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Anettt [7]
4 years ago
14

Match the width with the given area and length of a rectangle.

Mathematics
1 answer:
torisob [31]4 years ago
3 0

Answer:

5 cm A = 44cm; <u>w</u><u> </u><u>=</u><u> </u><u>8</u><u>.</u><u>8</u><u> </u><u>cm</u>

7.5 cm A = 48.75 cm2; <u>w</u><u> </u><u>=</u><u> </u><u>6.5</u><u> </u><u>cm</u>

5 cm A = 31.25; <u>w</u><u> </u><u>=</u><u> </u><u>6</u><u>.</u><u>2</u><u>5</u>

9.25 cm A = 55.5 cm; <u>w</u><u> </u><u>=</u><u> </u><u>6</u><u> </u><u>cm</u>

Step-by-step explanation:

What you have to do to find the answer is to take the length which for the first problem is 5 cm and then multiply the 5 by each width until you get the area. Example; Length <em>x</em> Width = Area

Basically you use the length times all of the widths until you match up the area.

Hope I could help

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2 years ago
Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defect
ad-work [718]

Answer:

a) P (0 defective component) = 0.5

   P( 1 defective component ) = 0.35

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b) P( 0 ) = 0

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Step-by-step explanation:

p( no defective ) = 0.5

p( 1 is defective ) = 0.35

p( 2 is defective ) = 0.15

Given that 2 components are selected at random

<u>a) Given that neither component is defective </u>

Probability of 0 defective component = 0.5

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P( 2 defective component ) = 0.15

<u>b) Given that one of the two tested component is defective </u>

P( 0 defective ) = 0

P( 1 defective ) = P ( \frac{x=1}{x\geq 1} ) = p( x = 1 ) / 1 - P ( x = 0 )

                                        = ( 0.5 )^1 ( 0.5 )^0 /  1 - ( 0.5)^0 (0.5)^1

                                        = 0.5 * 1 / 1 - 0.5 = 0.5 / 0.5 =  1

p ( 2 defective ) = p( x = 3 ) / 1 - P ( x = 0 )

                         = ( 0.5 )^2 ( 0.5 )^0 /  1 - ( 0.5)^0 (0.5)^2

                         = 0.25 / 0.75 = 0.33

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