Modelling the acquisition of language. The percentage \( p(t) \) of children who can speak at least single words by the age of \( t \) months can be modeled by the equation, \( p(t)=100\left(1-\frac{12,200}{t^{4.48}}\right) \). Use Desmos \( { }^{1} \) to graph this equation. a. What is the interpretation of the \( t \)-intercept of the graph? b. What percentage of children can speak at least single words by the age of 12 months? c. By what age are \( 90 \% \) of children speaking at least single words?

**a. Interpretation of the \( t \)-intercept** The \( t \)-intercept is the value of \( t \) where the percentage \( p(t)=0 \). To find it, set \[ 100\left(1-\frac{12200}{t^{4.48}}\right)=0. \] Dividing both sides by 100, we have \[ 1-\frac{12200}{t^{4.48}}=0. \] This implies \[ \frac{12200}{t^{4.48}}=1 \quad \Longrightarrow \quad t^{4.48}=12200. \] Thus, the \( t \)-intercept is \[ t=(12200)^{\frac{1}{4.48}}. \] In context, this intercept represents the age (in months) at which the model predicts that essentially 0% of children have spoken at least a single word, marking the beginning of language acquisition according to the model. --- **b. Percentage of children speaking at least single words by 12 months** We substitute \( t=12 \) into the model: \[ p(12)=100\left(1-\frac{12200}{12^{4.48}}\right). \] *Step 1. Compute \( 12^{4.48} \):* Using logarithms, \[ \log_{10}(12^{4.48})=4.48\log_{10}(12). \] Since \(\log_{10}(12)\approx1.079\), \[ 4.48\log_{10}(12)\approx4.48\times1.079\approx4.832. \] Thus, \[ 12^{4.48}\approx10^{4.832}\approx68,\!200. \] *Step 2. Compute \( p(12) \):* Substitute the value into the equation, \[ p(12)=100\left(1-\frac{12200}{68200}\right)=100\left(1-0.179\right). \] \[ p(12)\approx100\times0.821=82.1. \] So, approximately \(82.1\%\) of children can speak at least single words by the age of 12 months. --- **c. Finding the age when \( 90\% \) of children speak at least single words** We set \( p(t)=90 \) and solve for \( t \): \[ 100\left(1-\frac{12200}{t^{4.48}}\right)=90. \] Divide both sides by 100: \[ 1-\frac{12200}{t^{4.48}}=0.9. \] This gives \[ \frac{12200}{t^{4.48}}=1-0.9=0.1. \] Rearrange: \[ t^{4.48}=\frac{12200}{0.1}=122000. \] Taking both sides to the power of \( \frac{1}{4.48} \), \[ t=(122000)^{\frac{1}{4.48}}. \] *Approximation using logarithms:* - First, compute \(\log_{10}(122000)\). Notice that \[ 122000=1.22\times10^5,\quad \log_{10}(122000)=\log_{10}(1.22)+5\approx0.086+5=5.086. \] - Now, \[ t\approx10^{\frac{5.086}{4.48}}\approx10^{1.135}\approx13.64. \] Thus, approximately \( t\approx13.6 \) months. Therefore, about \(90\%\) of children are predicted to be speaking at least single words by approximately \(13.6\) months.