Need help with a math problem please!?

Your picture is way too small.  When I do my best to enlarge it, it looks like the decay function is A(t) = A~0 × e^(-0.0244t)  { read as "A-sub-zero" }

(a) the decay rate of Strontium 90 is -2.44% per year

(b) after twenty years, the original 800 grams is now 491.08 grams

(c) 200 grams of Strontium 90 will be left after 56.82 years (rounded)

(d) the half-life of Strontium 90 is 28.41 years.

How to get these answers:

(a) multiply the exponent of e by 100%: -.0244 × 100% = -2.44%

(b) 800 × e^(-.0244 × 20) = 800 × e^(-.488) = 800 × 0.613852873… = 491.08…

(c) 200 = 800 × e^(-.0244t) ...

1/4 = e^(-.0244t) ... take natural logarithm

-1.386294317… = -.0244t

t = 56.81…

(d) half-life = -ln(2) ÷ decay rate = -0.69314718… ÷ -.0244 = 28.40767…

Commentary:

I found this question in a couple of places, all with some initial amount of Sr~60 and variations of "after xxx years" and "how long until xxx grams are left", etc.  All of them had the decay rate as -.0244.  Here's the kicker: that's wrong.  Strontium-90's actual half-life is 28.79 years, making its decay rate closer to -0.0241

Parting words:  Next time, bigger picture.

I think that says:

Strontium 90 is a radioactive material that decays according to the function

A(t) = A₀ e^(-0.0244t)

Where A₀ is the initial amount present and A is the amount present at time t (in years).  Assume that a scientist has a sample of 800 grams of strontium 90.

So if this is correct, our equation is:

A(t) = 800 e^(-0.0244t)

The decay rate?  I think that's just out of the exponent:  -0.0244

----

How much is left after 20 years?  Solve for A(20):

A(20) = 800 e^(-0.0244 * 20)

A(20) = 800 e^(-0.488)

A(20) = 800(0.6138529)

A(20) = 491.0823 g (rounded to 4DP)

----

When will 200 grams remain?  Solve for "t" when A(t) = 200:

A(t) = 800 e^(-0.0244t)

200 = 800 e^(-0.0244t)

0.25 = e^(-0.0244t)

ln(0.25) = -0.0244t

t = - ln(0.25) / 0.0244

t = 56.8153 years (rounded to 4DP)

----

What's the half-life?  Solve for t when A(t) = 400 (half of 800):

A(t) = 800 e^(-0.0244t)

400 = 800 e^(-0.0244t)

0.5 = e^(-0.0244t)

ln(0.5) = -0.0244t

t = - ln(0.5) / 0.0244

t = 28.4077 years (rounded to 4DP)

(Note that 200 grams is two half-lives (800 -> 400 and 400 -> 200), so the answer to part C is twice the time as part D, so the math works out to show it's correct)

even enlarging the text I still cannot read everything......a) (dA/dt) / A ; b) let the variable be 20 ; c) let A = 200 and solve for t {use ln function } ; d) let initial amount be 2  and A = 1 and find t

Illegible , too blurry !