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jekas [21]
3 years ago
11

Use the substitution and to rewrite the equations in the system in terms of the variables and . Solve the system in terms of u a

nd v . Then back substitute to determine the solution set to the original system in terms of x and y.
-3/x+4/y=11
1/x-2/y=-5
Mathematics
1 answer:
Tatiana [17]3 years ago
4 0

Answer:

x = -1 and y = 1/2

Step-by-step explanation:

Let u = 1/x, and v = 1/y

Then the pair of equations

-3/x + 4/y = 11

1/x - 2/y = -5

Can be written as

-3u + 4v = 11 .................................(1)

u - 2v = -5......................................(2)

From (2)

u = 2v - 5 .......................................(3)

Substituting (3) into (1)

-3(2v - 5) + 4v = 11

-6v + 15 + 4v = 11

-6v + 4v = 11 - 15

-2v = -4

v = 4/2 = 2

Substituting this value of v in (3)

u = 2v - 5

u = 2(2) - 5

= 4 - 5

= -1

That is

u = -1, v = 2

Since u = 1/x, and v = 1/y, we have

1/x = -1

=> x = -1

And

1/y = 2

=> y = 1/2

Therefore

x = -1 and y = 1/2

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The weights of items produced by a company are normally distributed with a mean of 4.5 ounces and a standard deviation of 0.3 ou
GalinKa [24]

Answer:

The right solution is:

(a) 0.8849

(b) 12.28%

(c) 4.9935

(d) 28004

Step-by-step explanation:

Given:

Mean,

= 4.5

Standard deviation,

= 0.3

(a)

P(x > 4.14)

As we know,

⇒ z = \frac{4.14-4.5}{0.3}

      =-1.20

then,

⇒ P(z>-1.20) = P(z

                          =0.8849

(b)

P(4.8 < x < 5.04)

= P(\frac{4.8-4.5}{0.3} < \frac{x-\mu}{\sigma} < \frac{5.04-4.5}{0.3} )

= P(1

= P(z

= 0.9641 -0.8413

= 0.1228

or,

= 12.28 (%)

(c)

P(x > x) = 0.05

z value will be,

= 1.645

⇒ 1.645 = \frac{x - 4.5}{0.3}

         x = 4.9935

(d)

P(x < 5.01)

⇒ z = \frac{x- \mu}{\sigma}

      =\frac{5.01-4.5}{0.3}

      =1.7

P(z < 1.70) = 0.9554

⇒ n = \frac{27875}{0.9954}

       =28004

3 0
3 years ago
Hi i need help with this
ICE Princess25 [194]
The range would be [-4, infinity) because range is the y-value. The lowest y-value is -4. The greatest y-value goes up to infinity.
7 0
3 years ago
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