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Dennis_Churaev [7]
3 years ago
12

What is the midpoint of CD? • (1/2,-5/2) •(1,-5/2) •(1/2,-3/2) •(1,-3/2)

Mathematics
2 answers:
kodGreya [7K]3 years ago
8 0

Answer:

\displaystyle \left(\frac{1}{2},-\frac{5}{2}\right)

Step-by-step explanation:

Given a segment defined by the points (x1,y1) and (x2,y2), the midpoint (xm,ym) is calculated as follows:

\displaystyle x_m=\frac{x_1+x_2}{2}

\displaystyle y_m=\frac{y_1+y_2}{2}

The segment CD is defined by the points in graph C(-3,-4) and D(4,-1). The midpoint is:

\displaystyle x_m=\frac{-3+4}{2}=\frac{1}{2}

\displaystyle x_m=\frac{1}{2}

\displaystyle y_m=\frac{-4-1}{2}=\frac{-5}{2}

\displaystyle y_m=-\frac{5}{2}

Thus, the midpoint is:

\mathbf{\displaystyle \left(\frac{1}{2},-\frac{5}{2}\right)}

polet [3.4K]3 years ago
7 0

Answer:

(1/2,-5/2)

Step-by-step explanation:

\frac{ - 3 + 4 }{2}  = \frac{1}{2}  \\   \frac{ - 4 +  - 1}{2}  =   \frac{ - 5}{2}

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Suppose the correlation between height and weight for adults is 0.80. What proportion (or percent) of the variability in weight
gulaghasi [49]

Answer: 64% of the variability in weight can be explained by the relationship with height.

Step-by-step explanation:

  • In statistics, Correlation coefficient is denoted by 'r' is a measure of the strength of the relationship between two variables.
  • Coefficient of determination, r^2, is a measure of variability in one variable can be explained variation in the other.

Here, r= 0.80

\Rightarrow\ r^2= (0.80)^2=0.64

That means 64% of the variability in weight can be explained by the relationship with height.

8 0
3 years ago
Show me how you solve it
julia-pushkina [17]

Answer:

5^20

Step-by-step explanation:

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u>

\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} }  =  {a}^{m - n} }

Therefore:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  }

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u><u>I</u>

\displaystyle \large{( {a}^{m} ) ^{n} =  {a}^{m \times n}  }

Thus:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  = {5}^{5 \times 4}   } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =   {5}^{20} }

7 0
3 years ago
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