Sudent A :
Average reading `=(3.01+2.99)/(2)=3.0g`
Student B:
Average reading `=(3.05+2.95)/(2)=3.0g`
For both the students A and B, average reading is close to the correct reading (i.e., 3.0g). Hence, both recorded accurate reading. But the reading recorded by student A are more precise as they differ only by`pm0.01`, whereas reading recorded by the student B are differ by `pm 0.05`. Thus, the result of student A are both precise and accurate.