Given:
Desmond deposits $ 50 monthly.
Yearly he deposits = $50×12 = $ 600
Rate of interest compounded monthly = 4.7%
To find the amount he will receive after 10 years and the rate of change the value of his account after 10 years.
Formula
where,
A be the final amount
P be the principal
r be the rate of interest
t be the time and
n be the number of times the interest is compounded.
Now,
Taking,
P = 600, r = 4.7, n = 12, t = 10 we get,
or, 
or, 
Now,
At starting he has $ 600
At the end of 10 years he will be having $ 959.1
So,
The amount of change in his account = $ (959.1-600) = $ 359.1
Therefore the rate of change = 
= 59.85%
Hence,
a) His account will contain $ 959.1 after 10 years.
b) The rate of change in his account is 59.85% after 10 years.
Answer:
b
Step-by-step explanation:
Part percent
—— = ————. I use that method a lot and it helped me. ANSWER:65%
Whole 100
Answer:
if I am correct 2/3 is bigger than 0.65
and I think 5.5 is smaller than 5 3/7
and I think the order goes 3/8 5/6 1/2
Answer:the car was traveling at a speed of 80 ft/s when the brakes were first applied.
Step-by-step explanation:
The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,
a = - 16ft/s^2
While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.
This means that it travelled a distance,
s = 200 feet
We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.
Since the car came to a stop, its final velocity, v = 0
Applying Newton's equation of motion,
v^2 = u^2 + 2as
0 = u^2 - 2 × 16 × 200
u^2 = 6400
u = √6400
u = 80 ft/s
