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LiRa [457]
3 years ago
10

Which of the students found the correct value of their expression? Select all that are true. None of the students Student 2 Stud

ent 1 Student 3 I will give you TEN BIG POINTS PLZ HELP.
Mathematics
1 answer:
Mariulka [41]3 years ago
6 0

Answer:

Step-by-step explanation:

Student 2 and student 1

You might be interested in
Moe, Curly, and Larry are brothers. Moe is the youngest brother, white Larry is the oldest brother. Their ages are consecutive i
damaskus [11]

Answer:

middle brother = x

older brother = x+1

younger brother = x-1

x-1+x+x+1=129

3x+0=129

3x=129

x = 43(middle brother)

43-1 = 42(younger brother)

43+1 = 44(elder brother)

Check:

43-1+43+43+1=129

42+43+44=129

129=129

4 0
3 years ago
Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,
Gre4nikov [31]

Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

        Mean, \bar X =  \frac{\sum X}{n}

                       =  \frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}  =  \frac{26}{8}  = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

  • 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, the number of observations is even, i.e. n = 8.

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

                   =  \frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.}  }{2}

                   =  \frac{2+3}{2}  = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

               Range = Highest value - Lowest value

                           = 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

        Inter-quartile range = Q_3-Q_1

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

     =  (\frac{8+1}{4} )^{th} \text{ obs.}

     =  (2.25 )^{th} \text{ obs.}

Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]

     =  0 + 0.25[0 - 0] = 0

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

     =  3(\frac{8+1}{4} )^{th} \text{ obs.}

     =  (6.75 )^{th} \text{ obs.}

Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]

     =  4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

<u>Now, we will calculate the mean, median, range and inter-quartile range for Team Y;</u>

Mean of Team Y data is given by the following formula;

        Mean, \bar Y =  \frac{\sum Y}{n}

                       =  \frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}  =  \frac{22}{8}  = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

  • 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, the number of observations is even, i.e. n = 8.

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

                   =  \frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.}  }{2}

                   =  \frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.}  }{2}

                   =  \frac{2+3}{2}  = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

               Range = Highest value - Lowest value

                           = 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

        Inter-quartile range = Q_3-Q_1

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

     =  (\frac{8+1}{4} )^{th} \text{ obs.}

     =  (2.25 )^{th} \text{ obs.}

Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]

     =  1 + 0.25[2 - 1] = 1.25

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

     =  3(\frac{8+1}{4} )^{th} \text{ obs.}

     =  (6.75 )^{th} \text{ obs.}

Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]

     =  4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

4 0
4 years ago
49 Which of the following is a measure
Mariulka [41]

ok

we are ask for measuring of center

there are 3 different measures

mode, when the data value occurs the most frequently

median, is the data value in the middle of a sorted list of data

mean is the average of the numbers

so the answer is A) mean and median

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2 years ago
X(2x+1)(5x-30)≤0<br><br> I'm confused help
elena-s [515]

Answer:

i hate math sorry can't help you with this try anther person cAn help ya

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