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OleMash [197]
3 years ago
13

Find three consecutive numbers whose sum shall equal 45.​

Mathematics
2 answers:
Maru [420]3 years ago
7 0

Answer:

Since the consecutive numbers, so the first one is x, the next one is x+1, and so on....I hope this will help

Step-by-step explanation:

x + (x + 1) + (x + 2) = 45

x + x + 1 + x + 2 = 45

3x + 3 = 45

3x = 42

x = 14

Now that we know x = 14, let's add one to it.

x + 1 = 14 + 1 = 15.

Now we have our middle number, which is 15. You can always check your answer by adding 14 + 15 + 16 = 45. Whoo!

frosja888 [35]3 years ago
3 0

Answer:

the three consecutive numbers are 14, 15, 16

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a) $42,580

b) $43,900

Step-by-step explanation:

Recall, "mode" refers to the value which occurs most frequently.

In this case, the question says that there are 2 modes,

this means $34,000 and $50,500 both share the spot for the most frequently appearing salary.

because there are only 5 employees (and hence 5 salaries), the only possible way that there are two modes is if there are two of each mode, leaving only the last salary unknown.

i.e  if we list the 5 salaries (in no particular order)

$34,000   $34,000  $50,500   $50,500   $ x

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Then (34,000 + 34,000 + 50,500 + 50,500 + x) / 5 = 42,100

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Now we know all the values, we can rearrange them in increasing value:

$34,000   $34,000  $41,500   $50,500   $50,500  

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= ($34,000  + $34,000 + $43,900 +  $50,500  + $50,500  ) / 5

= $42,580 (answer for part a)

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Answer:

\huge\boxed{\sf (x*4) + (x*5) = 54}

Step-by-step explanation:

\sf x(4+5) = 54\\\\Applying \ distributive \ property\\\\(x*4) + (x*5) = 54\\\\Distributive\ Property\ is:\\\\A(B+C) = (A*B)+(A*C)\\\\\rule[225]{225}{2}

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