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MArishka [77]
3 years ago
12

3) 14 pennies to 35 pennies​

Mathematics
1 answer:
Leno4ka [110]3 years ago
5 0

Answer:

<h2>2/5</h2>

Step-by-step explanation:

14/35

Find the GCD of numerator and denominator

GCD of 14 and 35 is 7

Divide both the numerator and denominator by the GCD

14 ÷ 7

35 ÷ 7

2/5

I'm always happy to help :)

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Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers
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Dafna11 [192]

Answer:

A. Sample Space = S = {(9, 18),(9, 19),(9, 20),(9, 21),(10, 18),(10, 19),(10, 20),(10, 21),(11, 18),(11, 19),(11, 20),(11, 21),(12, 18),(12, 19),(12, 20),(12, 21)}

B.  {(9, 18),(10, 18),(11, 18),(12, 18)}

C. {(9, 18),(9, 19),(10, 18)}

D.  {(9, 18),(9, 19),(10, 18),(11, 18),(12, 18)}

E. B(complement) ∩C =  {(9, 19)}

Step-by-step explanation:

The Sample Space would contain each element of the box A associated with each element of the box B.

A. Sample Space = S = {(9, 18),(9, 19),(9, 20),(9, 21),(10, 18),(10, 19),(10, 20),(10, 21),(11, 18),(11, 19),(11, 20),(11, 21),(12, 18),(12, 19),(12, 20),(12, 21)}

B.  Let the Outcomes in the event B, the event that the second resistor has a resistance less than 19 be denoted by J then

J=  {(9, 18),(10, 18),(11, 18),(12, 18)}

C. Let the outcomes in the event C, the event that the sum of the resistances is equal to 28 be denoted by L then

L =  {(9, 18),(9, 19),(10, 18)}

D. The  outcomes in B∪C contains all the elements of B and C

B∪C=J∪L= {(9, 18),(9, 19),(10, 18),(11, 18),(12, 18)}

E. B complement contains those elements of the Sample Space which are not the elements of Set B.

B complement= S-B=  {(9, 19),(9, 20),(9, 21),(10, 19),(10, 20),(10, 21),(11, 19),(11, 20),(11, 21),(12, 19),(12, 20),(12, 21)}

B(complement) ∩C contains those elements of B and C which are common to both B complement and C.

B(complement) ∩C =  {(9, 19)}

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3 years ago
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