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3 years ago

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]

113.04 ft2

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

Read 2 more answers

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. -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

-8 =j

8 0

3 years ago

Twenty-five less than three times a number

ra1l [238]

Let's say n = the unknown number

then, 3n-25 would be your answer

3 0

3 years ago

Read 2 more answers

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 and 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 or 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