Question

Help?

Answer 1

@suppa_lova

Answer 2

The half–life of rubidium–89 is 15 minutes. If the initial mass of the isotope is 250 grams, how many grams will be left after 100 minutes?

Answer 3

a. 2.46 grams
b. 0.0098 grams
c. 0.90 grams
d. 225.31 grams

Answer 4

Will fan and medal!

Answer 5

@Michele_Laino

Answer 6

A?

Answer 7

here, we have to apply this equation:
\[m\left( t \right) = {m_0}\exp \left( { - \frac{t}{\tau }} \right)\]
where \(\tau\) is the half-life

Answer 8

so we insert 15, right?

Answer 9

we have to do this computation:
\[\huge m\left( {100} \right) = 250 \cdot {e^{\left( { - \frac{{100}}{{15}}} \right)}} = ...?\]

Answer 10

uhhh -5000e/3?

Answer 11

please wait, I'm doing the computation above...

Answer 12

Okay no problem

Answer 13

I got \(m(100)=0.318\) grams

Answer 14

nevertheless it is not an option of yours!

Answer 15

please wait, I'm working on your question...

Answer 16

I'm going to close this question because it's slowing my computer down but I'll still be on this waiting :)

Answer 17

I'm going to close this question because it's slowing my computer down but I'll still be on this waiting :)

Answer 18

sorry, in Italy what you call half-life correspond to a slightly different thing, so the right computation, is:
\[\huge m\left( {100} \right) = 250 \cdot {e^{\left( { - \frac{{100}}{{21.64}}} \right)}} = ...?\]

Answer 19

ohhh

Answer 20

−3140.34407169?

Answer 21

That's what I got.

Answer 22

here \(21.64\) is the result of this computation:
\[\frac{{15}}{{\ln 2}} = 21.64\]

Answer 23

So now what?

Answer 24

I got this result: \(m(100)=2.46\) grams

Answer 25

please check

Answer 26

I got the same answer

Answer 27

it is impossible, please believe me!

Answer 28

The problem?

Answer 29

step by step:
\[{e^{\left( { - \frac{{100}}{{21.64}}} \right)}} = 0.009842238\]

Answer 30

\[\huge {e^{\left( { - \frac{{100}}{{21.64}}} \right)}} = 0.009842238\]

Answer 31

therefore:
\[\huge \begin{gathered}
m\left( {100} \right) = 250 \cdot {e^{\left( { - \frac{{100}}{{21.64}}} \right)}} = \hfill \\
\hfill \\
= 250 \cdot 0.009842238 = 2.46 \hfill \\
\end{gathered} \]

Answer 32

Oh! I see now.

Answer 33

:)

Answer 34

Could you possibly help me with another?

Answer 35

ok!

Answer 36

The bacterium, cholera, reproduces at an exponential rate which can be modeled through the continuous growth model. If the value for k is 1.38 and we initially start with 2 bacteria, what is the equation for cholera growth, where t is the number of hours?

a. A(t) = 2e^1.38×t
b. A(t) = 2e^1.38×2t
c. A(t) = –2e^1.38×t
d. A(t) = 2e^–1.38×t

Answer 37

your problem can be modeled with a function which represents an exponential growth. At \(t=0\) such function has to be equal to \(2\)
So, what can you conclude?

Answer 38

that it isn't c?

Answer 39

right!
options C. and D. don't represent exponential growth, since they contain a minus sign

Answer 40

So we're left with A and B.

Answer 41

right!

Answer 42

Is it A?

Answer 43

correct!

Answer 44

I figured it would be A because we start out with 2 bacteria.

Answer 45

also option B. starts with 2 bacteria

Answer 46

nevertheless I think that option A, is the right option

Answer 47

Okay, thank you!

Answer 48

:)