Answer:
t = ln(0.5)/-r
Step-by-step explanation:
The decay rate parameter is missing. I will assume a value of 4% per day.
The exponential decay is modeled by the following equation:
A = A0*e^(-r*t)
where A is the mass after t time (in days), A0 is the initial mass and r is the rate (as a decimal).
At half-life A = A0/2, then:
A0/2 = A0*e^(-0.04*t)
0.5 = e^(-0.04*t)
ln(0.5) = -0.04*t
t = ln(0.5)/-0.04
t = 17.33 days
In general the half-life time is:
t = ln(0.5)/-r
Answer:
3/8
Step-by-step explanation:
1. We list all codes that start with 0 and have 0 as second digit:
2. We list all codes that start with 0 and have 1 as second digit:
3. We list all codes that start with 1 and have 0 as second digit:
4. We list all codes that start with 1 and have 1 as second digit:
5. We find the number of codes which have exactly 2 zeros: 6 codes
6. The total number of outcomes is: 16 codes
7. We find the probability that the number of codes has exactly 2 zeros: 6/16 3/8
Answer:
(-2_5) and (1, -4) is right
Cute one!
<span>
</span>Summarizing:
<span>sec(acot(tan(asin(sin(pi/3)))) .... use asin(sin(x))=x
</span>=sec(acot(tan(pi/3)))
=sec(acot(sqrt(3))) ......... use acot(x)=atan(1/x)
=sec(atan(1/sqrt(3)))
=sec(atan(sqrt(3)/3)) .... evaluate atan(sqrt(3)/3), use unit circle
=sec(pi/6)
=1/cos(pi/6)...... evaluate cos(pi/6), use unit circle
=1/(sqrt(3)/2)
=2/sqrt(3) .... now rationalize
=2sqrt(3)/3
