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EleoNora [17]
3 years ago
11

LCM of 96,144,126 find it​

Mathematics
1 answer:
lidiya [134]3 years ago
6 0

Answer:

2016

Step-by-step explanation:

To solve this problem, let us find the prime factors of the given numbers:

             96

        8   x    12

  2 x 2 x 2      2 x 2 x 3

              2⁵ x 3

             144

       12     x      12

  2 x 2 x 3       2 x 2 x 3

              2⁴  x 3²

                126

        2       x       63

      2                  7 x 9

     2                 3 x 3 x 7

            2 x 3² x 7

So, the lowest common multiple  =  2⁵ x 3² x 7  = 2016

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3 years ago
Choose correct way to write 7 cents using the dollar symbol.
OverLord2011 [107]
$.07. You do not write $.7 because that would mean 70 cents and should be written with a 0 at the end anyways. You do not write any dollar amount with a cents sign at the end, because the dollar sign replaces it. You write $.07 because 7 cents means 7/100 (7 one hundredths) so the seven must be written in the hundredths spot. Remember, the hundredth spot is the second number after a decimal.
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Which of the following types of statements can justify the steps of a proof?
Bond [772]

Answer:

Step-by-step explanation:

c

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3 years ago
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Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

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Marizza181 [45]

Answer:

$54.20

Step-by-step explanation:

minute=m

total monthly cost = p

39.99 + .49m = p

39.99 + .49(29) =p

39.99 + 14.21 =p

54.2 =p

3 0
3 years ago
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