Home / Digital SAT Math Practice Questions – Advanced : Two-variable data: models and scatterplots

Digital SAT Math Practice Questions – Advanced : Two-variable data: models and scatterplots

SAT MAth Practice questions – all topics

  • Problem-solving and Data Analysis Weightage: 15%  Questions: 5-7
    • Ratios, rates, proportional relationships, and units
    • Percentages
    • One-variable data: distributions and measures of centre and spread
    • Two-variable data: models and scatterplots
    • Probability and conditional probability
    • Inference from sample statistics and margin of error
    • Evaluating statistical claims: observational studies and Experiments

SAT MAth and English  – full syllabus practice tests

  Question  Hard


The scatterplot shows the relationship between two variables, \(x\) and \(y\). An equation for the exponential model shown can be written as
$
y = a(b)^x,
$
where \(a\) and \(b\) are positive constants. Which of the following is closest to the value of \(b\)?
A. 0.83
B. 1.83
C. 18.36
D. 126.35

▶️Answer/Explanation

Ans: A

  Question  Hard


The scatterplot shows the relationship between two variables, \(x\) and \(y\), for data set E. A line of best fit is shown.
Data set F is created by multiplying the \(y\)-coordinate of each data point from data set E by \(\mathbf{3.9}\).
Which of the following could be an equation of a line of best fit for data set F?
A. \(y = 46.8 + 5.9x\)
B. \(y = 46.8 + 1.5x\)
C. \(y = 12 + 5.9x\)
D. \(y = 12 + 1.5x\)

▶️Answer/Explanation

Ans: A

  Question  Hard

The scatterplot shows the relationship between the length of time y, in hours, a certain bird spent in flight and the number of days after January 11, x.

What is the average rate of change, in hours per day, of the length of time the bird spent in flight on January 13 to the length of time the bird spent in flight on January 15?

▶️Answer/Explanation

Ans: 4.5,9/2
January 13 is 2 days after January 11 → \(x = 2\)
From graph, \(y = 6\) hours
January 15 is 4 days after January 11 → \(x = 4\)
From graph, \(y = 15\) hours
Use the average rate of change formula.
$
\text{Rate} = \frac{\text{Change in } y}{\text{Change in } x} = \frac{15 – 6}{4 – 2} = \frac{9}{2} = \boxed{4.5 \text{ hours/day}}
$

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