direction is 0.9505 radians east of north.
Step-by-step explanation:
The position vector of the ship:
r= 1 i + 0 j
The position vector of the ship:
s= 6 i + 5 j
The vector joining the ship to the rock is:
t = r + s
t = (1 i + 0 j) + (6 i + 5 j)
t = 7 i + 5 j
The bearing of the rock to the ship is:
Θ= 
= 0.9505 radians
