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elixir [45]
3 years ago
6

In triangle RSW, line segment RW is a median and is equal to 27 cm. What is the length of RX?

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
4 0

Answer: LAST OPTION.

Step-by-step explanation:

By definition the line segment that goes from a vertex of the triangle to the midpoint of the oposite side, is called "Median of a triangle".

The medians of a triangle intersect at point called "The centroid of the triangle" and this divides each median in a ratio 2:1.

In this case, you can notice that the Centroid of the given triangle is the point "X".

Based on the explained before, we can write the following porportion:

\frac{RX}{XW}= \frac{2}{1}

Solving for "XW":

XW=\frac{RX}{2}

Since the lenght of "RX" is 27 centimeters, you know that;

RX+XW=27

Substituting XW=\frac{RX}{2} into RX+XW=27 and solving for "RX", we get that its lenght is:

RX+\frac{RX}{2}=27\\\\\frac{3}{2}RX=27\\\\RX=(27)(\frac{2}{3})\\\\RX=18\ cm

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If 5y = 6x + 16 and 9x = 7y-20, what is the value of 3x - 2y?
Law Incorporation [45]

The value of 3x-2y is -4. From the equation 5y=6x=16 and 9x=7y-20 the values of (x,y) is (4,8).

Step-by-step explanation:

        The given equations are,

                 5y=6x+16.......................(1)

                 9x=7y-20......................(2)

                3x-2y...............................(3)

Step:1

    Modify the equations (1) and (2) in (x+y=constant) format,

                6x-5y=-16.....................(4)

                9x-7y=-20....................(5)

Step:2

     Equation (4) is multiplied by 7     ( ∵ 7×Eqn(4))

                42x-35y=-112..............(6)

Step:3

     Equation (5) is multiplied by 5     (∵ 5×Eqn(5))

               45x-35y=-100.............(7)

Step:4

    Subtracting Equation (6) and (7)

               42x-35y=-112

              45x-35y=-100

    ( - )

     (42x-45x)+(35y-35y)=(100-112)

     (-3x)+(0)=(-12)

    x=4

Step:5

    Substitute X value in Equation (1),

             5y=6x+16...................(1)

     where, x=4,

             5y= (6×4)+16

             5y=40

            y= 8

Step: 6

      Check for solution, substitute x and y values in equation (2).

                9x=7y-20.......................(2)

               (9×4)=(7×8)-20

              36=36

Step:7

      From given, 3x-2y

      where, X=4 and y=8

              3x-2y = (3×4)-(2×8)

                        = 12-16

            3x-2y = -4

Result:

      The value for 3x-2y is -4. The values of (x,y) are (4,8).

   

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1. A researcher tested the diastolic blood pressure of 15 marathon runners and 15 non-runners. The mean for the runners was 75.9
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Answer:

Calculated value t =0.0109 < t = 1.701 at 28 degrees of freedom at 0.05 level of significance.

running a good way to lower a person's blood pressure

Step-by-step explanation:

<u>Step :1</u>

A researcher tested the diastolic blood pressure of 15 marathon runners and 15 non-runners

The first sample size is n₁ =15

The second sample size is n₂ =15

Given data the mean for the runners was 75.9 mm Hg with an SS of 1,500

The mean of the first sample x₁⁻ = 75.9 mm

The mean of the second sample x₂⁻   = 80.3mm

The standard deviation of the first sample (S₁) = 1,500

The standard deviation of the second sample (S₂) = 8

<u>Step 2</u>:-

<u>Null hypothesis </u>:- H₀ :  x₁⁻ = x₂⁻

<u>Alternative hypothesis</u>:- H₁ :  x₁⁻ ≠ x₂⁻

<u>level of significance</u> :- α= 0.05

The test statistic t = \frac{x_{1}^- -x_{2}^-  }{S\sqrt{\frac{1}{n_{1} } +\frac{1}{n_{2} } } }

where S^{2} =\frac{n_{1}S_{1} ^2+n_{2}S_{2} ^2 }{n_{1}+n_{2}-2}

n₁ =15 ,n₂ =15 x₁⁻ = 75.9 mm ,x₂⁻   = 80.3mm and (S₁) = 1,500 and (S₂) = 8

substitute all values in above equation, we get

S^{2} =\frac{15X(1500) ^2+15X(8) ^2 }{15+15-2}

s^2 = 1,205,391.42

Standard deviation = √1,205,391.42 = 1097.903

<u>Step 3</u>:-

The test statistic

                 t = \frac{x_{1}^- -x_{2}^-  }{S\sqrt{\frac{1}{n_{1} } +\frac{1}{n_{2} } } }

x₁⁻ = 75.9 mm ,x₂⁻   = 80.3mm, n₁ =15 ,n₂ =15 and S = 1097.903

The test statistic value t = -0.01097

modulus t = 0.0109

Calculated value t =0.0109

The degrees of freedom γ=n₁+n₂ -2 = 15+15 -2 =28

From t- distribution table

From tabulated value t = 1.701 at 28 degrees of freedom at 0.05 level of significance.

Calculated value t =0.0109 < t = 1.701 at 28 degrees of freedom at 0.05 level of significance.

Therefore we accepted null hypothesis.

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