Question

Could someone assist me with part C?

Answer 1

I got 335.

Answer 2

can u show ur work?

Answer 3

Sure.

Answer 4

P(t)= 52.2e^-0.0118t

1= 52.2e^-0.0118t
1/52.2 = e^-0.0118t
Ln( 1/52.2)= -0.0118t

t= 335.1764826
t=335

Answer 5

u should put 3 in place of p(t) in stead of 1, and I got r to be equal to 0.0176 nearly equal to yours

Answer 6

0.01176*

Answer 7

Oh my gosh! I see now.

Answer 8

I was following an example and it says 1 million and mine says 3 million.

Answer 9

Now im getting 242

Answer 10

yeah, but you approached the problem correctly, it wud be around 163 or 164

Answer 11

what base of log are you using*

Answer 12

Are you asking about Ln?

Answer 13

yes... Is it ln or Log that you are using

Answer 14

Ln

Answer 15

It's correct. @Destinyyyy

Answer 16

Yes I know. I already got it wrong. I was asking for clarification. Sometimes Pearson is incorrect. And you forgot the - sign.

Answer 17

no I didn't I divided it by 0.01176 instead of -0.01176 so, it came out to be -242.89 in stead of 242.89. but it is +ve, u r right