Answer:
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
Step-by-step explanation:
Problems of normally distributed samples are 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:
The percentage of cockroaches weighing between 77 grams and 83 grams
This is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 83. So
X = 83
has a pvalue of 0.7734
X = 77
has a pvalue of 0.2266
0.7734 - 0.2266 = 0.5468
Rounded to the nearest whole number, 55%
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
Answer:
If my memory for this is correct, the answer should be:
1. 717.60
2. 9311.20
3. 1067.40
4. 380.60
Step-by-step explanation:
Answer:
- 280 student tickets
- 520 adult tickets
Step-by-step explanation:
You may recognize that you are given two relationships between two unknowns. You can write equations for that.
You are asked for numbers of adult tickets and of student tickets. It often works well to let the values you're asked for be represented by variables. We can choose "a" for the number of adult tickets, and "s" for the number of student tickets. Then the problem statement tells us the relationships ...
a + s = 800 . . . . . . 800 tickets were sold
12.50a + 7.50s = 8600 . . . . . . . revenue from sales was 8600
(You are supposed to know that the revenue from selling "a" adult tickets is found by multiplying the ticket price by the number of tickets: 12.50a.)
___
You can solve these two equations any number of ways. One way is to do it by <em>elimination</em>. We can multiply the first equation by 12.50 and subtract the second equation:
12.50(a +s) -(12.50a +7.50s) = 12.50(800) -(8600)
5s = 1400 . . . . simplify. (The "a" variable has been eliminated.)
s = 280 . . . . . . divide by 5
Then the number of adult tickets can be found from the first equation:
a + 280 = 800
a = 520
280 student tickets and 520 adult tickets were sold.
