Answer
a) y | p(y)
25 | 0.8
100 | 0.15
300 | 0.05
E(y) = ∑ y . p(y)
E(y) = 25 × 0.8 + 100 × 0.15 + 300 × 0.05
E(y) = 50
average class size equal to E(y) = 50
b) y | p(y)
25 | 
100 | 
300 | 
E(y) = ∑ y . p(y)
E(y) = 25 × 0.4 + 100 × 0.3 + 300 × 0.3
E(y) = 130
average class size equal to E(y) = 130
c) Average Student in the class in a school = 50
Average student at the school has student = 130
Answer:
56
Step-by-step explanation:
10y+6
10(5)+6
10*5=50
50+6=56
Answer:
33/4
Step-by-step explanation:
Answer:
1.6962 = 1.70 (2 dp)
0.4247 = 0.425 to 3 dp
0.007395 = 0.007 to 3 dp
0.007395 = 0.0074 to 4 dp
32549 = 32500 to 3 significant figures
32549 = 32550 to 4 significant figures
909520 = 910000 to 3 significant figures
909520 = 909500 to 4 significant figures
Step-by-step explanation:
1.6962 = 1.70 (2 dp)
0.4247 = 0.425 to 3 dp
0.007395 = 0.007 to 3 dp
0.007395 = 0.0074 to 4 dp
32549 = 32500 to 3 significant figures
32549 = 32550 to 4 significant figures
909520 = 910000 to 3 significant figures
909520 = 909500 to 4 significant figures
The linear function is: g = 3n-2. Plug in values from -2 to 4 into n. Like so, G = 3(-2)-2 = g = -8. So, n(x) = -2 and g(y) = -8. And so on, and so on.
Basically, your n values are the x values.
The number you get out of the equation will be your y value.
