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Andrei [34K]
3 years ago
8

Jose thinks that Paula’s Printing will always be cheaper. Defend or refute Jose’s thinking using the table, the equations, or an

y other mathematics you can come up with.
Mathematics
1 answer:
svet-max [94.6K]3 years ago
5 0

It's impossible to answer this without the tables, charts, etc.

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The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below
nlexa [21]

Answer:

The true statement about Class B is that Class B has a smaller median and the same inter quartile range.

Step-by-step explanation:

We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;

Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40

Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42

1) <u>Firstly, we will calculate Median for Class A;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

  • If n is odd, then the formula for calculating median is given by;

                     Median  =  (\frac{n+1}{2})^{th} \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                     Median  =  \frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}

Here, number of observation is odd, i.e. n = 17.

So,  Median  =  (\frac{n+1}{2})^{th} \text{ obs.}

                     =  (\frac{17+1}{2})^{th} \text{ obs.}

                     =  (\frac{18}{2})^{th} \text{ obs.}

                     =  9^{th} \text { obs.} = 27

Hence, the median of class A is 27.

2) <u>Now, we will calculate Median for Class B;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

  • If n is odd, then the formula for calculating median is given by;

                     Median  =  (\frac{n+1}{2})^{th} \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                     Median  =  \frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}

Here, number of observation is odd, i.e. n = 17.

So,  Median  =  (\frac{n+1}{2})^{th} \text{ obs.}

                     =  (\frac{17+1}{2})^{th} \text{ obs.}

                     =  (\frac{18}{2})^{th} \text{ obs.}

                     =  9^{th} \text { obs.} = 26

Hence, the median of class B is 26.

3) Now, we will calculate the Inter quartile range for Class A;

Inter quartile range = Upper quartile - Lower quartile

                                = Q_3-Q_1

SO,  Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}

             =  (\frac{17+1}{4})^{th} \text{ obs.}

             =  (\frac{18}{4})^{th} \text{ obs.}

             =  4.5^{th} \text{ obs.}

             =  4^{th}  \text{ obs.} + 0.5[5^{th} \text{ obs.} -  4^{th} \text{ obs.}]

             =  21+ 0.5[22-  21]

             =  21.5

Similarly,  Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}

             =  3(\frac{17+1}{4})^{th} \text{ obs.}

             =  (\frac{54}{4})^{th} \text{ obs.}

             =  13.5^{th} \text{ obs.}

             =  13^{th}  \text{ obs.} + 0.5[14^{th} \text{ obs.} -  13^{th} \text{ obs.}]

             =  34+ 0.5[35-  34]

             =  34.5

Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.

4) Now, we will calculate the Inter quartile range for Class B;

Inter quartile range = Upper quartile - Lower quartile

                                = Q_3-Q_1

SO,  Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}

             =  (\frac{17+1}{4})^{th} \text{ obs.}

             =  (\frac{18}{4})^{th} \text{ obs.}

             =  4.5^{th} \text{ obs.}

             =  4^{th}  \text{ obs.} + 0.5[5^{th} \text{ obs.} -  4^{th} \text{ obs.}]

             =  18+ 0.5[20-  18]

             =  19

Similarly,  Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}

             =  3(\frac{17+1}{4})^{th} \text{ obs.}

             =  (\frac{54}{4})^{th} \text{ obs.}

             =  13.5^{th} \text{ obs.}

             =  13^{th}  \text{ obs.} + 0.5[14^{th} \text{ obs.} -  13^{th} \text{ obs.}]

             =  30+ 0.5[34-  30]

             =  32

Therefore, Inter quartile range for Class B = 32 - 19 = 13.

Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.

4 0
4 years ago
3x+4y=12 6x+2y=6 rewrite each equation and soap intercept form
miskamm [114]

  • Move 4y to the front so it should be 4y=12-3x
  • Divide the equation like how I did in the picture above then you should have y=3-3/4x
  • Then you switch it to y=-3/4x+3 and that's your answer.

Now for 6x+2y=6

  • Move 2y to the front so it looks like 2y=6-6x
  • Then you divide the equation by 2 like in the photo above and that should give you y=3-3x
  • Finally you switch the equation to y=-3x+3 because we are using the Slope Intercept Form. (Y=mx+b). That's your answer.

Sorry if my handwriting is bad.

5 0
3 years ago
The product of two positive numbers is 35 and their difference is 2. What is the sum of the two numbers.
tigry1 [53]
7×5=35
7-5=2
The two numbers are 7 and 5
4 0
3 years ago
Read 2 more answers
How many linear pairs are formed when two non parallel lines intersect?
Art [367]
The answer is either 4 or 2
6 0
3 years ago
Mike deposited $6500 into two saving accounts bearing simple interest. One of the accounts has an interest rate of 3% while the
ololo11 [35]
0.03x+0.06 (6500-x)=225
Solve for x
0.03x+390-0.06x=225
0.03x-0.06x=225-390
-0.03x=-165
X=165÷0.03
X=5500 invested at 3%

6500-5500=1000 invested at 6%
7 0
4 years ago
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