Answer:
the probability that a sample of the 35 exams will have a mean score of 518 or more is <em> 0.934 </em>or<em> 93.4%</em>.
Step-by-step explanation:
This is s z-test because we have been given a sample that is large (greater than 30) and also a standard deviation. The z-test compares sample results and normal distributions. Therefore, the z-statistic is:
(520 - 518) / (180/√35)
= 0.0657
Therefore, the probability is:
P(X ≥ 0.0657) = 1 - P(X < 0.0657)
where
- X is the value to be standardised
Thus,
P(X ≥ 0.0657) = 1 - (520 - 518) / (180/√35)
= 1 - 0.0657
= 0.934
Therefore, the probability that a sample of the 35 exams will have a mean score of 518 or more is <em>0.934 or 93.4%</em>.
Answer: 6
Step-by-step explanation:
Given Data:
Number of people = 6
Amount of Gold to be shared = 40lbs
Subtract 2/3 from your answer.
Therefore
If six people are to share 40lbs of gold evenly
= 40lbs / 6
Divide through by 2
= 20lbs / 3
= 20lbs / 3 - 2/3
= 18/3
= 6
Each person would get approximately 6lbs of gold
Answer:
A. Initially, there were 12 deer.
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. After 15 years, there will be 410 deer.
D. The deer population incresed by 30 specimens.
Step-by-step explanation:

The amount of deer that were initally in the reserve corresponds to the value of N when t=0


A. Initially, there were 12 deer.
B. 
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. 
C. After 15 years, there will be 410 deer.
D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:
ΔN=N(15)-N(10)
ΔN=410 deer - 380 deer
ΔN= 30 deer.
D. The deer population incresed by 30 specimens.