This exam was adminstered in June 2022. These answers were not posted until they were unlocked on the NY Regents website or were posted elsewhere on the web.
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Algebra 2 June 2022
Part III: Each correct answer will receive 4 credits. Partial credit can be earned. One computational mistake will lose 1 point. A conceptual error will generally lose 2 points (unless the rubric states otherwise). It is sometimes possible to get 1 point for a correct answer with no correct work shown.
State the number of solutions to the equation f(x) = g(x).
33. On the set of axes below, graph y = f(x) and y = g(x) for the given functions.
g(x) = 2x - 5 
Answer:
Use your calculator and the table of values for the cubic function. You should be able to graph the linear function from the slope and y-intercept.
YOU MUST GRAPH THE FUNCTIONS. NOT SOMETHING THAT ONLY **LOOKS LIKE** THE FUNCTIONS YOU SAW ON YOUR CALCULATOR.
You'd be surprised (or maybe you wouldn't) how lazy some students were when it came to graphing this problem. The most charitable thing I could say was that they were "careless" about it.
There is no reason to have a mistake in g(x). It's a simple linear function, starting at (0,-5) with a slope of 2 (up 2, over 1). And down 2, to the left 1.
For the cubic function, the five points on the graph below should be on your graph. And the extremes, just have lines with arrows. Do NOT make them vertical.
There are 3 solutions to the equation f(x) = g(x) because there are three places where the graphs intersect. You didn't have to explain, you just had to say "three". If your graph was incorrect and you only had two, you could respond "two" and get credit for that -- you would've lost the point on your graph.
According to the rubric, if you said "3" and your graph didn't show three, you still got a point because you might have solved it algebraically. Or it might've been a wild guess.
Determine, to the nearest tenth of a second, the time it takes this pendulum to complete one
swing.
Another Foucault pendulum at the United Nations building takes 9.6 seconds to complete one
swing. Determine, to the nearest tenth of a meter, the length of this pendulum.
34. A Foucault pendulum can be used to demonstrate that the Earth rotates. The time, t, in seconds,
that it takes for one swing or period of the pendulum can be modeled by the equation t = 2π √(L/g)
where L is the length of the pendulum in meters and g is a constant of 9.81 m/s2.
The first Foucault pendulum was constructed in 1851 and has a pendulum length of 67 m.
Answer:
First of all, the year is not needed for any calculations. (Yes, that needs to be said.)
The first part is straight calculation.
t = 2π √(L/g) = 2π √(67/9.81) = 16.420363... = 16.4
This was worth one point. If you rounded incorrectly, you lost that point.
For part two, you had to work backward to solve:
t = 2π √(L/g)
9.6 = 2π √(L/9.81)
9.6 / (2π) = √(L/9.81)
(9.6 / (2π))2 = L/9.81
(9.6 / (2π))2 * 9.81 = L
L = 22.900... = 22.9
This portion was worth three credits. One computational or rounding error cost 1 point. One conceptual error costs 2 points. Both would lose all three credits.
The biggest mistakes I saw had to do with 9.6/2π instead of 9.6/(2π). This was a calculator issue, so it should have been scored as a computational error, especially if the equation was written on the paper. Many students rounded their numbers in the middle of the solution. DON'T DO THAT.
A conceptual error would be, for example, squaring the equation after the second line above, but not squaring the 2π to 4π2.
35. In order to decrease the percentage of its residents who drive to work, a large city launches a campaign to encourage people to use public transportation instead. Before starting the campaign, the city's Department of Transportation uses census data to estimate that 65% of its residents drive to work. The Department of Transportation conducts a simulation, shown below, run 400 times based on this estimate. Each dot represents the proportion of 200 randomly selected residents who drive to work.
Use the simulation results to construct a plausible interval containing the middle 95% of the data. Round your answer to the nearest hundredth.
One year after launching the campaign, the Department of Transportation conducts a survey
of 200 randomly selected city residents and fmds that 122 of them drive to work. Should the
department conclude that the city's campaign was effective? Use statistical evidence from the
simulation to explain your answer.
Answer:
95% of the data falls within 2 standard deviations from the mean. The Mean is 0.651 and the Std Deviation is 0.034.
0.651 - 2 * 0.034 = 0.583, or about 0.58
0.651 + 2 * 0.034 = 0.719 or about 0.72
The interval would be from 0.58 to 0.72.
For the second part, note that 122/200 = 0.61. Since that number is in the 95% confidence interval, then the decrease was not statistically significant. So the campaign was not effective.
36. Solve the system of equations algebraically:
y + 5 = 2x
Answer:
Student 2x - 5 for y in the first equation (the circle) and solve for x.
x2 + y2 = 25
x2 + (2x - 5)2 = 25
x2 + 4x2 - 20x + 25 = 25
5x2 - 20x = 0
5x(x - 4) = 0
5x = 0 or x - 4 = 0
x = 0 or x = 4
y + 5 = 2(0)
y = -5
(0, -5)
y + 5 = 2(4)
y = 3
(3, 4)
End of Part III
How did you do?
More to come. Comments and questions welcome.
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