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1 year ago

Rotate the following figure 90° counterclockwise about the origin: l: (-3,-4), J: (-4, 0), K: (-3, O), L: (0, -2)

Mathematics

1 answer:

Juliette [100K]1 year ago

5 0

Let's begin by identifying key information given to us:

$I(-3,-4),J(-4,0),K(-3,0),L(0,-2)$

We are to rotate the figure 90 degrees counterclockwise about the origin. The transformation is mapped by the formula:

$\begin{gathered} A\mleft(x,y\mright)\rightarrow A^{\prime}\mleft(-y,x\mright) \\ I(-3,-4)\rightarrow I^{\prime}(4,-3) \\ J\mleft(-4,0\mright)\rightarrow J^{\prime}(0,-4) \\ K\mleft(-3,0\mright)\rightarrow K^{\prime}(0,-3) \\ L\mleft(0,-2\mright)\rightarrow L^{\prime}(2,0) \end{gathered}$

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3 years ago

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The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

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3 years ago

How do you add vectors? Add the vector <3,4> to the vector that goes 7 units at an angle of 2π/3.

lawyer [7]

Answer:

The sum of the two vectors is the vector <-0.5 , 4+3.5 $\sqrt{3}$ >

Step-by-step explanation:

The horizontal component (x) of a vector whose magnitude is b units and its direction is Ф° is b cos Ф

The vertical component (y) of a vector whose magnitude is b and its direction is Ф is b sin Ф

The vector is

∵ The vector goes 7 units at an angle $\frac{2\pi }{3}$

- That means its magnitude is 7 and its direction is $\frac{2\pi }{3}$

∴ x = 7 cos( $\frac{2\pi }{3}$ )

∴ y = 7 sin( $\frac{2\pi }{3}$ )

∵ cos( $\frac{2\pi }{3}$ ) = $-\frac{1}{2}$

∵ sin( $\frac{2\pi }{3}$ ) = $\frac{\sqrt{3}}{2}$

- Substitute them in x and y

∴ x = (7)( $-\frac{1}{2}$ )

∴ x = -3.5

∴ y = (7)( $\frac{\sqrt{3}}{2}$ )

∴ y = 3.5 $\sqrt{3}$

∴ The vector is <-3.5 , 3.5 $\sqrt{3}$ >

Now lets add the vectors by adding xs and ys components

∵ <3 , 4> + <-3.5 , 3.5 $\sqrt{3}$ > = <3 + -3.5 , 4 + 3.5 $\sqrt{3}$ >

∴ <3 , 4> + <-3.5 , 3.5 $\sqrt{3}$ > = <-0.5 , 4+3.5 $\sqrt{3}$ >

∴ The sum of the two vectors is the vector <-0.5 , 4+3.5 $\sqrt{3}$ >

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4 years ago

How many pieces of ribbon can you cut from 10 meters, if each piece is 4 cm long?

malfutka [58]

Answer:

answer:250 pieces

1 metre=100cm so

10metres =100*10 =1000cm

each piece is 4 CM long, so.,

no. of pieces=1000/4

by dividing by 2....

=500/4

again by dividing by2...

=250/1=250 pieces

so 250 pieces of ribbon can be cuttted from 10 metre long ribbon.

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3 years ago