Let x represent amount invested in the higher-yielding account.
We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be
.
We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.
, where,
I = Amount of interest,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
We are told that interest rates are 6% and 10%.


Amount of interest earned from lower-yielding account:
.
Amount of interest earned from higher-yielding account:
.

Let us solve for x.



Therefore, the man invested $30,000 at 10%.
Amount invested in the lower-yielding account would be
.
Therefore, the man invested $60,000 at 6%.
No Jane is not correct she has 25 cents where john has 30 cents
The population of the town in 1960 is 48.80 thousands
<h3>How to determine the population in 1950?</h3>
The equation of the model is given as:
f(t) = 42e^(0.015t)
1960 is 10 years after 1950.
This means that:
t = 10
Substitute t = 10 in f(t) = 42e^(0.015t)
f(10) = 42e^(0.015 * 10)
Evaluate
f(10) = 48.80
Hence, the population of the town in 1960 is 48.80 thousands
Read more about exponential functions at:
brainly.com/question/11464095
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Answer:
I am not 100% about this, but I think that Z=46 Y=30 and X=45
Step-by-step explanation: