Answer:
0.49
0.51
Explanation:
Probability that bulb is defective :
Let :
b1 = box 1 ; b2 = box 2 ; b3 = box 3
d = defective
P(defective bulb) = (p(b1) * (d|b1)) + (p(b2) * p(d|b2)) + (p(b3) * p(d|b3))
P(defective bulb) = (1/3 * 10/20) + (1/3 * 7/15) + (1/3 * 5/10))
P(defective bulb) = 10/60 + 7/45 + 5/30
P(defective bulb) = 1/6 + 7/45 + 1/6 = 0.4888
= 0.49
P(bulb is good) = 1 - P(defective bulb) = 1 - 0.49 = 0.51
<h3>
Option A</h3>
In a Series circuit with two identical loads, the voltage across each load will be: the same
<h3><u>
Explanation:</u></h3>
A series circuit is one with total the loads in a row. There is barely ONE path for electricity to pass. If this circuit was a series of flashbulbs, and one left out, the left bulbs would switch off. T
he current in a series circuit is universally the same and the voltage over the circuit is the amount of the unique voltage drops over each component. The voltage referred to as a series circuit is equivalent to the amount of the individual voltage drops.
Sound requires a medium to travel. Sound cannot travel in vacuum in space while light can. Therefore, the receiver from earth receives only light sent from moon, not sound.
Explanation:
- Sound is created by vibrations that requires medium to travel. The medium can be solid, liquid, or gas. Whatever it is, a medium is required.
- In space, there is no solid, liquid, or gas. In other words, it is vacuum. Sound waves cannot travel through vacuum.
- Light waves, on the other hand, can travel through vacuum. Hence, if the transmitter on moon emits sound waves and light waves, receiver from earth picks only the light waves.
Answer:
The required rated input apparent power is 14.98 HP
Explanation:
Real power input = real power output ÷ efficiency
real power output = 12 HP
efficiency = 90% = 0.9
Real power input = 12 ÷ 0.9 = 13.33 HP
Apparent power input = real power input ÷ power factor = 13.33 ÷ 0.89 = 14.98 HP