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inysia [295]
3 years ago
10

Biggest 3,5,7,8 digit even number Smallest 3,5,7,8 digit even number

Mathematics
1 answer:
Kryger [21]3 years ago
6 0

A number is even if it ends with an even digit, so 8 must be the last one.

Now, the digits to the left are more worth than those on the right, so if you want a big number you have to stack the greatest digits to the left, otherwise if you want a small number you have to stack the smallest digits to the left.

So, the greatest number is 7538, the smaller is 3578.

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Can someone help me with this problem? Thank you!
svlad2 [7]

<u>113.04 ft2</u>

Step-by-step explanation:

r=d/2=12/2=6

A= pi*r*r=3.14*6*6

A=113.04

8 0
2 years ago
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An international company has 27,700 employees in one country. if this represents 33.6% of the company's employees, how many empl
Yuri [45]
27,700 is .336 how many are all 100%

27,700 / .336 = X/ 1
82,440 employees
4 0
3 years ago
Solve for j.<br>-13j - 20 = -8j + 20​
aleksley [76]

Answer:

j = -8

Step-by-step explanation:

-13j - 20 = -8j + 20​

Add 13 j to each side

-13j+13j - 20 = -8j+13j + 20

-20 = 5j+20

subtract 20 from each side

-20-20 = 5j +20-20

-40 = 5j

Divide by 5

-40/5 = =5j/5

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8 0
3 years ago
Twenty-five less than three times a number
ra1l [238]
Let's say n = the unknown number


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3 0
3 years ago
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You spin each spinner once. On the spinners, R represents red, B represents blue, G represents green, Y represents yellow, and P
NeTakaya

Answer:

1/6

1/12

1/2

Step-by-step explanation:

There are three possible outcomes in the left spinner, and four possible outcomes in the right spinner.  So there are a total of 3×4=12 possible combinations.  We can show that by making a grid:

\left[\begin{array}{cccc}&R&B&G\\R&RR&BR&GR\\B&BR&BB&BG\\P&PR&BP&GP\\Y&RY&BY&GY\end{array}\right]

Of these 12 combinations, 2 show both spinners landing on the same color (RR and BB).  So the probability is 2/12 = 1/6.

There is only 1 outcome in which the first spinner lands on R <em>and</em> the second spinner lands on P (PR), so the probability is 1/12.

There are 6 outcomes in which the first spinner lands on R <em>or</em> the second spinner lands on P (RR, BR, PR, RY, BP, GP).  So the probability is 6/12 = 1/2.

7 0
3 years ago
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