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Colt1911 [192]
3 years ago
12

What is the probability of getting 3 queens in a 5-card hand using a standard deck of cards?

Mathematics
1 answer:
Oxana [17]3 years ago
4 0

Answer:

1.2%

Step-by-step explanation:

I did math

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Look at the figure rectangle PQRS, find m angel P
Aliun [14]
144 degrees, q=s, so 3a=4a-12. a=12, and 12*12= 144
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3 years ago
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Which linear inequality is represented by the graph? y ≥1/3 1/3x – 4 y ≤ x – 4 y ≤1/3x + 4 y ≥1/3x + 4
jenyasd209 [6]

Answer:

Step-by-step explanation:

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A puppeteer is making a triangular hat for a puppet. If two of the three angles of
Burka [1]

Answer:

The triangle is an isosceles triangle and the measure of the third angle is 120°

Step-by-step explanation:

the other nd 2 angles in the triangle both measures 30 degrees, the triangle angle sum theorem states that all the angles in a triangle add up to 180 degrees. Add the first two angles together and subtract it from 180. That's

180 - 60 and it equals 120

5 0
2 years ago
Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S
Sophie [7]

Answer:

Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that A=J^{-1}AJ. Thus, A↔A.

<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that A=P^{-1}BP. In this equality we can perform a right multiplication by P^{-1} and obtain AP^{-1} =P^{-1}B. Then, in the obtained equality we perform a left multiplication by P and get PAP^{-1} =B. If we write Q=P^{-1} and Q^{-1} = P we have B = Q^{-1}AQ. Thus, B↔A.

<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have A=P^{-1}BP and from B↔C we have B=Q^{-1}CQ. Now, if we substitute the last equality into the first one we get

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
3 years ago
Which table represents a linear function?
Delvig [45]

Answer:

I would go with the first one.

Step-by-step explanation:

The first one is the only one that increases by the same number. In other words, its slope is constant. A linear function has a constant slope. You can tell that it increases by 1/2 every time.

6 0
3 years ago
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