When you work out it helps clear your thoughts and think of nothing. It makes you feel good.
Question:
The operations manager for a well-drilling company must recommend whether to build a new facility, expand his existing one, or do nothing. He estimates that long-run profits (in $000) will vary with the amount of precipitation (rainfall) as follows:
Alternative Precipitation
Low Normal High
Do nothing -100 100 300
Expand 350 500 200
Build new 750 300 0
If he feels the chances of low, normal, and high precipitation are 30 percent, 20 percent, and 50 percent respectively, What is EVPI (Expected value of Perfect Information)?
A. $140,000
B. $170,000
C. $285,000
D. $305,000
E. $475,000
Answer:
D. $170,000
Explanation:
The expected long run profits are for
Low Normal High
Do nothing -100*0.3 100*0.2 300*0.5 = 140
Expand 350*0.3 500*0.2 200*0.5 = 305
Build new 750*0.3 300*0.2 0*0.5 = 285
Therefore the expected long run profits are
$140,000
$305,000
$285,000
Based on his selected option being either to build new or to expand, the most profitable option is to expand
=$305,000
EVPI = EPPI-EMV =$170,000
Answer:
Weight is used most often for measuring solid whereas volume is used most often for measuring liquid.
Explanation:
Weight is used most often for measuring solid, because solids have definite shape. Weight is usually expressed in Newton (N) because it is a function of mass and gravity. ( weight = mass x gravity).
Whereas volume is used most often for measuring liquid, usually expressed in cubic meter (m³) because liquids have no definite shape, rather they occupy the volume of their container.
Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
Let say for every 5 s of time interval the speed will remain constant
so it is given as
v(mi/h) 16 21 23 26 33 30 28
now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s
so here we will have
v(ft/s) 23.5 30.8 33.73 38.13 48.4 44 41.1
now for each interval of 5 s we will have to find the distance cover for above interval of time
so here it will cover 1298.1 ft distance in 30 s interval of time
