Answer:
a. The percentage of vehicles who pass through this construction zone who are exceeding the posted speed limit =90.82%
b. Percentage of vehicles travel through this construction zone with speeds between 50 mph and 55 mph= 2.28%
Step-by-step explanation:
We have to find
a) P(X>40)= 1- P(x=40)
Using the z statistic
Here
x= 40 mph
u= 44mph
σ= 3 mph
z=(40-44)/3=-1.33
From the z-table -1.67 = 0.9082
a) P(X>40)=
Probability exceeding the speed limit = 0.9082 = 90.82%
b) P(50<X<55)
Now
z1 = (50-44)/3 = 2
z2 = (55-44)/3= 3.67
Area for z>3.59 is almost equal to 1
From the z- table we get
P(55 < X < 60) = P((50-44)/3 < z < (55-44)/3)
= P(2 < z < 3.67)
= P(z<3.67) - P(z<2)
= 1 - 0.9772
= 0.0228
or 2.28%
Find m∠BOC, if m∠MOP = 110°.
Answer:
m∠BOC= 40 degrees
Step-by-step explanation:
A diagram has been drawn and attached below.
- OM bisects AOB into angles x and x respectively
- ON bisects ∠BOC into angles y and y respectively
- OP bisects ∠COD into angles z and z respectively.
Since ∠AOD is a straight line
x+x+y+y+z+z=180 degrees

We are given that:
m∠MOP = 110°.
From the diagram
∠MOP=x+2y+z
Therefore:
x+2y+z=110°.
Solving simultaneously by subtraction

x+2y+z=110°.
We obtain:
x+z=70°
Since we are required to find ∠BOC
∠BOC=2y
Therefore from x+2y+z=110° (since x+z=70°)
70+2y=110
2y=110-70
2y=40
Therefore:
m∠BOC= 40 degrees
The quantity remaining will be
.. 448*(1/2)^(24/6) = 448/16 = 28 . . . . grams
144 divided by 260 is 60%