Answer:
y=Ae^(1.25t)
Step-by-step explanation:
From the expression y=Ae^kt
After two days of the experiment, y = 49 million, t=2
After four days of the experiment, y= 600.25 million, t=4
A is the amount of bacteria present at time zero and t is the time after the experiment (in days)
At t=2 and y =49
49=Ae^2k…………….. (1)
At t=4 and y = 600.25
600.25=Ae^4k………… (2)
Divide equation (2) by equation (1)
600.25/49=(Ae^4k)/(Ae^2k )
12.25=e^2k
Take natural log of both sides
ln(12.25) =2k
2.505 =2k
k=1.25
The exponential equation that models this situation is y=Ae^(1.25t)
What is the digit at the hundred thousands place? It is 2. Is the digit to the right of it less than 5? No, 5 is 5. So replace the digits to the right of 2 with 0, and change the 2 to 3.
Answer: 300,000
Answer:
x-7-5/x+4
So answer C
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
1116 / 36 = 31
hope this helps
Can you be more specific about what is P(5)? The probability would probably be around five though.
