5 divided by 1/2 equal to 10 becuase you need to find how many 1/2 rope can you get from 5
Let C equal bags of cookies
Let M equal money earned
2.75c=m
Probably not enough sorry if it isn't
Answer:
Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Step-by-step explanation:
We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.
So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;
Z =
~ N(0,1)
where,
= average age of the random sample of horses with colic = 12 yrs
= average age of all horses seen at the veterinary clinic = 10 yrs
= standard deviation of all horses coming to the veterinary clinic = 8 yrs
n = sample of horses = 60
So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(
12)
P(
12) = P(
) = P(Z
1.94) = 1 - P(Z < 1.94)
= 1 - 0.97381 = 0.0262
Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Answer:
The answer is "Option A and Option B".
Step-by-step explanation:
In question 1:
In all cases, the entire population is measured so that the actual medium discrepancy could be measured as well as an interval of trust cannot be used.
This issue would be that she calculated the ages with all representatives of both classes, such that she measured a whole population. It's not necessary.
In question 2:
When the p-value is 0.042. At 90% trust and 92% trust level 11 (p-value below 0.10 and 0.08) are not included. however the biggest confidence level of 92%. Consequently, the largest trust level where the 11 is Not included in the trust interval is 92% trust.