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algol [13]
3 years ago
11

Given the linear systems (a) x1+2x2 = 2 and 3x1+7x2 = 8 (b) x1+2x2 = 1 and 3x1+7x2 = 7 solve both systems by incorporating the r

ight-hand sides into a 2x2 matrix B and computing the reduced row echelon form of (A|B) = [ 1 2 | 2 1 | 3 7 | 8 7 ]
Mathematics
1 answer:
Colt1911 [192]3 years ago
6 0

Answer:

  • (x1, x2) = (-2, 2)
  • (x1, x2) = (-7, 4)

Step-by-step explanation:

The augmented matrix is shown after the indicated row operations. The last line shows the solutions.

  \left[\begin{array}{cc|cc}1&2&2&1\\3&7&8&7\end{array}\right] \quad\text{given}\\\\\left[\begin{array}{cc|cc}1&2&2&1\\0&1&2&4\end{array}\right] \quad\text{subtract 3r1 from r2}\\\\\left[\begin{array}{cc|cc}1&0&-2&-7\\0&1&2&4\end{array}\right] \quad\text{subtract 2r2 from r1}

The solution to the first system is (x1, x2) = (-2, 2).

The solution to the second system is (x1, x2) = (-7, 4).

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djverab [1.8K]
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4 0
3 years ago
Simplify.
MakcuM [25]

Answer:

you just need to simplify all the answers so 9 and 4 - 6 + 2 + 3 you need to keep like adding all the answers together and whatever one gets you to what it's telling you then that's the answer

Step-by-step explanation:

I'm sorry if this doesn't make sense but I hope it helps

8 0
2 years ago
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katen-ka-za [31]

Answer:

(-1/3, 3/4)

Step-by-step explanation:

9x + 8y = 3

6x - 12y = -11

Let's solve the system by eliminating x. We need the coefficients of x to be additive inverses, so they will add to zero eliminating x. The LCM of 9 and 6 is 18. Let's multiply both sides of the first by 2 and both sides of the second equation by -3.

18x + 16y = 6

-18x + 36y = 33

The coefficients of x are 18 and -18, which add to zero. Now we add these two equations.

52y = 39

y = 39/52

y = 3/4

Now we substitute y with 3/4 in the first equation and solve for x.

9x + 8y = 3

9x + 8(3/4) = 3

9x + 6 = 3

9x = -3

x = -3/9

x = -1/3

Solution: (-1/3, 3/4)

4 0
2 years ago
Find the probabilities for a standard normal random variable Z.
vampirchik [111]

Answer:

Step-by-step explanation:

find the attachment showing std normal curve symmetrical about y axis.

Equal probabilities on either side of the mean thus the total probability to the right of mean is 0.50

From the table we can find that

a) P(Z>2.5) = 0.5- area lying between 0 and 2.5

= 0.5-0.4938 =0.0062

b) P(1.2<z<2.2) = F(2.2)-F(1.2)

= 0.9861-0.3849

=0.6012

4 0
3 years ago
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
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