BuilderBoi

\(30 \times 1.4 = \color{brown}\boxed{42}\)

\(30\) is the original price, and \(1.4\) is the markup.

There are a total of \(36\) possibilities (\(6\) for the first, \(6\) for the second). The only way the two numbers will be different is if they are the same, or if they differ by one. 

There are \(6\) possible ways to roll the same number twice, and there is \(10\) ways to roll two numbers that differ by one.

This means that the probability is \(\color{brown}\boxed {5\over 9}\)

This is a math forum, not a biology forum lol.

A) This forum is not meant to do your homework for you

B) We need the image to solve this problem

When it means "without using each digit more than once", I think it means that you can't use each block more than once. Although, it doesn't make sense why they would give you 2 blocks of each. 

In this case, the only possible answers would be \(\color{brown}\boxed{0, 1, 10 }\).  

The maximum value is \(18.9 - {1764\over-16.8} = 123.9\). Subsitute the equation like this:\(123.9=-4.2t^2+42t+18.9\). Solve for \(t\), and you get \(\color{brown}\boxed 5\)

For the original sequence: \({\text {sum}\over n } = 10\)

When you add the terms \(20\) and \(30\), we have the equation: \({{\text{sum} + 50}\over{n+2}} = 12\)

These two are a equations form a system, and you can solve for \(n\) from here.

Good Luck!

\({1 \over 2^3} \times {1 \over 2} \times {4 \choose 1} = {4 \over 16} = \color{brown}\boxed {1\over 4}\)

Explanation:

There is a \({1 \over 2} \times {1 \over 2} \times {1 \over 2} \) chance of the successes (\(3\) tails)

There is a \(1 \over 2\) chance of failure (\(1\) head)

There are \(4 \choose 1\) ways to order the failure (the same as \(4 \choose 3\), which counts the ways to order the successes.)

\({1 \over 3} \times 24 = \color{brown}\boxed 8\) 

After 162 hours, there will be \(105 \times 2 = 210\) bacteria.

After 324 hours, there will be \(210 \times 2 = \color{brown}\boxed{420}\) bacteria