Adding in some of answer 3

docs/ss2-24-final/index.html

@@ -361,17 +361,16 @@ <h1 class="title"> </h1>
 <p><br></p>
 <hr />
 <h2 id="problem-2">Problem 2</h2>
-<p><strong>[Eligible for midterm redemption]</strong></p>
 <p>Consider a dataset of 4 values, <span class="math inline">y_1 &lt;
-y_2 &lt; y_3 &lt; y_4</span>, with a mean of 6.<br />
-Let <span class="math inline">Y_\text{abs}(h) = \frac{1}{4} \sum_{i =
+y_2 &lt; y_3 &lt; y_4</span>, with a mean of 6.</p>
+<p>Let <span class="math inline">Y_\text{abs}(h) = \frac{1}{4} \sum_{i =
 1}^4 |y_i - h|</span> represent the mean absolute error of a constant
 prediction <span class="math inline">h</span> on this dataset of 4
 values.</p>
 <p>Similarly, consider another dataset of 3 values, <span
 class="math inline">x_1 &lt; x_2 &lt; x_3</span>, that also has a mean
-of 6.<br />
-Let <span class="math inline">X_\text{abs}(h) = \frac{1}{3} \sum_{i =
+of 6.</p>
+<p>Let <span class="math inline">X_\text{abs}(h) = \frac{1}{3} \sum_{i =
 1}^3 |x_i - h|</span> represent the mean absolute error of a constant
 prediction <span class="math inline">h</span> on this dataset of 3
 values.</p>
@@ -408,9 +407,14 @@ <h2 class="accordion-header" id="heading2_1">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p>Without an absolute value sign, the empirical risk is minimized when
-<span class="math display">h</span> is as small as possible, so <span class="math inline">h^* = -\infty</span>.</p>
-<p>TODO</p>
+<p><span class="math inline">Z(h) = \frac{1}{7} \sum_{i = 1}^7 (h -
+z_i)</span> is minimized when <span class="math inline">h</span> is as
+small as possible. If <span class="math inline">h</span> is smaller than
+<span class="math inline">z_i</span> then it will make a negative risk!
+This means the smaller <span class="math inline">h</span> is the smaller
+the difference will be!</p>
+<p>When looking at our answers available the smallest risk would be
+<span class="math inline">h^* = -\infty</span>.</p>
 </div>
 </div>
 </div>
@@ -441,19 +445,22 @@ <h2 class="accordion-header" id="heading2_2">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p><span class="math display">y_3</span>, the median of the dataset.</p>
-<p>TODO</p>
+<p><span class="math inline">y_3</span>, the median of the dataset.</p>
+<p>Recall <span class="math inline">h^*</span> for <span class="math inline">T_\text{abs}(h)</span> is the median of the
+dataset!</p>
+<p>Our dataset is: <span class="math inline">x_1, y_1, y_2, y_3, y_4,
+x_2, x_3</span>. Our median is <span class="math inline">y_3</span>,
+which means <span class="math inline">h^* = y_3</span>.</p>
 </div>
 </div>
 </div>
 </div>
 <p><br></p>
+<h3 id="problem-2.3">Problem 2.3</h3>
 <p>Suppose the slope of <span class="math inline">T_\text{abs}(h)</span>
 is <span class="math inline">-\frac{1}{7}</span> at some <span
 class="math inline">h_p</span>. <em>Hint: think about what values of
 <span class="math inline">h</span> could have this slope.</em></p>
-<p><br></p>
-<h3 id="problem-2.3">Problem 2.3</h3>
 <p>Suppose the dataset is now modified by moving the <span
 class="math inline">\{x_i\}</span> such that<br />
 <span class="math inline">y_1 &lt; y_2 &lt; y_3 &lt; y_4 &lt; x_1 &lt;
@@ -474,19 +481,20 @@ <h2 class="accordion-header" id="heading2_3">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p><span class="math display">-\frac{3}{7}</span> Explanation: slope of
-<span class="math display">T_{abs}(h)</span> (number of points to the
-left of h - number of points to the right of h) / 7. If the slope of
-<span class="math display">T_{abs}(h)</span> was originally <span class="math display">-\frac17</span>, there must have been four points
-to the right of h and 3 points to the left of h, meaning <span class="math display">y_2&lt;h&lt;y_3</span>. It follows that in the
+<p><span class="math display">-\frac{3}{7}</span></p>
