Question

the amount of radioactive material in an ore sample is given by the exponential function $a(t) = 150(3.4)^{-0.5t}$, where $a(t)$ is the amount present, in grams, of the sample $t$ months after the initial measurement. how much was present 7 months later? there was/were approximately \boxed{} gram(s) present 7 months later. (simplify your answer. type an integer or decimal rounded to the nearest hundredth as needed.)

Explanation:

Step1: Substitute \( t = 7 \) into the function

Given \( A(t)=150(3.4)^{-0.5t} \), when \( t = 7 \), we have \( A(7)=150(3.4)^{-0.5\times7}=150(3.4)^{-3.5}\).

Step2: Simplify the exponent

First, \( - 0.5\times7=-3.5\). Then, using the property \( a^{-n}=\frac{1}{a^{n}}\), so \( (3.4)^{-3.5}=\frac{1}{(3.4)^{3.5}}\). And \( (3.4)^{3.5}=(3.4)^{3}\times(3.4)^{0.5}\). \( (3.4)^{3}=3.4\times3.4\times3.4 = 39.304\), \( (3.4)^{0.5}=\sqrt{3.4}\approx1.8439\). Then \( (3.4)^{3.5}\approx39.304\times1.8439\approx72.48\).

Step3: Calculate the value of \( A(7) \)

\( A(7)=150\times\frac{1}{72.48}\approx2.07\)

Answer:

\( 2.07\)