5. linear forms: a candle was 10 inches tall when lit. it melts at a rate of 1 inch every 3 hours. which form best describes this?
a slope - intercept form because initial value and rate are given.
b standard form
c point - slope form because a data point is provided.
d point - slope because only the slope is given.
6. data strength: a data set shows x - values increasing steadily, but y - values fluctuate randomly between 2 and 12. how strong is the linear relationship?
a. strong
b moderate
c weak
d not linear
7. function transformation: if the original line is f(x), which point is on the line f(x - 2)=4? (assume the graph has a point at (4,4))
a (6,4)
b (2,4)
c (4,6)
d (4,2)
8. line logic: a line has a negative slope and a positive y - intercept. select the point that cannot be on this line
a (-2,10)
b (6,5)
c (5,10)
d (5, - 2)
9. residuals: which residual plot indicates a strong linear model?
a a clear u - shape.
b points fanning out.
c randomly scattered points near the x - axis.

Question

Accepted Answer

### Question 5: Linear Forms # Brief Explanations: The candle's initial height (10 inches) is the y - intercept, and the rate of melting (1 inch every 3 hours) is the slope. The slope - intercept form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. Since we have the initial value ($b$) and the rate ($m$), the slope - intercept form is appropriate. Standard form is $Ax+By = C$, point - slope form is $y - y_1=m(x - x_1)$ (which requires a point $(x_1,y_1)$ and slope, not just initial value and rate). So option A is correct. # Answer: A. Slope - intercept form because initial value and rate are given. ### Question 6: Data Strength # Brief Explanations: For a linear relationship, as $x$ increases, $y$ should show a consistent (increasing or decreasing) pattern. Here, $x$ increases steadily but $y$ fluctuates randomly between 2 and 12. A weak linear relationship occurs when there is a very loose association between $x$ and $y$. Since $y$ is random, the linear relationship is weak. It's not “not linear” because $x$ is increasing steadily, but the $y$ variation is random, so the linear association is weak. # Answer: C. Weak ### Question 7: Function Transformation # Explanation: ## Step 1: Recall function transformation rule For a function $y = f(x - h)$, the graph of $f(x)$ is shifted $h$ units to the right. We know that $f(4)=4$ (since the point $(4,4)$ is on $y = f(x)$). ## Step 2: Find $x$ for $f(x - 2)$ We want to find $x$ such that $x-2 = 4$. Solving for $x$, we add 2 to both sides of the equation $x-2=4$, so $x=4 + 2=6$. When $x = 6$, $f(6 - 2)=f(4)=4$. So the point $(6,4)$ is on the line $f(x - 2)=4$. # Answer: A. $(6,4)$ ### Question 8: Line Logic # Brief Explanations: A line with a negative slope and positive y - intercept has the equation $y=mx + b$ where $m<0$ and $b > 0$. Let's analyze each point: - For point A $(- 2,10)$: If $x=-2$, $y = mx + b$, $10=-2m + b$. Since $m<0$, $-2m>0$, and $b>0$, this is possible. - For point B $(6,5)$: $5 = 6m + b$. Since $m<0$, $6m<0$, and $b>0$, $6m + b$ can be 5 (e.g., $m=-0.5$, $b = 8$, $y=-0.5x + 8$, when $x = 6$, $y=-3 + 8 = 5$). - For point C $(5,10)$: $10=5m + b$. Since $m<0$, $5m<0$, and $b>0$, $5m + b=10$ would require $b=10 - 5m$. But if $m<0$, $10-5m>10$, and $b>0$ but for a negative slope line, as $x$ increases, $y$ should decrease. When $x = 5$, if $x$ is positive and increasing from a smaller positive $x$, $y$ should be less than the y - intercept. If $x = 0$, $y=b>0$, and as $x$ increases (since $m<0$), $y$ decreases. So $y = 10$ when $x = 5$ is not possible. - For point D $(5,-2)$: $ - 2=5m + b$. Since $m<0$, $5m<0$, and $b>0$, it's possible (e.g., $m=-2$, $b = 8$, $y=-2x + 8$, when $x = 5$, $y=-10 + 8=-2$). So the point that cannot be on the line is $(5,10)$. # Answer: C. $(5,10)$ ### Question 9: Residuals # Brief Explanations: A residual plot for a strong linear model should have residuals (the differences between observed and predicted values) that are randomly scattered near the $x$ - axis. A U - shape in residuals indicates a non - linear relationship (the model is not capturing the curvature). Points fanning out indicate that the variance of residuals is not constant (heteroscedasticity), which is not a sign of a strong linear model. So a residual plot with randomly scattered points near the $x$ - axis indicates a strong linear model. # Answer: C. Randomly scattered points near the x - axis.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 5: Linear Forms

Step 1: Recall function transformation rule

Step 2: Find \(x\) for \(f(x - 2)\)

Answer:

Question 6: Data Strength