Please Help Quick ASAP Hurry This is Pre-Algebra 1
2 answers:
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Answer:
25.5 days
Step-by-step explanation:
Mean number of days (μ) = 22 days
Standard deviation (σ) = 6 days
Z-score for the 72nd percentile (according to tabulated values) = 0.583
The z-score for any number of days, X, is determined by:
The value of X that is greater than 72% of the trial times is:
Therefore, 72% of all of these types of trials are completed within 25.5 days.
Answer:
(0,-4)
(3,0)
Step-by-step explanation:
Let start at the orgin.
This is a linear equation since the equation is in the form of
where m is the slope and b is the y intercept.
Since we starting at the orgin, and b is our y intercept.
Our first point is
(0,-4).
since the slope is 4/3.
We would rise 4 from the y value and run 3 to the x value.
In other words, to find your second point, go up 4 units from the first point and move to the right 3 units.
So our next point is at
(3,0).
U can continously go up 4 units and move 3 units to the right to find other points.
1) Mean: 6.5
2) Standard deviation: 3.2
Step-by-step explanation:
1)
A uniform random distribution is the distribution of a variable that can assume n different values, and each of these values is equally likely to occur.
This means that the probability of occurring of each of these values is the same:
For a uniform random distribution, the mean of the distribution is simply equal to the median value of the distribution, which is exactly at the center of the distribution itself. It can be calculated as
where
a is the lowest value of the distribution
b is the highest value of the distribution
Here the distribution consists of all values ranging from 1 to 12, so:
a = 1
b = 12
Therefore the mean value is
2)
For a uniform random distribution, it can be dimostrated that the variance of the distribution is
For the distribution in this problem, we have:
a = 1
b = 12
Therefore the variance is
The standard deviation is the square root of the variance, therefore for this distribution it is:
Learn more about mean and standard deviation:
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