Help me, please!!!!! CORRECT ANSWER GETS BRAINLIEST!!!!!!

Mathematics

1 answer:

katen-ka-za [31]4 years ago

5 0

Hello,

Here is your answer:

The proper answer to your question is..... 200,000=4x-20

When your doing questions like these make sure you pay attention to the question.

Your answer is 200,000=4x-20

If you need anymore help feel free to ask me!

Hope this helps!

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Angus Cut His Neighbors' Lawns For The Summer. By The End, He Had $325 In His Lawn-cutting Bank Account And He Paid A Total $75

rodikova [14]

Answer:

C. N = 400 / 10

Step-by-step explanation:

Lawn-cutting Bank Account = $325

Amount paid for gas = $75

Total amount earned = Lawn-cutting Bank Account + Amount paid for gas

= $325 + $75

= $400

Amount earned per every lawn he mowed = $10

How Many Lawns Did He mow?

Let N = number of lawns he mowed

N = Total amount earned / Amount earned per every lawn he mowed

N = 400 / 10

N = 40 lawns

5 0

3 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:

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