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AleksandrR [38]
2 years ago
10

Offering career academies in high schools has become more popular during the past 30 years because they help students prepare fo

r work and postsecondary education. A principal at a large high school with a Science, Technology, Engineering, and Mathematics (STEM) Academy is interested in determining whether the status of a student is associated with level of participation in advanced placement (AP) courses. Student status is categorized as (1) STEM for students in the STEM program or (2) regular. A simple random sample of 200 students in the high school was taken and each student was asked two questions:
Are you in the STEM Academy?
In how many AP courses are you currently enrolled?
The responses of the 200 students are summarized in the table.

Level of Participation in Advanced Placement (AP) Courses Student Status
STEM Regular Total
No AP courses 17 31 48
One AP course 38 70 108
Two or more AP courses 20 24 44
Total 75 125 200
Part A: Calculate the proportion of STEM students who participate in at least one AP course and the proportion of regular students in the sample who participate in at least one AP course.

Part B: Is participating in two or more AP courses independent of student status?

Part C: Describe a method that could have been used to select a simple random sample of 200 students from the high school.

Part D: Is there any reason to believe there is bias in the method that you selected? Why or why not?

Part E: The responses of the 200 students are summarized in the segment bar graph shown.


Compare the distributions and what the graphs reveal about the association between level of participation in AP courses and student status among the 200 students in the sample. (5 points)

Part F: Do these data support the conjecture that student status is related to level of participation in AP courses? Give appropriate statistical evidence to support your conclusion. (10 points)
Mathematics
1 answer:
lidiya [134]2 years ago
4 0

The proportion of STEM students who participate in at least one AP course is 0.19.

<h3>How to calculate proportion</h3>

It can be deduced that the proportion of STEM students who participate in at least one AP course will be:

= 38/200

= 0.19

The proportion of regular students in the sample who participate in at least one AP course will be:

= 70/200

= 0.35

Also, participating in two or more AP courses is independent of student status. This is because the p value is more than the 0.05.

A method that could have been used to select a simple random sample of 200 students from the high school is by writing all the registration numbers of the students in a container an randomly picking.

There is bias in the sampling because the convenience sampling is used. This doesn't give everyone an equal chance.

In conclusion, status is not related to level of participation in AP courses.

Learn more about proportion on:

brainly.com/question/19994681

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Answer:

Step-by-step explanation:

We have to perform one sample proportion test. To do so we have to proceed through following steps.

(1) We have to collect data about vehicle sales in last 20 years.

(2) Using random number table or random number generator or otherwise we have to select sample of vehicles from the list of sale. In this step, we have to keep an eye on the fact that sample is to be selected from all years (all of 20). Using stratified random sampling by dividing vehicles sold over different years is an effective way to do so. That means, we may divide vehicles sold into 20 homogeneous strata over 20 years and select sample from each strata.

(3) After selecting sample of vehicles we have to communicate to corresponding vehicle owners to gather information about that particular vehicle.

(4) We have to note number of vehicles which are still on the road.

Suppose, we have selected  n vehicles as sample and after communicating we found that \tiny r of those are still on the road.

We have to test for null hypothesis H_0:p=0.80

against the alternative hypothesis H_1 \neq 0.80

Our test statistic is given

z=\frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n} }

here,

\hat p =\frac{r}{n}

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z=\frac{r/n-0.80\sqrt{n} }{0.40} \\\\=2.5(\frac{r}{n} -0.8)\sqrt{n}

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In the diagram what is m
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What is the product of -2x^3+x-5 and x^3-3x-4 ? (a) Show your work
saul85 [17]

Answer:

-3x3 - 2x - 4

Step-by-step explanation:

 ((((2•(x3))+x)-5x3)-3x)-4

 (((2x3 +  x) -  5x3) -  3x) -  4

Pulling out like terms :

4.1     Pull out like factors :

  -3x3 - 2x - 4  =   -1 • (3x3 + 2x + 4)

Polynomial Roots Calculator :

4.2    Find roots (zeroes) of :       F(x) = 3x3 + 2x + 4

Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  3  and the Trailing Constant is  4.

The factor(s) are:

of the Leading Coefficient :  1,3

of the Trailing Constant :  1 ,2 ,4

Let us test ....

  P    Q    P/Q    F(P/Q)     Divisor

     -1       1        -1.00        -1.00    

     -1       3        -0.33        3.22    

     -2       1        -2.00        -24.00    

     -2       3        -0.67        1.78    

     -4       1        -4.00        -196.00    

     -4       3        -1.33        -5.78    

     1       1        1.00        9.00    

     1       3        0.33        4.78    

     2       1        2.00        32.00    

     2       3        0.67        6.22    

     4       1        4.00        204.00    

     4       3        1.33        13.78    

Polynomial Roots Calculator found no rational roots

Final result :

 -3x3 - 2x - 4

Processing ends successfully

plz mark me as brainliest :)

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GIVEAWAY!! QUICK QUESTION!!!
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Answer:

<u>-2</u>

Step-by-step explanation:

<u>Given</u> :

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<u>Solving</u> :

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  • ⇒ -2k
  • ⇒ <u>-2 is the coefficient</u>
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