First we need to calculate annual withdrawal of each investment
The formula of the present value of an annuity ordinary is
Pv=pmt [(1-(1+r)^(-n))÷(r)]
Pv present value 28000
PMT annual withdrawal. ?
R interest rate
N time in years
Solve the formula for PMT
PMT=pv÷[(1-(1+r)^(-n))÷(r)]
Now solve for the first investment
PMT=28,000÷((1−(1+0.058)^(−4))
÷(0.058))=8,043.59
The return of this investment is
8,043.59×4years=32,174.36
Solve for the second investment
PMT=28,000÷((1−(1+0.07083)^(
−3))÷(0.07083))=10,685.63
The return of this investment is
10,685.63×3years=32,056.89
So from the return of the first investment and the second investment as you can see the first offer is the yield the highest return with the amount of 32,174.36
Answer d
Hope it helps!
Power
= Work done / Time
= 15000 N * 3 m / 50 s
= 45000/50 N-m/s
= 900 W
It may sound like an extremely low value, but 3.0 / 50 s is also a very slow speed.
Answer:
The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350
Step-by-step explanation:
Here we have, pile of money ≥ $200
Amount in put the left pocket = $50
Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100
Amount put in right pocket = ≥ $150 - $100 ≥ $50
Total amount remaining with Jeri = $50 +≥ $50 ≥ $100
Also original pile - $200 < $100
Therefore where maximum amount given away to have more money = $200 we have
2/3× (original pile - 50) = $200
Maximum amount for original pile = $350
Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.
Answer:
They are parallel because their slopes are equal
Step-by-step explanation:
Multiply the cost per hour by 60 to find the cost for a groundskeeper
cost = 12 × 60
cost = 720
The cost for each groundskeeper is $720
Find the cost for seven groundkeepers
Multiply cost for each groundskeeper by 7
cost = 720 × 7
cost = 5,040
The monthly labor cost for seven groundskeepers is $5,040
