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denis-greek [22]
4 years ago
6

Solve by elimination 5x+4y = 12 3x - 3y = 18 Please help solve and provide steps.

Mathematics
2 answers:
kramer4 years ago
6 0

5x + 4y = 12   ⇒   3(5x + 4y = 12)   ⇒   15x + 12y = 36

3x - 3y = 18   ⇒    4(3x - 3y = 18)   ⇒   <u> 12x - 12y</u> =  <u>72  </u>

                                                              27x         = 108

                                                              \frac{27x}{27}   =  \frac{108}{27}

                                                                   x        =   4

5x + 4y = 12   ⇒   5(4) + 4y = 12   ⇒  20 + 4y = 12   ⇒   4y = -8   ⇒   y = -2

Answer: x = 4, y = -2


SOVA2 [1]4 years ago
4 0
First you solve for y.
5x+ y= 12 so Y= 12-5x
And plug it into the next equation
3x- 3(12-5x) = 18
Distribute
3x-36+15x=18
Add 36 to both sides
3x+ 15x= 54
18x= 54 ... divide
x= 3
And you can plug in x to solve for Y
5(3) + y = 12
15+ y = 12 ...subtract 15 from both sides
y= -3
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Answer:

The 90%  confidence level is  19.15<  L  <   20.85

Step-by-step explanation:

From the question we are told that

     The sample size is  n =  64

     The mean age is  \= x  =  20  \ years

      The standard deviation  is   \sigma  =  4 \ years

 

Generally  the degree of freedom for this data set is mathematically represented as

        df  = n -  1

substituting values

        df  = 64 -  1

        df  = 63

Given that the level of confidence is  90%  the significance level is mathematically evaluated as

          \alpha  =  100 - 90

         \alpha  =10 %  

         \alpha   = 0.10

Now   \frac{\alpha }{2}  =  \frac{0.10}{2}  = 0.05

Since we are considering a on tail experiment

The  critical value for half of  this significance level at the calculated  degree of freedom is obtained from the critical value table as

           t_{df, \frac{ \alpha}{2}   } = t_{63,  0.05   } =  1.669

   The margin for error is mathematically represented as

          MOE  =  t_{df ,  \frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

substituting values  

          MOE  = 1.699  *   \frac{4 }{\sqrt{64} }

         MOE  = 0.85

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

           \= x - MOE  <  L  <  \= x  + E

substituting values  

         20  - 0. 85 <  L  <   20  + 0.85

         19.15<  L  <   20.85

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