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Bezzdna [24]
3 years ago
12

(08.05)

Mathematics
2 answers:
Crank3 years ago
7 0
If you use the first answer, the equation becomes 2x + 2y = 16.  You could subtract this from equation Q to eliminate x.

The second answer makes equation P -2x - 2y = -16.  You could add this to equation Q to eliminate x.  You can use both of these methods to eliminate the x term, but if you can only choose one, you should choose the second option.
snow_tiger [21]3 years ago
6 0
Answer is <span>−2(x + y = 8)
cause

</span><span>−2(x + y = 8) (multiply -2 to 1st equation)
-2x - 2y = - 16
</span><span> 2x + 5y = 24
</span>--------------------add
 0 + 3y = 8 (eliminate x to solve for y)
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breadth = x - 1

perimeter = 40

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40= 2( 2x + 3 + x - 1)

40= 2(3x - 2)

40 = 6x- 4

40+4 =6x

44/6 =x

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You paint 1 2 12 wall in 1 4 14 hour. At that rate, how long will it take you to paint one wall
Rudiy27
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Read 2 more answers
you and your go shopping for gift wrap one day. You buy 3 rolls of wrapping paper and 5 bows for a total of $32. Your friend get
Marysya12 [62]
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3(9) + 5(1)= 32
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8 0
3 years ago
Read 2 more answers
X + ky = 1
Marina CMI [18]
X + k y = 1
k x + y = 1  / * ( - k )
----------------
      x + k y = 1
- k² x - k y = - k
--------------------
x - k² x = 1 - k
x ( 1 - k² ) = 1 - k
x = ( 1 - k ) / ( 1 - k² ) = ( 1 - k ) / ( 1 - k ) ( 1 + k )
y = 1 - k( 1 - k )/( 1 - k² )
y = ( 1 - k ) / ( 1 - k² ) = ( 1 - k ) / ( 1 - k ) ( 1 + k )
a ) For  k = - 1 this system has no solution.
b ) For k ≠ - 1 and k ≠ 1, the system has unique solution:
( x , y ) = ( 1/ (1 + k) ,  1/( 1 + k ) ).
c ) For k = 1, there are infinitely many solutions.

3 0
2 years ago
This problem has been solved!See the answerA municipal bond service has three rating categories (A, B, and C). Suppose that in t
mariarad [96]

Answer:

a. \frac{35}{51}

b. \frac{51}{100}

c. \frac{1}{5}

Step-by-step explanation:

Suppose cities represented by C', suburbs represented by S and rural represented by R,

Let x be the total number of bonds issued throughout the US,

According to the question,

n(A) = 70% of x = 0.7x,

n(B) = 10% of x = 0.1x,

n(C) = 20% of x = 0.2x,

n(A∩C') = 50% of n(A) = 0.5 × 0.7x = 0.35x,

n(A∩S) = 20% of n(A) = 0.2 × 0.7x = 0.14x,

n(A∩R) = 30% of n(A) = 0.3 × 0.7x = 0.21x,

n(B∩C') = 40% of n(B) = 0.4 × 0.1x = 0.04x,

n(B∩S) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(B∩R) = 30% of n(B) = 0.3 × 0.1x = 0.03x,

n(C∩C') = 60% of n(C) = 0.6 × 0.2x = 0.12x,

n(C∩S) = 15% of n(C) = 0.15 × 0.2x = 0.03x,

n(C∩R) = 25% of n(C) = 0.25 × 0.2x = 0.05x,

n(C') = n(A∩C')  + n(B∩C')  + n(C∩C')  = 0.35x + 0.04x + 0.12x = 0.51x

n(S) = n(A∩S) + n(B∩S) + n(C∩S) = 0.14x + 0.03x + 0.03x = 0.20x

a. The probability that it will receive an A rating, if a new municipal bond is to be issued by a city,

P(\frac{A}{C'})=\frac{P(A\cap C')}{P(C')}=\frac{0.35x/x}{0.51x/x}=\frac{0.35}{0.51}=\frac{35}{51}

b. The proportion of municipal bonds are issued by cities = \frac{n(C')}{x}

=\frac{0.51x}{x}

=\frac{51}{100}

c. The proportion of municipal bonds are issued by suburbs = \frac{n(S)}{x}

=\frac{0.20x}{x}

=\frac{20}{100}

=\frac{1}{5}

3 0
2 years ago
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