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SIZIF [17.4K]
3 years ago
5

If square HIJK is dilated by a scale factor of

Mathematics
2 answers:
WITCHER [35]3 years ago
3 0
C. point K is (3,-3) and the dilation moves from 3 to 1 so you divide both X and Y coordinates by 3 to get (1,-1)
hope this helps.

Sav [38]3 years ago
3 0

Answer:

C) (1, -1)  

Step-by-step explanation:

When a figure is dilated, the coordinates of each point are multiplied by the scale factor.  The scale factor in this problem is 1/3; this means that every point gets multiplied by 1/3.

The coordinates of K are (3, -3).  Applying the dilation, we have

1/3(3, -3) = (1/3*3, 1/3*-3) = (1, -1)

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A go-kart's speed is 607,200 ft per hour. What is the speed in miles per hour?
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The points (0, -6) and (-10, 4) lie on the graph of a linear function.
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A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o
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Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is  p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑_{x=0 b( x, n, p )

= ³∑_{x=0 b( x, 10, 0.6 )

= ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.6)^x( 1 - 0.6 )^{10-x

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

b)

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.3 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.3)^x( 1 - 0.3 )^{10-x

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.4 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.4)^x( 1 - 0.4 )^{10-x

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑_{x=4 b( x, n, p )

= ¹⁰∑_{x=4 b( x, 10, 0.5 )

= ¹⁰∑_{x=4 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - ³∑_{x=0 \left[\begin{array}{ccc}10\\x\\\end{array}\right](0.5)^x( 1 - 0.5 )^{10-x

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

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