250 a study of the annual population of the red - winged blackbird in ft. mill south carolina shows the…

250 a study of the annual population of the red - winged blackbird in ft. mill south carolina shows the population (b(t)) can be represented by the function (b(t)=750(1.16)^{t}) where the (t) represents the number of years since the study began. in terms of the monthly rate of growth, the population of red - winged blackbirds can be best approximated by the function 1 (b(t)=750(1.012)^{t}) 2 (b(t)=750(1.012)^{12t}) 3 (b(t)=750(1.16)^{12t}) 4 (b(t)=750(1.16)^{\frac{t}{12}}) 260 iridium - 192 is an isotope of iridium and has a half - life of 73.83 days. if a laboratory experiment begins with 100 grams of iridium - 192, the number of grams (a) of iridium - 192 present after (t) days would be (a = 100(\frac{1}{2})^{\frac{t}{73.83}}) which equation approximates the amount of iridium - 192 present after (t) days? (a = 100\frac{73.83}{2})

250 a study of the annual population of the red - winged blackbird in ft. mill south carolina shows the population (b(t)) can be represented by the function (b(t)=750(1.16)^{t}) where the (t) represents the number of years since the study began. in terms of the monthly rate of growth, the population of red - winged blackbirds can be best approximated by the function 1 (b(t)=750(1.012)^{t}) 2 (b(t)=750(1.012)^{12t}) 3 (b(t)=750(1.16)^{12t}) 4 (b(t)=750(1.16)^{\frac{t}{12}}) 260 iridium - 192 is an isotope of iridium and has a half - life of 73.83 days. if a laboratory experiment begins with 100 grams of iridium - 192, the number of grams (a) of iridium - 192 present after (t) days would be (a = 100(\frac{1}{2})^{\frac{t}{73.83}}) which equation approximates the amount of iridium - 192 present after (t) days? (a = 100\frac{73.83}{2})

Answer

Explanation:

Step1: Convert annual - growth rate to monthly - growth rate

Let the annual - growth rate formula be $B(t)=B_0(1 + r_a)^t$, where $r_a$ is the annual growth rate and $t$ is the number of years. We want to find a monthly - growth rate formula $B(t)=B_0(1 + r_m)^{12t}$, where $r_m$ is the monthly growth rate. Given $B_0 = 750$ and $r_a=0.16$, we set $(1 + r_a)^t=(1 + r_m)^{12t}$. Taking the $t$ - th root of both sides (assuming $t\neq0$), we get $1 + r_a=(1 + r_m)^{12}$. Then $1 + r_m=(1 + r_a)^{\frac{1}{12}}$. Since $r_a = 0.16$, $1 + r_m=(1.16)^{\frac{1}{12}}\approx1.012$. So the population function in terms of months is $B(t)=750(1.012)^{12t}$.

Step2: Analyze the half - life formula

The general formula for radioactive decay is $A = A_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $A_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. Given $A_0 = 100$ grams and $T_{1/2}=73.83$ days, the amount of Iridium - 192 present after $t$ days is $A = 100(\frac{1}{2})^{\frac{t}{73.83}}$.

Answer:

  1. $B(t)=750(1.012)^{12t}$
  2. $A = 100(\frac{1}{2})^{\frac{t}{73.83}}$