The total area under the curve of a Normal Distribution in Statistics is equal to 1, representing the entirety of the probability for all possible outcomes.
In the field of Statistics, when working with a Normal Distribution, the total area under the curve is equal to 1. This concept is crucially important as it represents the entirety of the probability for all possible outcomes, which in a Normal Distribution should amount to 100% or '1' when expressed as a decimal. The values on the x-axis represent the outcomes and the area under the curve for a given range represents the probability of outcomes in that range.
The curve of a Normal Distribution is symmetric, meaning half of the total area (0.5) is to the left of the mean and the other half (0.5) is to the right of the mean.
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m = 0.7
m = 1.5
m = 3
m = 1.5 is the solution to the equation.
An equation is a mathematical statement that is made up of two expressions connected by an equal sign. For example, 3x – 5 = 16 is an equation. Solving this equation, we get the value of the variable x as x = 7.
1.6m-4.8=-1.6m
Subtract 1.6m to both sides to get
-4.8=-3.2m
Divide both sides by -3.2to get
x = 1.5
m = 1.5 is the solution to the equation.
The solution of an equation is the set of all values that, when substituted for unknowns, make an equation true. For equations having one unknown, raised to a single power, two fundamental rules of algebra, including the additive property and the multiplicative property, are used to determine its solutions.
To learn more about An equation, refer
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Answer:15 seniors served on the student council during their freshman year, 14 seniors served during their sophomore year, 16 seniors served during their junior year, and 3 seniors have never served before.
Step-by-step explanation:
Using inclusion and exclusion principles, we find that 2 seniors served on the student council during each of the four years in high school.
The problem can be solved using the Principle of Inclusion and Exclusion (PIE), a common technique in combinatorial mathematics. First, we add the number of seniors serving in their freshman, sophomore, and junior years: 3 (never served) + 10 (junior) + 9 (sophomore) + 11 (freshman) giving us 33.
Then, we subtract the number of seniors who served during both sophomore and junior years, freshman and junior years, and freshman and sophomore years: 33 - 5 (sophomore and junior) - 6 (freshman and junior) - 4 (freshman and sophomore). This results in 18.
However, from the initial condition we know that there are 20 seniors in total. Therefore, the two 'extra' seniors must have served all four years in high school. Thus we find that 2 seniors served on the student council during each of the four years in high school.
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