What is the written form of 30.55

What is 12 to the 2nd power

Alex Ar [27]

Answer: 144

Step-by-step explanation: The problem shown here can be read as "10 to the 2nd," oe "12 to the 2nd power."

The 12 is called the base and the 2 that is raised up is called the exponent.

The exponent of 2 tells us that we multiply our base 2 times.

So we have 12 · 12 which is 144.

The numbers that we multiply together are called factors so in this problem, we have 2 factors of 12.

8 0

2 years ago

James did not get the PlayStation that he wanted for his birthday, but instead, he got $80. He decided that he will save the $80

Vika [28.1K]

W=14 that’s because 360-80=280 then you have to divide that by 20. And you get 14 weeks.

6 0

2 years ago

Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea

elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) = P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725

There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways to select exactly 3 aces is:

$4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192$

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

$P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}$

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

$P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052$