+<p>Slope of <span class="math inline">T_{abs}(h)</span> is equal to
+<span class="math inline">\frac{1}{7} * \text{(number of points to the
+left of h - number of points to the right of h)}</span>. If the slope of
+<span class="math display">T_{abs}(h)</span> was originally <span class="math display">-\frac{1}{7}</span>, there must have been four
+points to the right of h and 3 points to the left of h, meaning <span class="math display">y_2&lt;h&lt;y_3</span>. It follows that in the
 modified dataset there are 5 points to the right of h and 2 points to
-the left of h, meaning the slope of <span class="math display">T_{abs}(h)</span> must be <span class="math display">-\frac37</span></p>
-<p>TODO</p>
+the left of h, meaning the slope of <span class="math display">T_{abs}(h)</span> must be <span class="math display">-\frac{3}{7}</span></p>
 </div>
 </div>
 </div>
 </div>
 <p><br></p>
+<h3 id="problem-2.4">Problem 2.4</h3>
 <p><em>The following information is repeated from the previous page, for
 your convenience.</em></p>
 <p>Consider a dataset of 4 values, <span class="math inline">y_1 &lt;
@@ -512,8 +520,6 @@ <h1 class="title"> </h1>
 is <span class="math inline">-\frac{1}{7}</span> at some <span
 class="math inline">h_p</span>. <em>Hint: think about what values of
 <span class="math inline">h</span> could have this slope.</em></p>
-<p><br></p>
-<h3 id="problem-2.4">Problem 2.4</h3>
 <p>Suppose the dataset is now modified by repeating each value <span
 class="math inline">y_i</span> such that it now contains <span
 class="math inline">x_1, y_1, y_1, y_2, y_2, y_3, y_3, y_4, y_4</span>,
@@ -535,6 +541,8 @@ <h2 class="accordion-header" id="heading2_4">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
+<p>There are two answers based on if you believed <span class="math inline">x_3</span> and <span class="math inline">x_4</span>
+was still in the dataset.</p>
 <p>Correct case 1: assumes that <span class="math display">x_3</span>
 and <span class="math display">x_4</span> are still in the dataset and
 finds the answer to be <span class="math display">-\frac{1}{11}</span></p>
@@ -628,9 +636,12 @@ <h2 class="accordion-header" id="heading3_1">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p>Correctly explains that vectors are linearly independent for any
-<span class="math display">\alpha \neq 1</span>.</p>
-<p>TODO</p>
+<p>The vectors are linearly independent for any <span class="math inline">\alpha \neq 1</span>.</p>
+<p>To be linearly independent it means there is not a linear combination
+between any of the vectors. We can see that if we add <span class="math inline">\vec v_1</span> and <span class="math inline">\vec
+v_2</span> it looks almost like <span class="math inline">\vec
+v_3</span>, so as long as we can make it so $<span class="math inline">\alpha \neq 1</span> then the vectors will be
+independent.</p>
 </div>
 </div>
 </div>
@@ -654,10 +665,15 @@ <h2 class="accordion-header" id="heading3_2">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p>Correctly notes there are no values of <span class="math display">\alpha</span> for which <span class="math display">\vec{v}_1, \vec{v}_3</span> are orthogonal, as the
-first component containing <span class="math display">\alpha</span> is
-multiplied by 0.</p>
-<p>TODO</p>
+<p>We know for <span class="math inline">\vec v_1</span> and <span class="math inline">\vec v_3</span> to be orthogonal their dot product
+should equal zero.</p>
+<p><span class="math display">\vec v_1 \cdot \vec v_3 = (0)(\alpha) +
+(1)(1) + (1)(2) = 0</span></p>
+<p>There are no values of <span class="math inline">\alpha</span> for
+which <span class="math inline">\vec{v}_1, \vec{v}_3</span> are
+orthogonal. We can see <span class="math inline">(0)(\alpha) = 0</span>,
+which means that we cannot manipulate <span class="math inline">\alpha</span> in any way to make the vectors
+orthogonal.</p>
 </div>
 </div>
 </div>
