Answer:
unit single fraction...... i believe
Step-by-step explanation:
Answer:
In the next 31 years
Step-by-step explanation:
In this question, we are tasked with calculating the number of years it will take for a school’s population to reach a certain amount of it is increasing at a certain rate.
We can use a mathematical exponential approach to solve this.
Let’s say;
P = I( 1 + r)^t
where P is the projected population at 926
I is the initial population at 502
r is the percentage change or rate at 2% which is same as 2/100 = 0.02
and t is time
Plugging these values, we have
926 = 502(1 + 0.02)^t
Divide through by 502
1.845 = 1.02^t
take the logarithm of both sides
log 1.845 = tlog 1.02
t = log 1.845/log 1.02
t = 30.9 = 31 years
The first number is 170
The second number is 171
Step-by-step explanation:
Let the first number be x^3 and the second number be x^3+1
The sum of the two is 341
X^3+x^3+1=341
2x^3+1=341
Substrate 1 from both sides
2x^3=341-1
Answer:
$2000 is invested at 8%
$2000 is invested at 14%
Step-by-step explanation:
A total of $4000 is invested part at 8% and the remainder at 14%
Annual interest is $440.
Simple interest formula;
I = P × R × T
Where I is the interest, P is the principal, R is the rate and T is the time.
P = $4000
R = 8% and 14%
T = 1 year
I = $440
Let's say $a is invested at 8% and;
$b is invested at 14%
Then,
($a ×
× 1 ) + ($b ×
× 1) = $440
and
$a + $b = $4000
This forms a simultaneous equation;
0.08a + 0.14b = 440 ... (i)
a + b = 4000 ... (ii)
Multiplying (i) by 1 and (ii) by 0.08 we get;
0.08a + 0.14b = 440 ... (i)
0.08a + 0.08b = 320 ... (ii)
Subtracting (i) - (ii) we get;
0 + 0.06b = 120
0.06b = 120
b = 120 ÷ 0.06 = 2000
So amount invested at 14% ($b) = $2000 and,
The amount invested at 8% ($a) = $4000 - $2000 = $2000