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aliya0001 [1]
3 years ago
9

Which linear equation is represented by the table?

Mathematics
1 answer:
pentagon [3]3 years ago
6 0

Answer:

On E. the first answer is = -2.5x-15 and the solution is (-6,0)

Step-by-step explanation:

I just guessed this on E. and got it right.

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The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie
Fittoniya [83]

Answer:

The confidence interval is  25.16  < \mu < 26.85

Step-by-step explanation:

From  the question we are given a data set

     25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is  

       \= x  = \frac{\sum x_i}{n}

where is the sample size with values  n =  10

         \= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}

          \= x = 26

The standard deviation is evaluated as

             \sigma  =  \sqrt{\frac{\sum (x-\= x)}{n} }

substituting values

     = \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }

                  \cdot \ \cdot \ \cdot  \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }

     \sigma  =  1.625

The  confidence level is given as 90% hence the level of significance is calculated as

    \alpha  =  100 -90

    \alpha  =10%

   \alpha = 0.10

Now the critical values of \frac{\alpha }{2} is obtained from the normal distribution table as

      Z_{\frac{\alpha }{2} } =  1.645

The reason  we are obtaining  the critical values of \frac{\alpha }{2} instead of  \alpha is because  we are considering two tails of the area under the normal curve

  The margin of error is evaluated as

            MOE =  Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

substituting values

           MOE =  1.645 *  \frac{1.625 }{\sqrt{10} }

           MOE = 0.845

The  90%, two sided confidence interval is mathematically evaluated as

           \= x  - MOE  < \mu < \= x  + MOE

           26  - 0.845  < \mu < 26  + 0.845

           25.16  < \mu < 26.85

Given that the lower and the upper limit is greater than  25 then we can assure the manufactures  that the battery life exceeds 25 hours

5 0
3 years ago
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