Are these two separate questions?
There will be 1.078x students next year and equation is number of students in next year = x + 7.8% of x
<h3><u>Solution:</u></h3>
Given, There are "x" number of students at helms.
The number of students increases by 7.8% each year which means if there "x" number of students in present year, then the number of students in next year will be x + 7.8% of x
Number of students in next year = number of students in present year + increased number of students.
Thus there will be 1.078x students in next year
Round off 1,563,385 to the nearest million. A. 1,600,000 B. 1,563,400 C. 2,000,000 D. 1,563,000
iren2701 [21]
1,563,385 ...when rounded to the nearest million is 2,000,000.
Answer:
N = 920(1+0.03)^4t
Step-by-step explanation:
According to the given statement a car repair center services 920 cars in 2012. The number of cars serviced increases quarterly at a rate of 12% per year after 2012.
Rate is 12 % annually
rate in quarterly = 12/4= 3%
We will apply the compound interest equation:
N=P( 1+r/n)^nt
N= ending number of cars serviced.
P= the number of cars serviced in 2012,
r = interest rate
n = the number of compoundings per year
t= total number of years.
Number of compoundings for t years = n*t = 4t
Initial number of cars serviced=920
The quarterly rate of growth = n=4
r = 3%
The growth rate = 1.03
Compound period multiplied by number of years = 920(1.03)^4t
Thus N = 920(1+0.03)^4t
N = number of cars serviced after t years...
