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egoroff_w [7]
3 years ago
12

A frog is sitting exactly in the middle of a board that is five feet long. Every ten seconds he jumps one foot to the left or on

e foot to the right, at random. You want to use simulation to estimate the probability that he jumps off the board in sixty seconds or less. (For example, if he jumps LLRLRL, he is still on the board, on the left-most square. If he jumps LLRLL, he has jumped off the board in fifty seconds.)
Required:
Describe how you would use the digits of a random number table to carry out this simulation.
Mathematics
1 answer:
mart [117]3 years ago
4 0

Answer:

Since the Frog makes a Jump every 10 seconds

   ( using the digits on a random number table )

  • It will take 7 jumps with 2 jumps in reverse direction i.e. either left or right for the frog to be off the board in 60 seconds
  • it will take 3 jumps in similar direction as well to get the frog off the board
  • also 5 jumps with 1 jump in the opposite direction is what it will take to get the frog off the board as well within the time frame of 60 seconds

Step-by-step explanation:

A frog sitting exactly in the middle of a board which is 5ft long : frog is 2.5 ft from the edge of the board

every 10 seconds the frog jumps ; L or R

If he jumps ; LLRLRL ( still on Board on the left-most square )

if he jumps ; LLRLL ( He is off the Board in fifty seconds )

Since the Frog makes a Jump every 10 seconds

   ( using the digits on a random number table )

  • It will take 7 jumps with 2 jumps in reverse direction i.e. either left or right for the frog to be off the board in 60 seconds
  • it will take 3 jumps in similar direction as well to get the frog off the board
  • also 5 jumps with 1 jump in the opposite direction is what it will take to get the frog off the board as well within the time frame of 60 seconds
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The ​half-life of a radioactive element is 130​ days, but your sample will not be useful to you after​ 80% of the radioactive nu
gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt} =-\lambda N

\Rightarrow \frac{dN}{N} =-\lambda dt

Integrating both sides

\int \frac{dN}{N} =\int\lambda dt

\Rightarrow ln|N|=-\lambda t+c

When t=0, N=N_0 = initial amount

\Rightarrow ln|N_0|=-\lambda .0+c

\Rightarrow c= ln|N_0|

\therefore ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t.......(1)

                            \frac{N}{N_0}=e^{-\lambda t}.........(2)

Logarithm:

  • ln|\frac mn|= ln|m|-ln|n|
  • ln|ab|=ln|a|+ln|b|
  • ln|e^a|=a
  • ln|a|=b \Rightarrow a=e^b
  • ln|1|=0

130 days is the half-life of the given radioactive element.

For half life,

N=\frac12 N_0,  t=t_\frac12=130 days.

we plug all values in equation (1)

ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130

\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130

\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130

\rightarrow -ln|2|=-\lambda \times 130

\rightarrow \lambda= \frac{-ln|2|}{-130}

\rightarrow \lambda= \frac{ln|2|}{130}

We need to find the time when the sample remains 80% of its original.

N=\frac{80}{100}N_0

\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t

\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t

\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t

\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}

\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}

\Rightarrow t\approx 42

We can use the sample about 42 days.

7 0
4 years ago
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