The half-life of radioactive cobalt is 5.27 years. Suppose that a nuclear accident has left the level of cobalt radiation in a c
ertain region at 100 times the level acceptable for human habitation. How long will it be until the region is again habitable?
1 answer:
Answer:
35 years
Step-by-step explanation:
The proportion p that remains after y years is ...
p = (1/2)^(y/5.27)
In order for 1/100 to remain (the level decays from 100 times to 1 times), we have ...
.01 = .5^(y/5.27)
log(0.01) = y/5.27·log(0.5) . . . take logs
y = 5.27·log(0.01)/log(0.5) ≈ 35.01 ≈ 35 . . . . years
You might be interested in
1 2/3 = 5/3
2 1/3 = 7/3
2 = 2
5/3 * 7/3 * 2 = 70/9 cubic inches = 0.127455 liters
Answer:
Step-by-step explanation:
volume of reqd. figure=10×5×1+3×3×1=50+9=59 cm³
Gross pay = 765
federal withholding tax = 68
Social security tax = 765 * 6.2% = 47.43
Medicare tax = 765 * 1.45% = 11.09
State tax (22% of federal tax) = 68 * 22% = 14.96
765 - 68 - 47.43 - 11.09 - 14.96 = 623.52
<span>Lauren's net pay is 623.52 so the answer is D.</span>
Answer:
yes it is
Step-by-step explanation:
plug x and y in
3 * 5 - 10 = 5 <== And this is true
