A student earned grades of , , , , and . Those courses had the corresponding numbers of credit hours , , , , and . The grading system assigns quality points to letter grades as follows: A4; B3; C2; D1; F 0. Compute the grade point average (GPA) as a weighted mean and round the result with two decimal places. If the Dean's list requires a GPA of 3.00 or greater, did this student make the Dean's list

grade point average (GPA) as a weighted mean = 3.14

Yes, student makes the Dean's list.

Step-by-step explanation:

Since the data is not given I will explain the question with a relevant example:

For example

Data:

Students grades are:  A, C, B, A, D

The corresponding number of credit hours: 3, 3, 3, 4, 1

The grading system assigns quality points to letter grades as:

A = 4

B = 3

C = 2

D = 1

F = 0

To find:

grade point average (GPA) as a weighted mean

If the Dean's list requires a GPA of 3.00 or greater, did this student make the Dean's list?

Solution:

Weighted Mean =   Σx w / Σw

Here

Using the earned grades of A, C, B, A, D and corresponding quality points to these letter grades we get:

A = 4

C = 2

B = 3

A = 4

D = 1

So the values in x are:

x

4

2

3

4

1

Now the number of credit hours are represented as weights w :

w

3

3

3

4

1

In order to calculate weighted mean first multiply x with w

x w

4 * 3 = 12

2 * 3 = 6

3 * 3 = 9

4 * 4 = 16

1 * 1 = 1

The sum of x w is :

Σx w = 12 + 6 + 9 + 16 + 1 = 44

Σx w  =  44

Now compute the sum of w

Σw = 3 + 3 + 3 + 4 + 1 = 14

Σw = 14

Putting the values in the weighted mean formula:

Weighted Mean =   Σx w / Σw

                           =  44 / 14

                           = 3.1429

Weighted Mean =   Σx w / Σw = 3.14

Since the Dean's list requires a GPA of 3.00 or greater and the grade point average (GPA) as a weighted mean of the student is 3.14 so the student makes the Dean's list because his/her GPA is higher than 3.00

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

i dont know men

step-by-step explanation: