Professor Cramer determines a final grade based on attendance, two papers, three major tests, and a final exam. Each of these activities has a total of 100 possible points. However, the activities carry different weights. Attendance is worth 3%, each paper is worth 9%, each test is worth 17%, and the final is worth 28%. (a) What is the average for a student with 81 on attendance, 90 on the first paper, 92 on the second paper, 86 on test 1, 71 on test 2, 93 on test 3, and 62 on the final exam

78.67

Step-by-step explanation:

The computation of the average for a student is shown below:

= Different Weights × different activities

= (3% × 81) + (9% × 90) + (9% × 92) + (17% × 86) + (17% × 71) + (17% × 93) + (28% × 62)

= 2.43 + 8.1 + 8.28 + 14.62 + 12.07 + 15.81 + 17.36

= 78.67

Hence, the average of the student is 78.67

We simply applied the above formula

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}