Question

I WILL MARK BRAINLIEST TO MORE THAN ONE PERSON WHO HELPS

Answer 1

Step-by-step explanation:

Given

No. of employees $N(t)=500(0.02)^{0.7^{t}}$

(a)When the company opens i.e. at t=0

$N(0)=500(0.02)^{1}=10$

There are 10 employees at the beginning

(b)when N=100

$100=500(0.02)^{0.7^{t}}\\0.2=0.02^{0.7^{t}}$

Taking log both sides

$\ln (0.2)=0.7^t \cdot \ln (0.02)$

$0.7^t=\frac{\ln (0.2)}{\ln (0.02)}=0.4114$

again taking log

$t\times \ln (0.7)=\ln (0.4114)$

$t=\dfrac{\ln (0.4114)}{\ln (0.7)}=2.49$

(c)when t tends to $\infty$, $(0.7)^t\rightarrow 0$

So, $N(\infty)=500\cdot(0.02)^0=500\times1=500$

i.e. there can be 500 employees at max