96% of 25 is 24 hope this helps
Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
Step-by-step explanation:
It often works well to write an exponential expression as ...
value = (initial value)×(growth factor)^(t/(growth period))
(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...
P(t) = 100·2.3^t
__
(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours
__
(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t
P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour
__
(d) We want to find t such that ...
P(t) = 10000
100·2.3^t = 10000 . . . substitute for P(t)
2.3^t = 100 . . . . . . . . divide by 100
t·log(2.3) = log(100)
t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000
Answer:
Step-by-step explanation:
Volumes of two spheres A and B = 648 cm³ and 1029 cm³
Things to remember:
1). Scale factor of two objects =
[
and
are the radii of two circles]
2). Area scale factor = 
3). Volume scale factor = 
Volume scale factor Or Volume ratio = 
![\frac{r_1}{r_2}=\sqrt[3]{\frac{648}{1029} }](https://tex.z-dn.net/?f=%5Cfrac%7Br_1%7D%7Br_2%7D%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B648%7D%7B1029%7D%20%7D)
![\frac{r_1}{r_2}=\frac{6(\sqrt[3]{3})}{7(\sqrt[3]{3})}](https://tex.z-dn.net/?f=%5Cfrac%7Br_1%7D%7Br_2%7D%3D%5Cfrac%7B6%28%5Csqrt%5B3%5D%7B3%7D%29%7D%7B7%28%5Csqrt%5B3%5D%7B3%7D%29%7D)

Therefore, scale factor =
≈ 6 : 7
Area scale factor Or area ratio = 
= 
≈ 36 : 49
Volume scale factor or Volume ratio = 
= 
≈ 216 : 343