Thursday, May 23, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


22. Consider f(x) = 4x2 + 6x - 3, and p(x) defined by the graph below

The difference between the values of the maximum of p and minimum of f is

(1) 0.25
(2) 1.25
(3) 3.25
(4) 10.25

Answer: (4) 10.25
The minimum point for f(x) occurs on the axis of symmetry, which is x = -b/(2a)
x = -(6)/((2)(4)) = -6/8 = -.75
The maximum of f(x) is f(-.75) = 4(-.75)^2 + 6(-.75) - 3 = -5.25
The maximum of p(x) is 5.
The difference is 5 - 5.25 = 10.25

If you had graph f(x), the minimum point wouldn't be in the table of values, but you could use the functions to find it. However, one you see that the minimum is below zero, there is only one possible answer because the other three are too small.





23. The scores on a mathematics college-entry exam are normally distributed with a mean of 68 and standard deviation 7.2. Students scoring higher than one standard deviation above the mean will not be enrolled in the mathematics tutoring program. How many of the 750 incoming students can be expected to be enrolled in the tutoring program?

(1) 631
(2) 512
(3) 238
(4) 119

Answer: (1) 631
68.27% percent of the data is within one standard deviation of the mean. That means 34.135% score within one standard deviation above as well as below. Since 50 + 34 = 84, 84 percent of the incoming students will be enrolled in the tutoring program.
.84135 * 750 = 631.
Depending upon the number of decimals you used, you may have gotten 630, which makes (1) the best choice.





24. How many solutions exist for 1 / (1 - x2) = -|3x - 2| + 5?

(1) 1
(2) 2
(3) 3
(4) 4

Answer: (4) 4
Fastest solution is to graph both the left and right side of the equation and check for the number of intersections.

End of Part I.



Comments and questions welcome.

More Algebra 2 problems.

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