@@ -681,9 +697,21 @@ <h2 class="accordion-header" id="heading3_3">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p>Correctly notes <span class="math display">\alpha = -2</span> for
-this question, as this makes the dot product of <span class="math display">\vec{v_1} \cdot \vec{v_3} = 0</span>.</p>
-<p>TODO</p>
+<p><span class="math inline">\alpha = -2</span></p>
+<p>We know for <span class="math inline">\vec v_2</span> and <span class="math inline">\vec v_3</span> to be orthogonal their dot product
+should equal zero.</p>
+<p>The dot product is:</p>
+<p><span class="math display">
+\vec v_2 \cdot \vec v_3 = (1)(\alpha) + (0)(1) + (1)(2)
+</span></p>
+<p>So we can do:</p>
+<p><span class="math display">\begin{align*}
+0 &amp;= (1)(\alpha) + (0)(1) + (1)(2)\\
+0 &amp;=\alpha + 0 + 2\\
+-2 &amp;= \alpha
+\end{align*}</span></p>
+<p>We can clearly see when <span class="math inline">\alpha = -2</span>
+then the dot product will equal zero.</p>
 </div>
 </div>
 </div>
@@ -716,9 +744,10 @@ <h2 class="accordion-header" id="heading3_4">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p>Correct: answered yes because any vector in <span class="math display">\mathbb{R}^3</span> is in the span of 3 linearly
-independent vectors in <span class="math display">\mathbb{R}^3</span>.</p>
-<p>TODO</p>
+<p>Yes</p>
+<p>If <span class="math inline">\alpha = 3</span> and all three vectors
+are independent from one another then any vector in <span class="math inline">\mathbb{R}^3</span> is in the span of 3 linearly
+independent vectors in <span class="math inline">\mathbb{R}^3</span>.</p>
 </div>
 </div>
 </div>
@@ -742,7 +771,7 @@ <h3 id="problem-3.5">Problem 3.5</h3>
 8
 \end{bmatrix}</span> onto <span class="math inline">\vec{v}_1</span>?
 Provide your answer in the form of a vector. Show your work, and put
-your answer in a .</p>
+your answer in a box.</p>
 <div id="accordionExample" class="accordion">
 <div class="accordion-item">
 <h2 class="accordion-header" id="heading3_5">
@@ -756,15 +785,20 @@ <h2 class="accordion-header" id="heading3_5">
 <header id="title-block-header">
 <h1 class="title"> </h1>
 </header>
-<p>Correctly projects <span class="math inline">\begin{bmatrix}3 \\ 5 \\
-8\end{bmatrix}</span> onto <span class="math display">\vec{v_1}</span>:
-<span class="math display">\frac{\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix}
-\cdot \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}}{\begin{bmatrix}0 \\ 1 \\
-1 \end{bmatrix}\cdot \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}}
-\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} =
-\frac{13}{2}  \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} =
-\begin{bmatrix}0 \\ 6.5 \\ 6.5 \end{bmatrix}</span></p>
-<p>TODO</p>
+<p><span class="math inline">\begin{bmatrix}0 \\ 6.5 \\ 6.5
+\end{bmatrix}</span></p>
+<p>We follow the equation <span class="math inline">\frac{\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix} \cdot
+\vec v_1}{\vec v_1 \cdot \vec v_1} \vec v_1</span> to find the
+projection of <span class="math inline">\begin{bmatrix}3 \\ 5 \\
+8\end{bmatrix}</span> onto <span class="math inline">\vec{v_1}</span>:</p>
+<p><span class="math display">\begin{align*}
+\frac{\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix} \cdot \begin{bmatrix}0 \\
+1 \\ 1 \end{bmatrix}}{\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}\cdot
+\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}} \begin{bmatrix}0 \\ 1 \\ 1
+\end{bmatrix} &amp;= \frac{13}{2}  \begin{bmatrix}0 \\ 1 \\ 1
+\end{bmatrix}\\
+&amp;= \begin{bmatrix}0 \\ 6.5 \\ 6.5 \end{bmatrix}
+\end{align*}</span></p>
 </div>
 </div>
 </div>
@@ -1931,6 +1965,7 @@ <h1 class="title"> </h1>
 <p>And predicts Mature</p>
 <p>If a student does not receive this rubric item, they should only
 receive <strong>one</strong> of the other rubric items.</p>
+<p>$P() = </p>
 <p>TODO</p>
 </div>
 </div>

