Em, um experimento uma barra de ferro, quando submetida a um processo de aqueciamento, revelos alteracbes em suas dimensbes e caracteristicas fisicas. Rase fenbmeno de expanslo trmica, observado durante o equecimento da barra. proporcionos ams sire de mudanças mensuriveis nas suas propriedodes estruturais. As informsçles coletadas encontramse no quadro que segue. Temperatura ( .( )^circ mathrm(C))(mathrm(x)) & 10 & 15 & 20 & 25 & 30 Comprimento(mm) (y) & 100 & 100 & 101 & 1011 & 1014 Fonte: elaborado pela sutore. Qual 6 o coeficiente angular da reta de regresilo? Dados: [ b=((x^2-x^2-x^2.)/(+x^2)-x^(2) ] Asainale a alternativa correta.

To solve this problem, we need to calculate the angular coefficient (slope) of the regression line based on the given data. The slope formula for a linear regression is derived from the least squares method and can be expressed as:<br /><br />\[<br />b = \frac{n \sum(xy) - \sum x \sum y}{n \sum(x^2) - (\sum x)^2}<br />\]<br /><br />Where:<br />- \( n \) is the number of data points,<br />- \( x \) represents the independent variable (temperature in this case),<br />- \( y \) represents the dependent variable (length in this case),<br />- \( \sum xy \) is the sum of the product of corresponding \( x \) and \( y \) values,<br />- \( \sum x \) is the sum of all \( x \) values,<br />- \( \sum y \) is the sum of all \( y \) values,<br />- \( \sum x^2 \) is the sum of the squares of \( x \) values.<br /><br />## Step 1: Extract the data<br />From the table provided:<br />- \( x \) (Temperature in \( ^{\circ}C \)): \( 10, 15, 20, 25, 30 \)<br />- \( y \) (Length in mm): \( 100, 100, 101, 1011, 1014 \)<br /><br />## Step 2: Calculate the necessary sums<br />We calculate the following:<br />1. \( \sum x = 10 + 15 + 20 + 25 + 30 = 100 \)<br />2. \( \sum y = 100 + 100 + 101 + 1011 + 1014 = 2326 \)<br />3. \( \sum x^2 = 10^2 + 15^2 + 20^2 + 25^2 + 30^2 = 100 + 225 + 400 + 625 + 900 = 2250 \)<br />4. \( \sum xy = (10 \cdot 100) + (15 \cdot 100) + (20 \cdot 101) + (25 \cdot 1011) + (30 \cdot 1014) \)<br /> \[<br /> \sum xy = 1000 + 1500 + 2020 + 25275 + 30420 = 59215<br /> \]<br />5. \( n = 5 \) (number of data points).<br /><br />## Step 3: Apply the slope formula<br />Substitute the calculated values into the formula for \( b \):<br />\[<br />b = \frac{n \sum(xy) - \sum x \sum y}{n \sum(x^2) - (\sum x)^2}<br />\]<br />Substituting:<br />\[<br />b = \frac{5(59215) - (100)(2326)}{5(2250) - (100)^2}<br />\]<br />Simplify the numerator:<br />\[<br />5(59215) = 296075, \quad (100)(2326) = 232600<br />\]<br />\[<br />\text{Numerator} = 296075 - 232600 = 63475<br />\]<br />Simplify the denominator:<br />\[<br />5(2250) = 11250, \quad (100)^2 = 10000<br />\]<br />\[<br />\text{Denominator} = 11250 - 10000 = 1250<br />\]<br />Finally, calculate \( b \):<br />\[<br />b = \frac{63475}{1250} = 50.78<br />\]<br /><br />## Step 4: Identify the correct alternative<br />The angular coefficient of the regression line is approximately \( 50.78 \).