We can solve this problem by stating that the summation
of momentums must be zero. That is, the momentum at one end of the pivot must
be equal to the momentum at the other end.
Momentum 1 = Momentum 2
Since Momentum is the product of Force and Distance, and
with this rule created, we can say that:
F1 * d1 = F2 * d2
Where,
F1 = 60 lbs
d1 = 2 inches from the pivot
F2 = unknown X
d2 = 3 inches from the pivot
Substituting to the equation to find for F2:
60 lbs * 2 inches = F2 * 3 inches
F2 = 40 lbs
<span>Therefore 40 lbs of upward force must be pushed to open
the valve.</span>
<span>Answer: B</span>
At the time of her grandson's birth, a grandmother deposits $12,000.00 in an account that pays 2% compound monthly. What will be that value of the account at the child's twenty-first birthday, assuming that no other deposits or withdrawls are made during the period.
---
A(t) = P(1+(r/n))^(nt)
---
A(21) = 12000(1+(0.02/12))^(12*21)
---
A(21) = 12000(1.5214)
---
A(21) = #18,257.15
The number of lights switched on in a room and the room's brightness.
Answer:
7.2 volts
Step-by-step explanation:
3% of 240 is ...
0.03 × 240 = 3 × 2.40 = 7.20
The maximum allowable drop on a 240-volt circuit is 7.2 volts.
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
