## Monday, August 26, 2019

### August 2019 Algebra 1 Regents Parts 3 & 4

The following are some of the multiple questions from the August 2019 New York State Common Core Algebra I Regents exam.

### August 2019 Algebra I, Part III

Each correct answer is worth up to 4 credits. Partial credit can be given. Work must be shown or explained.

33. On the set of axes below, graph the following system of inequalities:

2x + y > 8
y - 5 < 3x

Determine if the point (1,8) is in the solution set. Explain your answer.

Rewrite the equations in slope-intercept form

y > -2x + 8
y - 5 < 3x + 5

Note that the first is greater than or equal to, so it will be a solid line that is shaded above.
The second is less than, so it will be a broken or dashed line that is shaded below.
If you are not sure which side to shade, substitute the point (0, 0) in the inequality. If you get a true statement as a result, then shade that side. If the statement is false, shade the other side.
IMPORTANT: this test doesn't work if the line goes through (0, 0). In that case, pick any other point not on the line. It will work for any point, but (0, 0) is the easiest one to use!

The first line (solid) has a slope of -2 and a y-intercept of 8.
The second line (broken) has a slope of 3 and a y-intercept of 5.

The point (1, 8) is NOT part of the Solution Set because it is on the broken line, which is a boundary, but not a part of the solution.
Important: if you have a graphing error, then your response to this part of the question should match your graph to get the point.

34. On the day Alexander was born, his father invested \$5000 in an account with a 1.2% annual growth rate. Write a function, A(t), that represents the value of this investment t years after Alexander’s birth.

Determine, to the nearest dollar, how much more the investment will be worth when Alexander turns 32 than when he turns 17.

For the first part, you have to write a formula.
For the second part, you have to use the formula twice for when he is 32 and 17, and subtract the difference.

A(t) = 5000 (1.012)t

A(32) - A(17) = 5000 (1.012)32 - 5000 (1.012)17 = 1199.91758695
\$1200.

You can put that entire equation in your calculator. You do not need to evaluate the parts separately.

35. Stephen collected data from a travel website. The data included a hotel’s distance from Times Square in Manhattan and the cost of a room for one weekend night in August. A table containing these data appears below.

 Distance From Times Square (city blocks) (x) 0 0 1 1 3 4 7 11 14 19 Cost of a Room (dollars) (y) 293 263 244 224 185 170 219 153 136 111

Write the linear regression equation for this data set. Round all values to the nearest hundredth.

State the correlation coefficient for this data set, to the nearest hundredth.

Enter the data into two lists, L1 and L2.
Make sure DiagnosticOn has been set. (If you don't know, just do select it from the Menu.)
Press STAT to get to the statistics menu and arrow over to CALC. Choose option 4, LinReg(ax+b).
You will get a = -7.76, b = 246.34 and r = -.88, rounded to the nearest hundredth.

The linear regression is y = -7.76x + 246.34

The correlation coefficient is r = -0.88

In the context of this problem, the correlation coefficient suggest a strong negative correlation between the number of blocks from Times Square and the cost of a room. As you get farther away from Times Square, the cost of the room decreases.
(It's not enough to just say that it's a strong negative correlation.)

Coincidentally, this question is very similar to Question 35 of the June 2019 Regents. I barely had to do any editing to my solution (which, yes, I cut and pasted from the old post because all the formatting is in place).

36. A snowstorm started at midnight. For the first 4 hours, it snowed at an average rate of one-half inch per hour.
The snow then started to fall at an average rate of one inch per hour for the next 6 hours.
Then it stopped snowing for 3 hours.
Then it started snowing again at an average rate of one-half inch per hour for the next 4 hours until the storm was over.
On the set of axes below, graph the amount of snow accumulated over the time interval of the storm.

Determine the average rate of snowfall over the length of the storm. State the rate, to the nearest hundredth of an inch per hour.

Notice that the scale for the x-axis is 1 box = 1 hour, but the scale for the y-axis is 1 box = 1/2 inch of snow. Don't let that confuse you.
You can plot points and draw line segments just by counting the boxes. The thing to remember is if no snow falls, the accumulation doesn't change. Slope is 0 (flat line) for that time period.
To find the average, what would the slope be between the origin (0,0) and the highest point on the graph at the end of the storm, which should be (17, 10)?
(10 - 0) / (17 - 0) = 10 / 17 = 0.5882... = 0.59 inches per hour (to the nearest hundredth).

Again, if you made a graphing error, use whatever point you finished at.
For example, if you missed a point when it stopped for 3 hours (suppose you graphed 2), and you ended up at (16, 10), you would have done 10/16 = 0.625, which is 0.63. You get credit if you are consistent.

### August 2018 Algebra I, Part IV

A correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.

37. Allysa spent \$35 to purchase 12 chickens. She bought two different types of chickens. Americana chickens cost \$3.75 each and Delaware chickens cost \$2.50 each.

Write a system of equations that can be used to determine the number of Americana chickens, A, and the number of Delaware chickens, D, she purchased.

Determine algebraically how many of each type of chicken Allysa purchased.

Each Americana chicken lays 2 eggs per day and each Delaware chicken lays 1 egg per day. Allysa only sells eggs by the full dozen for \$2.50. Determine how much money she expects to take in at the end of the first week with her 12 chickens.

A + D = 12
3.75A + 2.50D = 35

You can solve by Sustitution (Substitute 12 - A for D) or Elimination (multiply the first equation by -2.50).

A + D = 12
3.75A + 2.50D = 35
3.75A + 2.50(12 - A) = 35
3.75A + 30 - 2.50 = 35
1.25A + 30 = 35
1.25A = 5
A = 4
4 + D = 12
D = 8
4 American and 8 Delaware

Or

A + D = 12
3.75A + 2.50D = 35
-2.50A - 2.50D = 30
1.25A + 30 = 35
1.25A = 5
A = 4
4 + D = 12
D = 8
4 American and 8 Delaware

The number of eggs laid per day are 4(2) + 8(1) = 16.
After one week, she will have 16 * 7 = 112 eggs
This makes 112 / 12 = 9.333.... dozen.
She can sell 9 dozen for \$2.50 each, which is 9 * 2.50 = \$22.50.

I would hate to have to follow through the work, checking for consistency, if the number of chickens was incorrect!

End of Exam

How did you do?