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2 years ago

21x =7? A. x=3 B. None of these C. x=147 D. x=28

Mathematics

1 answer:

Varvara68 [4.7K]2 years ago

6 0

Answer: B

Step-by-step explanation:

To isolate x, you divide both sides by 21

21x=7

$\frac{21x}{21}=\frac{7}{21}$

x = 7/21

Simplify by dividing both the numerator and denominator by 7

x = 1/3

Since, 1/3 is not an option, B is the correct answer

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3 years ago

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Ian earns £420 a week after a 5% rise. <br> What was his pay before?

belka [17]

It turns out that 5% of 400 is 20.

(1/20)*400=20.

So Ian was being paid £400 a week before he was given a pay rise.

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3 years ago

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Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

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3 years ago

0<2x<4 solve for the given variable

Papessa [141]

Answer:

0 < x < 2

Step-by-step explanation:

0 < 2x < 4

Divide 2 into all the parts.

0/2 < 2x/2 < 4/2

The x variable should have no coefficient.

0 < x < 2

8 0

3 years ago

in a city school of 1,200 students, 40% of the students are on the honor roll, 60% have a part-time job, and 22% are on the hono

Alexus [3.1K]

Let
A = event that the student is on the honor roll
B = event that the student has a part-time job
C = event that the student is on the honor roll and has a part-time job

We are given
P(A) = 0.40
P(B) = 0.60
P(C) = 0.22
note: P(C) = P(A and B)

We want to find out P(A|B) which is "the probability of getting event A given that we know event B is true". This is a conditional probability

P(A|B) = [P(A and B)]/P(B)
P(A|B) = P(C)/P(B)
P(A|B) = 0.22/0.6
P(A|B) = 0.3667 which is approximate

Convert this to a percentage to get roughly 36.67% and this rounds to 37%

Final Answer: 37%

4 0

3 years ago