Question

In a typing class, the average number of words per minute N typed after t weeks of lessons was found to be

N =157/1 + 5.4e^-0.12t
Find the time necessary to type (a) 50 words per minute and (b) 75 words per minute

Answer 1

For 50 wpm : 7.7 weeks approximately

For 75 wpm : 13.3 weeks approximately

Explanation:

Given the relationship between N, the typing speed and t, the number of weeks of lessons as:

`\mbox{\large N= (157)/(1+5.4e^(-0.12t))}`

To solve for t for a particular N

cross-multiply the left side with the right side denominator to get

`\mbox{\large 1+5.4e^(-0.12t ) = (157)/(N)}`

For N= 50 this works out to


`\mbox{\large 1+5.4e^(-0.12t ) = (157)/(50)}`

`\mbox{\large 5.4e^(-0.12t ) = (157)/(50) - 1}\\`

`\mbox{\large 5.4e^(-0.12t ) = (157-50)/(50) } = (107)/(5) = 2.14`

Divide by 5.4 both sides:

`\mbox{\large e^(-0.12t ) = (2.14)/(5.4)} = 0.3963\\`

Take natural logs on both sides

`\mbox{\large \ln(e^(-0.12t)) = \ln(0.3963)\\}`

`\mbox{\large \ln(e^(-0.12t)) = -0.12 \;\textrm{since $ln(e^x)$ = x}}` since ln(eˣ) = x

ln(0.3963) = -0.9256

Therefore

`t = (-0.9256)/(-0.12) = (0.9256)/(0.12) = {7.7133 \;weeks\approx \mbox{\large \boxed{7.7 \;weeks}}`

For N = 75, follow the same strategy and you will end up with

`1+5.4e^(-0.12t)=(157)/(75) \\\\\\1+5.4e^(-0.12t)= 2.0933\\\\5.4e^(-0.12t) = 2.0933 - 1 = 1.0933\\\\e^(-0.12t) = (1.0933)/(5.4) = 0.2025\\\\`

Taking logs on both sides

`-0.12t =-1.5971`

`t = (-1.5971)/(-0.12) = 13.3097 \;weeks \approx \boxed{\mbox{\large 13.3 \;weeks}}`

Answer 2

i hope this helpes

Explanation:

(a) To find the time necessary to type 50 words per minute, we need to solve the equation:

157/1 + 5.4e^-0.12t = 50

Rearranging the equation to isolate e^-0.12t and using the natural logarithm, we can find t:

ln(N - 157) = ln(5.4) - 0.12t

-ln(N - 157) = 0.12t - ln(5.4)

0.12t = ln(5.4) + ln(N - 157)

t = (ln(5.4) + ln(N - 157)) / 0.12

Plugging in N = 50, we get:

t ≈ 11.52 weeks

(b) To find the time necessary to type 75 words per minute, we follow the same steps as above:

157/1 + 5.4e^-0.12t = 75

ln(N - 157) = ln(5.4) - 0.12t

-ln(N - 157) = 0.12t - ln(5.4)

0.12t = ln(5.4) + ln(N - 157)

t = (ln(5.4) + ln(N - 157)) / 0.12

Plugging in N = 75, we get:

t ≈ 9.32 weeks