A college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student is ready for college-level mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 150150 students who completed this core set of courses results in a mean math score of 22.722.7 on the college entrance exam with a standard deviation of 3.93.9. Do these results suggest that students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 2222 on the math portion of the exam? Complete parts a) through d) below.

Yes, Students are scoring above 2222 on the math portion of the exam.

Step-by-step explanation:

We are given that a college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student is ready for college-level mathematics i.e., population mean is 22.

Null Hypothesis, : = 22 {Students are scoring equal to 2222 on the math portion of the exam}

Alternate Hypothesis, : > 22 {Students are scoring above 2222 on the math portion of the exam}

Also, a random sample of 150 students who completed this core set of courses results in a mean math score of 22.7 on the college entrance exam with a standard deviation of 3.93 i.e.;

Sample mean, = 22.7    and Sample standard deviation, s = 3.93

The Test statistics is given by;

                            Z = ~

Test Statistics = ~

                        = 2.1815

Since, we are not given with the significance level, so we assume it to be 5%. At 5% significance level t table gives critical value of 1.6578 at 149 degree of freedom. Since our test statistics is more than the critical value so we reject null hypothesis as our test statistics fall in the rejection region.

Therefore, we conclude that Students are scoring above 2222 on the math portion of the exam.

15 batches are the most you can make

step-by-step explanation:

a.   we have two

lines:   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 lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: 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.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!