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katovenus [111]
3 years ago
9

Number 2 please I don’t know how to do this

Mathematics
1 answer:
Paul [167]3 years ago
6 0
Hi shhssnn Gehretbdbd.
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What is<br> V7<br> in simplest radical form?<br> Enter your answer in the box.
MissTica

Answer:

ans: square root of whole term (7/11), I.e 1/2 power of (7/11)

Step-by-step explanation:

square root or radical can be written in power form as

7^(1/2) / 11^(1/2)

use laws of indices I.e

(7/11)^(1/2)

which is in simple radical form

7 0
2 years ago
Number them by order please
Igoryamba
1. D
2. B
3. A
Hope this help :)
5 0
2 years ago
Read 2 more answers
When decreases in one variable are accompanied by decreases in another variable, the variables are described as?
Oksi-84 [34.3K]
Directly proportional or positively correlated
7 0
3 years ago
Express your answer in scientific notation:<br> 8.8 * 10^7 + 3.1 * 10.8 =
Sindrei [870]
8.80000334
×
10
7
hope this helps
7 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

5 0
3 years ago
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