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1 year ago

Please answerNot to be un polite but all of you are ignoring me and I pay y’all to help me so I don’t get it

Mathematics

1 answer:

dalvyx [7]1 year ago

7 0

Solution

Take for example in the Ukraine Russia war,

Data and probability can be used to determine the number of death tolls.

Probability can be used to estimate likelihood of Russia making moves.

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Can you please help me on 2,3,4, and 5? I will mark the brainliest plus 12 points for answering.

Marianna [84]

Answer:

2) Exact

3) Exact

4) Exact

5) Approximation

Step-by-step explanation:

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3 years ago

If (50)x = 1, what are the possible values of x? Explain your answer

Lelechka [254]

Than you multiplie tha 50 by 1/50 so will get 50/50 what is equal 1

hope this is understandably sure right easy

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3 years ago

The electric heaters, in a 240 volt electric furnace, draw 40 amps of current. What is the total KW rating of the heaters?

expeople1 [14]

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Answer:

9.6 kW

Step-by-step explanation:

The product of volts and amps is watts. Here, that product is ...

(240 V)(40 A) = 9600 W

1000 W is 1 kW, so this quantity is 9.6 kW

This tells the power consumption. The "rating" may be different.

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3 years ago

How do you solve for x: -(4 - x) = 3/4 (x - 6)?

pav-90 [236]

I honestly don’t know I am just doing this because I need to ask a question hope I helped

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3 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 equivalence relation 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.

Reflexive: 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.

Symmetric: 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.

Transitive: 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 equivalence relation.

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3 years ago