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Doss [256]
4 years ago
15

Tax returns filled manually have a 20% chance of containing errors while tax returns filled electronically have a .05% chance of

containing the same if 2.7 million tax returns are filed each way how many more erroneous manually filed returns will there be than erroneous electronically filed returns
Mathematics
2 answers:
Travka [436]4 years ago
7 0

Answer:

538,650

Step-by-step explanation:

We must first find how many errors there will be if filed manually and if filed electronically

Manually: 2,700,000*20% or 2,700,000*.2

Answer: 540,000 errors

Electronically: 2,700,000*.05% or 2,700,000*.0005

Answer: 1,350 errors

We must then find the difference; 540,000-1,350=538,650

Gelneren [198K]4 years ago
4 0

Answer:

538,650.

Step-by-step explanation:

Number of erroneous manual returns = 20% of 2.7 million

= 0.20 * 2,700,00 = 540,000.

Number for electronically returned =  2,700,000 * 0.0005 = 1350.

Difference = 540,00 - 1350 = 538,650.

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If M is the set of all square of integers that are less than 100 and N is the set of all positive odd numbers that are under 15,
ollegr [7]

A set is simply the group of numbers.

  • <em>The set of M and N are: </em>M = \{1,4,9,16,25,36,49,64,81\}<em>  and </em>N= \{1,3,5,7,9,11,13\}<em>.</em>
  • <em>The set of M n N is: </em>M\ n\ N = \{1,9\}<em>.</em>
  • <em>The set of  M u N is: </em>M\ u\ N = \{1,3,4,5,7,9,11,13,16,25,36,49,64,81\}<em />

<u>(a) The set of M and N</u>

Given that:

M = squares of integers less than 100

N = Positive odd numbers less than 15

So, we have:

M = \{1,4,9,16,25,36,49,64,81\}

N= \{1,3,5,7,9,11,13\}

<u>(b) M n N and M u N</u>

M n N is the set of common elements between sets M and N.

So, we have:

M\ n\ N = \{1,9\}

M u N is the set of all elements in sets M and N.

So, we have:

M\ u\ N = \{1,3,4,5,7,9,11,13,16,25,36,49,64,81\}

Read more about sets at:

brainly.com/question/2332158

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