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avanturin [10]
3 years ago
10

A college entrance exam company determined that a score of 2222 on the mathematics portion of the exam suggests that a student i

s 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.
Mathematics
1 answer:
kolbaska11 [484]3 years ago
7 0

Answer:

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, H_0 : \mu = 22 {Students are scoring equal to 2222 on the math portion of the​ exam}

Alternate Hypothesis, H_0 : \mu > 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, Xbar = 22.7    and Sample standard deviation, s = 3.93

The Test statistics is given by;

                            Z = \frac{Xbar -\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

Test Statistics = \frac{22.7 -22}{\frac{3.93}{\sqrt{150} } } ~ t_1_4_9

                        = 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.

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Your friend says that the solutions to the inequality 5x - 14x < - 18 are x ≤2. Solve the inequality correctly. What error di
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<h3>Answer:  x > 2</h3>

Work Shown:

5x - 14x < - 18

-9x < - 18

x > -18/(-9)

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The inequality sign flips when dividing both sides by a negative number.

Here's another approach you could take.

5x - 14x < - 18

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0 < -18+9x

18 < 9x

9x > 18

x > 18/9

x > 2

It's a slightly longer pathway, but it avoids a sign flip when you divide both sides by the positive number 9.

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The length of a rectangle is 3 centimeters less than four times its width. Its area is 10 square centimeters. Find the dimension
Kay [80]

Answer:

W = 2 cm

L = 5 cm

Step-by-step explanation:

A rectangle is a four sided shape with 4 perpendicular angles. It has two pairs of parallel sides which are equal in distance: width and length. Its area, the amount of space inside it, can be found using the formula A = l*w. If the area is 10 cm² and the length is "3 cm less than 4 times the width" or 4w - 3, you can substitute and solve for w.

A = l*w

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Subtract 10 from both sides to make the equation equal to 0. Then solve the quadratic by quadratic formula.

4w² - 3w - 10 = 0

Substitute a = 4, b = -3 and c = -10.

w = \frac{3 +/- \sqrt{(-3)^2 - 4(4)(-10)} }{2(4)} = \frac{3 +/- \sqrt{9 +160)} }{8} =  \frac{3 +/- \sqrt{169} }{8} = \frac{3+/-13}{8}

There are two possible solutions which can be found.

3 + 13 / 8 = 16/ 8 = 2

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Since w is a side length or distance, it must be positive so w = 2 cm.

If the width is 2 cm then the length is 4(2) - 3 = 8 - 3 = 5 cm.

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