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3 years ago

1. Parallelogram WXYZ is shown on the coordinate plane ect... Added photo plz help!!

Mathematics

1 answer:

aliya0001 [1]3 years ago

5 0

Answer:

Step-by-step explanation:

12.718 units

Step-by-step explanation:

The coordinates of the vertices of parallelogram WXYZ are given to be W(0,-1), X(4,0), Y(3,-2) and Z(-1,-3).

So, the perimeter of the parallelogram will be 2(WX + XY) {Since opposite sides of parallelogram are same in length}

Now, length of WX = units, To find the units click this link :

https://tex.z-dn.net/?f=%5Csqrt%7B(-1)%5E%7B2%7D%20%2B%204%5E%7B2%7D%20%7D%20%3D%20%5Csqrt%7B17%7D%20%3D%204.123

And, length of XY = units, To find the units click this link :

https://tex.z-dn.net/?f=%5Csqrt%7B(4-3)%5E%7B2%7D%20%2B%20(0-(-2))%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B5%7D%20%3D%202.236

Therefore, the perimeter of the parallelogram WXYZ = 2(4.123 + 2.236) = 12.718 units. (Answer)

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Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

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$\mathbf{s_ 1 =16.11}$

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Step-by-step explanation:

Let $x_1$ and $x_2$ be the two variables that represents the battery life in hours for talking usage and battery life in hours for internet usage respectively.

The hypothesis can be formulated as:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

$H_a : \sigma_1^2 > \sigma_2^2$

The standard deviation for the battery usage for talking is :

$\bar x_1 = \dfrac{1}{n_1} \sum x_i \\ \\ \bar x_1 = \dfrac{1}{12}(35.8 +22.4+...+35.5) \\ \\ \bar x_1 = \dfrac{241.2}{12} \\ \\ \bar x_1 =20.1$

The standard deviation Is:

$s_ 1 = \sqrt{\dfrac{1}{n_1-1}\sum (x{_1i}-\bar x_i)^2}$

$s_ 1 = \sqrt{\dfrac{1}{12-1}\sum (35.8- 20.1)^2+ (35.5-20.1)^2}$

$s_ 1 = \sqrt{259.568}$

$\mathbf{s_ 1 =16.11}$

The standard deviation for the battery life usage for the internet is :

$\bar x_2 = \dfrac{1}{n_2} \sum x_{2i}$

$\bar x_2 = \dfrac{1}{10} (24.0+12.5+36.4+...+4.7})$

$\bar x_2 = \dfrac{115}{10}$

$\bar x_2 = 11.5$

Thus; the standard deviation is:

$s_2 = \sqrt{\dfrac{1}{n_2-1}(x_{2i}- \bar x_2)^2}$

$s_2 = \sqrt{\dfrac{1}{10-1}(24-11.5)^2+(4.7-11.5)^2}$

$s_2 = \sqrt{63.60}$

$\mathbf{s_2 = 7.98}$

4 0

3 years ago