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Ugo [173]
2 years ago
14

A total of 620 tickets were sold for the school play. They were either adult tickets or student tickets. There were 70 more stud

ent tickets sold than adult tickets.
How many adult tickets were sold?
Mathematics
2 answers:
Phoenix [80]2 years ago
8 0

Answer:

There were 240 adult tickets sold and 380 student tickets sold.

Step-by-step explanation:

How did I get this?

First, I divided 620 by 2. You get 310. Then I subtract 70 by on of the 310's. You get 240. I take that 70 that I subtracted and add it to the other 310, you get 380. (This may sound confusing so I will write it out)

620/2= 310

310-70= 240 adult tickets

310+70=380 student tickets

AlexFokin [52]2 years ago
4 0

Answer:

22

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

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t - Time, measured in days.

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The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

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The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

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Answer:

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Step-by-step explanation:

Incomplete question [See comment for complete question]

Given

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Time= \frac{65.94}{14} min

Time= 4.71 min\\

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