problems/ss2-24-final/ss2-24-final-q03.md

@@ -17,9 +17,9 @@ a box.
 
 # BEGIN SOLUTION
 
-Correctly explains that vectors are linearly independent for any $$\alpha \neq 1$$.
+The vectors are linearly independent for any $\alpha \neq 1$.
 
-TODO
+To be linearly independent it means there is not a linear combination between any of the vectors. We can see that if we add $\vec v_1$ and $\vec v_2$ it looks almost like $\vec v_3$, so as long as we can make it so $$\alpha \neq 1$ then the vectors will be independent.
 
 # END SOLUTION
 
@@ -32,9 +32,11 @@ orthogonal? Show your work, and put your answer in a box.
 
 # BEGIN SOLUTION
 
-Correctly notes there are no values of $$\alpha$$ for which $$\vec{v}_1, \vec{v}_3$$ are orthogonal, as the first component containing $$\alpha$$ is multiplied by 0.
+We know for $\vec v_1$ and $\vec v_3$ to be orthogonal their dot product should equal zero.
 
-TODO
+$$\vec v_1 \cdot \vec v_3 = (0)(\alpha) + (1)(1) + (1)(2) = 0$$
+
+There are no values of $\alpha$ for which $\vec{v}_1, \vec{v}_3$ are orthogonal. We can see $(0)(\alpha) = 0$, which means that we cannot manipulate $\alpha$ in any way to make the vectors orthogonal.
 
 # END SOLUTION
 
@@ -47,9 +49,25 @@ orthogonal? Show your work, and put your answer in a box.
 
 # BEGIN SOLUTION
 
-Correctly notes $$\alpha = -2$$ for this question, as this makes the dot product of $$\vec{v_1} \cdot \vec{v_3} = 0$$.
+$\alpha = -2$
 
-TODO
+We know for $\vec v_2$ and $\vec v_3$ to be orthogonal their dot product should equal zero.
+
+The dot product is:
+
+$$
+\vec v_2 \cdot \vec v_3 = (1)(\alpha) + (0)(1) + (1)(2)
+$$
+
+So we can do:
+
+\begin{align*}
+0 &= (1)(\alpha) + (0)(1) + (1)(2)\\
+0 &=\alpha + 0 + 2\\
+-2 &= \alpha
+\end{align*}
+
+We can clearly see when $\alpha = -2$ then the dot product will equal zero.
 
 # END SOLUTION
 
@@ -69,9 +87,9 @@ Is the vector $\begin{bmatrix}
 
 # BEGIN SOLUTION
 
-Correct: answered yes because any vector in $$\mathbb{R}^3$$ is in the span of 3 linearly independent vectors in $$\mathbb{R}^3$$.
+Yes
 
-TODO
+If $\alpha = 3$ and all three vectors are independent from one another then any vector in $\mathbb{R}^3$ is in the span of 3 linearly independent vectors in $\mathbb{R}^3$.
 
 # END SOLUTION
 
@@ -94,13 +112,18 @@ What is the projection of the vector $\begin{bmatrix}
 5 \\
 8
 \end{bmatrix}$ onto $\vec{v}_1$? Provide your answer in the form of a
-vector. Show your work, and put your answer in a .
+vector. Show your work, and put your answer in a box.
 
 # BEGIN SOLUTION
 
-Correctly projects $\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix}$ onto $$\vec{v_1}$$: $$\frac{\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix} \cdot \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}}{\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}\cdot \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}} \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} = \frac{13}{2}  \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix}0 \\ 6.5 \\ 6.5 \end{bmatrix}$$
+$\begin{bmatrix}0 \\ 6.5 \\ 6.5 \end{bmatrix}$
 
-TODO
+We follow the equation  $\frac{\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix} \cdot \vec v_1}{\vec v_1 \cdot \vec v_1} \vec v_1$ to find the projection of $\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix}$ onto $\vec{v_1}$:
+
+\begin{align*}
+\frac{\begin{bmatrix}3 \\ 5 \\ 8\end{bmatrix} \cdot \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}}{\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}\cdot \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}} \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} &= \frac{13}{2}  \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}\\
+&= \begin{bmatrix}0 \\ 6.5 \\ 6.5 \end{bmatrix}
+\end{align*}
 
 # END SOLUTION