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adoni [48]
4 years ago
15

Suppose that a compensation professional would like to calculate the median salary. he orders four salaries as follows: $20,000,

$22,000, $24,000, $26,000. what is the median salary for this data set?
Mathematics
1 answer:
Anastasy [175]4 years ago
8 0
The median means the middle number.
First, you put them in order from least to greatest.
Then, you look to see which number is in the middle.

Since there are only four numbers given, you have to go in between two of the numbers.

The median is in between 22,000 and 24,000.

20,000  22,000  _  24,000  26,000

The number that is equally between 22,000 and 24,000 is 23,000.

The median for the set of data is $23,000.

Hope this helps!
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On three examinations, you have grades of 85, 78, and 84. There is still a final examination, which counts as one grade In order
Greeley [361]

Answer:

  • It's not possible to earn an A in the course
  • I must have a 73 or MORE to earn a B in the course

Step-by-step explanation:

The average obtained at the end of the course will be:

\frac{85+78+84+x}{4} = av

Where x is the grade obtained in the final examination and av is the final average. To obtain an A, av has to be at least 90, av≥90, and to obtain an B, av has to be at least 80, av≥80

  • Is an A in the course possible?

So, if we get 100 on the final average:

x = 100,

av = (85+78+84+100)/4 = 86,75 and  86,75∠90.

<em>Answer: No, the higher grade obtained would be 86,75.</em>

  • What grade you must have in the final to earn a B in the course

To earn a B, av≥80:

av= \frac{85+78+84+x}{4 \ }\geq  80\\ \frac{247+x}{4}\geq80   \\247+x\geq 80*4\\ x\geq 320-247\\ x\geq 73

<em>Answer: I must have a 73 or MORE to earn a B in the course</em>

6 0
4 years ago
Using the set of data below, which statements are true? Check all that apply. 13, 11, 16, 12, 42, 8 The mean of this data set is
Andreas93 [3]

The correct statements about the given data are as follows;

The mean of this data set is 17.

The median is 14.5.

Given that

Using the set of data below.

13, 11, 16, 12, 42, 8

According to the question

Using the set of data below.

13, 11, 16, 12, 42, 8

<h3>Mean is the average of the given data.</h3>

The mean of the given data can be calculated as,

\rm Mean =\dfrac{Sum \ of \ all \ data}{Total \ number \ of \ observation}\\\\Mean=\dfrac{13+11+16+12+42+8}{6}\\\\Mean=\dfrac{102}{6}\\\\Mean=17

The mean of this data set is 17.

Medium; The medium is the average of the middle terms.

The medium of the given data can be calculated as,

\rm Median=\dfrac{16+12}{2}\\\\Median=\dfrac{28}{2}\\\\Median=14\\\\

The median is 14.

To know more about Mean click the link given below.

brainly.com/question/1241759

6 0
3 years ago
If Heidi is hired by company A, she would be paid $14.60 an hour. If she is hired by company B, she would be paid $16.60 an hour
Alinara [238K]

If Heidi only gets paid when she is talking to clients, she will be working

... 9 × (40 min/60 (min/h)) = 6 h

per day.

Her $2.00 difference in pay per hour will result in a difference in pay per day of

... $2.00/h × 6 h/day = $12/day

Company B will pay Heidi $12 more for a day's work.

5 0
4 years ago
The sum of three consecutive interferes is -54. Determine the smallest interger.
OlgaM077 [116]
If x is the smallest integer, then the next consecutive integer is x+1, and the third consecutive integer is x+2
Solve:
x+x+1+x+2=-54\\3x+3=-54\\3x=-57\\x=-19
>the smallest integer is -19
5 0
3 years ago
Question 9 of 10
harkovskaia [24]

Out of the two parallelograms shown, we can say that A. Both I and II; congruent diagonals property.

<h3>Which of the shapes is a rectangle?</h3>

Rectangles can be proven to be rectangles by the congruent diagonals property.

This states that rectangles should be able to have diagonal lines extending from all sides within the rectangle that are perpendicular to each other.

These lines should bisect each other. As shown in the daigram, both shapes have that quality and so are both rectangles.

Find out more on the congruent diagonals property at brainly.com/question/1851850.

#SPJ1

5 0
2 years ago
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