Question 383502: Students sit a statistics test and then are tested with equivalent tests, monthly after that. The average test result, T, after n months, was found to be given by T=17.4-4.6log(n+1). After how many months would the average mark be expected to be less than 12 marks?
The sales of a new electronic gadget are growing exponentially such that, 10 000 gadgets were sold in 2000 and 82 700 gadgets were sold in 2002. If this trend continues when will sales of the gadget reach over 500 000?
Please help Im so stuck and going bonkers!!!!!
Found 2 solutions by Theo, stanbon:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! first problem:
Students sit a statistics test and then are tested with equivalent tests, monthly after that. The average test result, T, after n months, was found to be given by T=17.4-4.6log(n+1). After how many months would the average mark be expected to be less than 12 marks?
T is the average test result (otherwise known as mark).
n is the number of months.
You want to know in how many months, T will be less than or equal to 12.
your formula to solve is T <= 12
Since T = 17.4 - 4.6*log(n+1), then you can substitute that formula for T to get:
17.4 - 4.6 * log(n+1) <= 12
You need to solve this equation.
subtract 17.4 from both sides of this equation to get:
-4.6 * log(n+1) <= (12-17.4)
simplify this to get:
-4.6*log(n+1) <= -5.4
divide both sides of this equation by -4.6 to get:
log(n+1) >= -5.4 / -4.6 (multiply both sides by -1 reverses inequality)
simplify this to get:
log(n+1) >= 1.173913043
your equation is now:
log(n+1) >= 1.173913043
if log(n+1 = x, then n+1 = the number whose log is x (the anti-log of x).
the number whose log is 1.173913043 is equal to 14.92495545 because the log of 14.92495545 = 1.173913043.
you get:
n+1 >= 14.92495545
subtract 1 from both sides of this equation to get:
n >= 13.92495545
that should be your answer.
your original equation that you tried to solve is:
17.4 - 4.6 * log(n+1) <= 12
this equation should be true when n >= 13.92495545
Let's take n = 14
we get:
17.4 - 4.6 * log(15) <= 12
this simplifies to 11.98998021 which is less than 12 to the equation is true.
so if n > n >= 13.92495545, the equation it true.
let's take n = 13.92495545
we get:
17.4 - 4.6 * log(14.92495545) <= 12
this simplifies to 12 which is equal to 12 so the equation is true.
so if n = 13.92495545, the equation is true.
let's take n = 14.
we get:
17.4 - 4.6 * log(*14) <= 12
this simplifies to 12.12781104 which is not <= 12 so the equation is false.
so if n > 13.92495545, the equation is false.
when n >= 13.92495545, the equation is true.
when n < 13.92495545, the equation is false.
this confirms the answer is correct.
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second problem:
The sales of a new electronic gadget are growing exponentially such that, 10 000 gadgets were sold in 2000 and 82 700 gadgets were sold in 2002. If this trend continues when will sales of the gadget reach over 500 000?
10,000 gadgets were sold in 2000
82,700 gadgets were sold in 2002.
the exponential growth rate would be using the formula:
f = p * (1+i)^n
i is the annual growth rate
n is the number of years
you get:
82,700 = 10,000 * (1+i)^2
divide both sides of this equation by 10,000 to get:
82,700 / 10,000 = (1+i)^2
take the square root of both sides of this equation to get:
+/- sqrt(82,700/10,000) = 1+1
since I know the answer will not be negative, we'll stick to the positive answer only.
solve for 1+i to get 1+i = 2.875760769
subtract 1 from both sides of this equation to get:
i = 1.875760769
that is the growth rate per year.
it is equivalent to 187.5760769% growth rate per year.
the formula for growth is again:
f = p * (1+i)^n
first you want to test if your growth rate is good.
go from 2000 to 2002 using that growth rate.
your formula is:
f = p * (1+i)^n
p = 10,000
i = 1.875760769
n = 2
you get:
f = 10,000 * (2.875760769)^2
solve for f to get f = 82,700
the growth rate is good.
start from 2002
formula is f = p * (1+i)^n
f = 500,000
p = 82,700
i = (2.875760769)
n is what you want to find.
formula becomes:
500,000 = 82,700 * (2.875760769)^n
divide both sides of the equation by 82,700 to get:
500,000 / 82,700 = (2.875760769)^n
take log of both sides to get:
log(500,000/82,700) = log((2.875760769)^n)
by laws of logarithms, log(x^y) = y*log(x), so you get:
log(500,000/82,700) = n * log((2.875760769)
divide both sides of this equation by log((2.875760769) to get:
log(500,000/82,700) / log((2.875760769) = n
solve for n to get n = 1.70345461
That's the number of years it will take from 2002 to reach 500,000.
That + 2 is the number of years it will take from 2000 to reach 500,000.
from 2000, you get 10,000 * (2.875760769)^3.70345461 = 500,000
from 2002, you get 82,700 * ((2.875760769)^1.70345461 = 500,000
the question was:
If this trend continues when will sales of the gadget reach over 500 000?
Those sales will reach over 500,000 in 1.70345461 years from 2002.
That puts you sometime in 2004.
.70345461 * 12 = 8.441455315 which means sometime in august of 2004.
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Students sit a statistics test and then are tested with equivalent tests, monthly after that. The average test result, T, after n months, was found to be given by T=17.4-4.6log(n+1). After how many months would the average mark be expected to be less than 12 marks?
---
Solve 17.4-4.6log(n+1)< 12 for "n".
4.6log(n+1) = 5.4
log(n+1) = 1.173913
n+1 = 10^1.173913
n = 13.92 months
===========================
The sales of a new electronic gadget are growing exponentially such that,
10 000 gadgets were sold in 2000 and 82,700 gadgets were sold in 2002. If this trend continues when will sales of the gadget reach over 500,000?
-----------------
Form: y = ab^x
Solve for a and b.
---------------------------
82700 = ab^2
10000 = ab^0
-----
a = 10000
---
b^2 = 82700/10000
b^2 = 8.27
b = 2.8757
-----
Equation:
y = 10,000*2.8757^x
---
Solve 10,000*2.8757^x > 500,000
2.8757^x > 50
x*log(2.8757) > log50
x > 3.7035 years
------
Year 2000 + 3.7 = year 2004
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Cheers,
Stan H.
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