Help please help help help
2 answers:
You might be interested in
The correct question is
Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4−x and y = 8-x^-1 intersect are the solutions of the equation 4−x = 8-x^-1<span>.
Part B: Make tables to find the solution to 4−x = </span>8-x^-1<span>. Take the integer values of x between −3 and 3.
Part C: How can you solve the equation 4−x = </span>8-x^-1 graphically?
Part A. We have two equations: y = 4-x and y = 8-x^-1
Given two simultaneous equations that are both to be true, then the solution is the points where the lines cross. The intersection is where the two equations are equal. Therefore the solution that works for both equations is when
4-x = 8-x^-1
This is where the two graphs will cross and that is the common point that satisfies both equations.
Part B
see the attached table
the table shows that one of the solutions is in the interval [-1,1]
Part C To solve graphically the equation 4-x = 8-x^-1
Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4−x and y = 8-x^-1 intersect are the solutions of the equation 4−x = 8-x^-1<span>.
Part B: Make tables to find the solution to 4−x = </span>8-x^-1<span>. Take the integer values of x between −3 and 3.
Part C: How can you solve the equation 4−x = </span>8-x^-1 graphically?
Part A. We have two equations: y = 4-x and y = 8-x^-1
Given two simultaneous equations that are both to be true, then the solution is the points where the lines cross. The intersection is where the two equations are equal. Therefore the solution that works for both equations is when
4-x = 8-x^-1
This is where the two graphs will cross and that is the common point that satisfies both equations.
Part B
see the attached table
the table shows that one of the solutions is in the interval [-1,1]
Part C To solve graphically the equation 4-x = 8-x^-1
We would graph both equations: y = 4-x and y = 8-x^-1
The point on the graph where the lines cross is the solution to the system of equations.
using a graph tool
see the attached figure N 2
the solutions are the points
(-4.24,8.24)
(0.24,3.76)
The probability of making a type II error is given by one minus the power of the hypothesis test.
In general for the alternative hypothesis ,
The power of a hypothesis test is given by:
The probability of <span>making a type ii error when μ = 31,000 is given by
</span>
In general for the alternative hypothesis ,
The power of a hypothesis test is given by:
The probability of <span>making a type ii error when μ = 31,000 is given by