Create a system of equations that includes one linear equation and one quadratic equation. part 1. show all work to solving your system of equations algebraically. part 2. graph your system of equations, and show the solution graphically to verify your solution.

When asked to create a system like this, the easiest way to do so is by setting up real basic equations. For the basis of this, let's use the following:

f(x) = x^2 (Quadratic)

f(x) = x (Linear)

Making them basic like this will allow us to solve rather easily. To solve a system where both are equal to f(x), you may set equal to each other to solve.

x = x^2 >subtract x from both sides

x^2 - x = 0 ---> factor out an x.

x(x - 1) = 0

Now set each part equal to 0 to solve.

x = 0

x - 1 = 0

x = 1

Now you know that the graph has solution sets at x = 0 and x = 1. If we put them back into either equation, you'll get out the ordered pairs of (0, 0) and (1, 1). When graphing these two, it'll be easy to see those two points.

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step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.