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Nastasia [14]
3 years ago
5

}{3}" alt="\frac{1}{2} +\frac{1}{3}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
iren [92.7K]3 years ago
4 0

Answer:

\frac{1}{2}  +  \frac{1}{3}  =  \frac{5}{6}

Step-by-step explanation:

1/2+1/3 L.C.M = 6

\frac{3 + 2}{6}  =  \frac{5}{6}

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In order to efficiently bid on a contract, a contractor wants to be 95% confident that his error is less than two hours in estim
Sonja [21]

Answer:

Step-by-step explanation:

Given that in order to efficiently bid on a contract, a contractor wants to be 95% confident that his error is less than two hours in estimating the average time it takes to install tile flooring.

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8 0
3 years ago
Alexander Litvinenko was poisoned with 10 micrograms of the radioactive substance Polonium-210. Since radioactive decay follows
koban [17]

Answer:

The amount of Polonium-210 left in his body after 72 days is 6.937 μg.

Step-by-step explanation:

The decay rate of Polonium-210 is the following:

N(t) = N_{0}e^{-\lambda t}     (1)

Where:

N(t) is the quantity of Po-210 at time t =?

N₀ is the initial quantity of Po-210 = 10 μg

λ is the decay constant  

t is the time = 72 d  

The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.    

First, we need to find the decay constant:

\lambda = \frac{ln(2)}{t_{1/2}}    (2)

Where t(1/2) is the half-life of Po-210 = 138.376 days

By entering equation (2) into (1) we have:

N(t) = N_{0}e^{-\frac{ln(2)}{t_{1/2}}*t}} = 10* \frac{99.498}{100}*e^{-\frac{ln(2)}{138.376}*72} = 6.937 \mu g    

Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.  

I hope it helps you!  

8 0
3 years ago
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