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Talja [164]
2 years ago
7

NEED THIS DONE IMMEDIATLY. ONLY CORRECT ANSWERS. 10 POINTS AND ILL MARK BRAINLIEST. Points A and B are to be mapped onto a numbe

r line according to two equations. The solution to the equation 2/3a = -24 is the coordinate of point a. The solution to the equation 30 = -B/0.2 is the coordinate of point B.
A) solve the first equation to find the coordinat if point A. Show your work
B) solve the second equation to find the coordinate of point B. Show your work.
C) determine the distance between points A and B. Explain.
Mathematics
1 answer:
Wewaii [24]2 years ago
8 0
A) -24÷(2/3)=-24×(3/2)=-36
B) 30×0.2=6. 6÷-1=-6
C)-36-(-6)=30 units
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.6988271188 as fraction
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Answer: 1747067797/2500000000

Step-by-step explanation:

Rewrite the decimal number as a fraction with 1 in the denominator

0.6988271188=0.6988271188/1

Multiply to remove 10 decimal places. Here, you multiply top and bottom by 10 to the 10 power = 10000000000

0.6988271188/1 * 10000000000/10000000000 * 6988271188/10000000000

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3 years ago
Find the slope of the line going through the pair of points: (11, 7), (-9, 8)
nekit [7.7K]

Answer:

1/-20 or -0.05

Step-by-step explanation:

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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
3 years ago
What is the measure of angle 1?
S_A_V [24]

Answer:

44

Step-by-step explanation:

The angle subtends the arc with measure of 88 degrees, so it has to be 88/2 which is 44.

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Write 46 over 74 in simplest form
deff fn [24]
\hbox{The GCF \rightarrow2}

\frac{46\div2}{74\div2}  \\  \\ \bold {\boxed{{\pmb{ \frac{23}{37} }}}}
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