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Aleonysh [2.5K]
4 years ago
9

Which constants could each equation be multiplied by to eliminate the x-variable using addition in this system of equations? 2 x

+ 3 y = 25. Negative 3 x + 4 y = 22.
1.The first equation can be multiplied by –3 and the second equation by 2.
2. The first equation can be multiplied by –4 and the second equation by 2.
3. The first equation can be multiplied by 3 and the second equation by 2.
4. The first equation can be multiplied by 4 and the second equation by –3.
Mathematics
2 answers:
lions [1.4K]4 years ago
8 0

Answer:

hes right

Step-by-step explanation:

Sever21 [200]4 years ago
4 0

Answer:

3. The first equation can be multiplied by 3 and the second equation by 2.

Step-by-step explanation:

2 x + 3 y = 25 (1)

-3 x + 4 y = 22 (2)

Multiply equation (1) by 3 and (2) by 2.

3(2x) = 6x

2(-3x) = -6x

When you add the two equations, x will get cancelled last

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3 0
4 years ago
The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 44,663 miles, with a standard
expeople1 [14]

Answer:

The probability is  P(| \= X- \mu|  < 199 ) = 0.733

Step-by-step explanation:

From the question we are told that

   The mean mileage of a tire is   \mu =  44663

   The standard deviation is  \sigma  =  2594

   The sample size is  n  =  209

 Generally the standard error of mean is mathematically represented as

    \sigma_{x}  =  \frac{\sigma}{\sqrt{n} }

=>  \sigma_{x}  =  \frac{2594}{\sqrt{209} }  

=>  \sigma_{x}  =   179.4      

Generally  the probability that the sample mean would differ from the population mean by less than 199 miles  is mathematically represented as

     P(| \= X- \mu| < 199 ) = P(|\frac{\= X - \mu}{ \sigma_{x}}|  <  \frac{199}{ 179.4})

\frac{\= X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ \= X )

=>  P(| \= X- \mu|  < 199 ) = P(|Z|  <  \frac{199}{ 179.4})

=>   P(| \= X- \mu|  < 199 ) = P(|Z|<  1.11)

=>   P(| \= X- \mu| < 199 ) = P(-1.11  \le  Z \le 1.11 )

=>   P(| \= X- \mu|  < 199 ) = P(Z  \le 1.11 ) - P( Z \le -1.11 )

From the z  table  the area under the normal curve to the left corresponding to 1.11 and -1.11 is  

        P(Z  \le 1.11 ) = 0.8665

and

       P(Z  \le - 1.11 ) = 0.1335

So

    P(| \= X- \mu| < 199 ) = 0.8665 - 0.1335

=>  P(| \= X- \mu|  < 199 ) = 0.733

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Answer:

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Step-by-step explanation:

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