The regular admission price for an amusement park is $72. For a group of 15, or more, the admission price is reduce by $25 per p
1 answer:
<h3>Answer: 20 people</h3>
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Explanation:
x = number of people
Without the discount, each person pays $72, so the total cost is 72x dollars if x people go through the gate individually.
Now applying the group discount, those x people pay 72-25 = 47 dollars each. Meaning the total cost is now 47x. Keep in mind that in order for the discount to kick in.
Subtracting the two amounts leads to 72x-47x = 25x. It is not a coincidence that we get a coefficient of 25 here.
Set this equal to 500 and solve for x
25x = 500
x = 500/25
x = 20
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If x = 20 people attend individually, then they pay 72*x = 72*20 = 1440 dollars.
If 20 people go as one group, they pay 47x = 47*20 = 940 dollars
We have a savings of 1440-940 = 500 dollars. This confirms the answer.
All 20 people must go as one group if they want to save $500.
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Answer:
I got which is 46.76537.... in decimal form.
Step-by-step explanation:
Answer:
a) The expected value of the sample mean weight is 20.4 pounds.
b)The standard deviation of the sample mean weight is 0.123.
c) There is a 14.46% probability the sample mean weight will be less than 20.27.
d) This value is .
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that
Consider the sample mean weight of 100 watermelons of this variety. This means that .
a. What is the expected value of the sample mean weight? Give an exact answer.
By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.
b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.
By the Central Limit Theorem, that is:
The standard deviation of the sample mean weight is 0.123.
c. What is the approximate probability the sample mean weight will be less than 20.27?
This is the pvalue of Z when .
Since we are working with the sample mean, we use instead of
in the Z score formula
has a pvalue of 0.1446.
This means that there is a 14.46% probability the sample mean weight will be less than 20.27.
d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?
This is the value of that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.
So we use
This value is .