Answer to Question #198830 in Algebra for Zainab

Question

An experiment was conducted to determine the effects of elapsed time on a person's memory. Subjects were asked to look at a picture which contained many different objects. After studying the pictures, they were asked to recall as so many of the objects as they could. At different time intervals following this, they would be asked to recall as many objects as they could. Based on the experiment, the following function was developed: = f(t) = 90 - 30 ln(t) where t ≥ 1

For this function represents the average percent recall as a function of time since studying the pictures (measured in hours). A value of = 50 would indicate that at the corresponding time t the average recall for the study group was 50 percent.

a) What is the average percent recall after 1 hour?

b) Find the expression for the rate of change in with respect to time.

c) What is the maximum percent and minimum percent recall?

Answer 1

Ans:-

 Based on the experiment, the following function was developed:

$\Rightarrow f(t) = 90 - 30 ln(t) \ \ where \ t ≥ 1$

(a) At $t=1 \ hour$

$f(1)= 90-30ln(1)=90-30\times0=90$

So Average percent recall after 1 hour is 90

(b) $f(t)=90-30ln(t)$

Differentiate with respect to t

$f'(t)=-30\times \dfrac{1}{t}$

Therefore the above equation will be  the expression for the rate of change in with respect to time.

(c) For the maximum percent and minimum percent recall

$f'(t)=-30 \times \dfrac{1}{t}$

Again differentiate with respect to t

$f''(t)=30\times \dfrac{1}{t^2}$

We know that $t ≥ 1$ So $f''(t)$ is always greater than so

We get minimum percentage recall which is $0 \ \%$ and Maximum percentage recall will be $90 \ \%$