Answer:
D 1/25
Step-by-step explanation:
6+8= 14
14/350 = 0.04
1/25= 0.04
Answer:
Step-by-step explanation:
g(x)=3(10-x)^2-8
=3(12)^2-8
=144*3-8
=432-8
=424
Answer:
-2*3
Step-by-step explanation:
If she had to refund $2 for each cup (-2) and she refunded 3 cups (3) she would lose a balance of -2*3.
Answer:A
Step-by-step explanation:
Answer:
The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0
Step-by-step explanation:
Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);


P(Xi≥19.0)=0.473

p=0.473
Yi~Ber(0.473)

Based on the Central Limit Theorem:

Then:


Based on the Central Limit Theorem:


Then:
the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0