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antoniya [11.8K]
3 years ago
15

If matrix A has dimensions 3 x 2 and matrix B has dimensions 3 x 2, can matrix A and matrix B be multiplied?

Mathematics
2 answers:
Charra [1.4K]3 years ago
8 0
Matrix A has 3 rows, 2 columns
Matrix B has 3 rows, 2 columns

The number of columns in matrix A (which is 2) does not match up with the number of rows in matrix B (which is 3). 

So the matrix multiplication A*B is undefined. We cannot multiply the matrices.
erik [133]3 years ago
6 0

Answer:

we can not multiply matrix A and B.

Step-by-step explanation:

We have been given that

dimension of matrix A = 3 x 2

dimension of matrix B = 3 x 2

Rule of multiplying two matrices:

Two matrix can be multiplied if the column of first matrix is equal to the row of second matrix.

column of first matrix = row of second matrix

Here, we have

Column of first matrix A = 2

Row of second matrix b = 3

Since,, column of matrix A ≠ Row of matrix B

Hence, we can not multiply matrix A and B.

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Option A. 5.25 units

Step-by-step explanation:

we know that

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\frac{PR}{XZ}=\frac{QR}{YZ}

substitute the given values

\frac{7}{XZ}=\frac{4}{3}

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Which one of the following is an irrational number?
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Answer:

sqrt(10)

Step-by-step explanation:

Since the second number ends and does not continue it wouldn't be the second one, and since the third is able to be predicted it wouldn't be the thrid one either and since the fourth one is 7 and is a whole number the only answer choice possible would be the first one.

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3 years ago
Jeanne has many nickels, dimes, and quarters in her wallet. She chooses 3 coins at random. What is the probability that all thre
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Answer:

Step-by-step explanation:

There isn't enough said about the distribution of coins in her wallet, but we'll just assume that the number is so large that any coin is equally likely to be drawn.

Stated another way, there are 27 possible outcomes of the three draws (3 x 3 x 3) and we'll assume each is equally likely.

PROBLEM 1:

This is a conditional probability question. We only have to consider the cases where she could have drawn 2 quarters and another coin. The possible draws are:

DQQ, NQQ, QDQ, QNQ, QQD, QQN or QQQ*.

That's 7 possible draws (with equal probability) and only 1* of them is a draw with 3 quarters.

Answer:

P(three quarters given two are quarters) = 1/7

PROBLEM 2:

Again, this is conditional probability. To help count the ways, let's instead count the ways to *not* draw any dimes. That means you have 2 choices for the first coin, 2 choices for the second coin and 2 choices for the third coin.

So 8 out of the 27 draws would *not* contain a dime. By subtracting, we can see that 19 of the draws *would* contain at least one dime.

Now think of the ways to create a draw consisting of one of each coin. We have the 3 different coins and they can be drawn in any order. That would be 3! or 6 ways.

If that isn't clear, let's list them all out:

DDD, DDN, DDQ, DND, DNN, DNQ*, DQD, DQN*, DQQ, NDD, NDN, NDQ*, NND, NQD*, QDD, QDN*, QDQ, QND*, QQD

There are 19 possible outcomes with at least one dime and exactly 6 of them have one of each type.

P(all different given at least one is a dime) = 6/19

3 0
3 years ago
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