Svetlana Popov earned the following grades last semester. Find her grade point average to the nearest tenth. Assume A = 4, B = 3, C = 2, D = 1, and F = 0.
Grade × Credits | Grade | Credits | Course |
3 × 4 = 12 | A (= 4) | 3 | Business Mathematics |
4 × 2 = 8 | C (= 2) | 4 | Retailing |
3 × 3 = 9 | B (= 3) | 3 | English |
2 × 4 = 8 | A (= 4) | 2 | Biology |
2 × 1 = \underline{\mathbf{\ \ 2}} | D (= 1) | \underline{\ 2} | Biology Lab |
39 | 14 | Totals |
Her grade point average is \frac{\mathbf{39}}{\mathbf{14}} = 2.79 = 2.8, which is a B–. Anything between 2.5 and 3.0 is a B– at her school.
This problem is solved using a scientific calculator as follows.
Note: Refer to Appendix B for calculator basics.