Let x be the page number then x+1 would be the other page number (because the pages are two facing pages which means they follow on consecutively)
Then
(x)(x+1) = 156
x^2 + x - 156 = 0
x = 12 or x = -13( but x can't be a negative number)
so x = 12
and the next page is 13
The difference between the mean of the sample and the mean of the population is A. 0.55
<h3>Calculations and Parameters:</h3>
Given the population data is:
0 3 3 1 3 2 2 2 5 4 3 3 3 0 1 2 0 2 3 3
The sample data is:
5 3 0 2 4
The mean of the population data is:
45/20= 2.25
The mean of the sample data is:
14/5= 2.8
Therefore, the difference between the mean of the sample and the population is:
2.8-2.25= 0.55
Read more about mean here:
brainly.com/question/20118982
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Answer:
21
Step-by-step explanation:
49 - 4 x 49 / 7 =
49 - 196 / 7 =
49 - 28 =
21
<em>Hope that helps!</em>
Answer:
A gestation length of 279 days represents the 18th percentile.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions 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 a certain breed of cattle, the length of gestation has a mean of 284 days and a standard deviation is 5.5 days.
This means that 
What length of gestation, rounded to the nearest whole number, represents the 18th percentile?
This is X when Z has a pvalue of 0.18. So X when Z = -0.915.




A gestation length of 279 days represents the 18th percentile.