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Gre4nikov [31]
3 years ago
6

Identify the linear function graph

Mathematics
1 answer:
crimeas [40]3 years ago
3 0

The answer is choice B in the upper right hand corner.

Choice B is the graph that is a straight line and it's a single unbroken straight line. There are no gaps or jumps or any kinds of bends like choice A shows. Choice A is an absolute value graph, which is the result of gluing together two linear functions. Overall, graph A is considered nonlinear. Choices C and D are nonlinear as well.

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Please simplify I need the extra credit all answers get brainiest.
antoniya [11.8K]

Answer:

3/10a^15b^7c^4

Step-by-step explanation:

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What is the z score associated with 75 percentile
ddd [48]

Answer:

First, determine how many standard deviations above the mean one would have to be to be in the 75th percentile. This can be found by using a z table and finding the z associated with 0.75. The value of z is 0.674. Thus, one must be .674 standard deviations above the mean to be in the 75th percentile

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Can someone Solve this math problem for me I'm having some trouble.<br><br>27×26+98÷25?<br><br>​
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Answer:

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Step-by-step explanation:

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Read 2 more answers
Solve for x. 3/x-1 - 1/x^-1=5/x-1<br><br> Please show all work
navik [9.2K]

3/(x-1) - 1/(x^2-1) = 5/(x-1)

Subtract 3/(x-1) from both sides

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-1/[(x-1)(x+1] = 2/(x-1)

Multiply by (x-1)(x+1) on both sides

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5 0
3 years ago
Read 2 more answers
Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar
Mamont248 [21]

Answer:

a

The 95% confidence interval is

   0.7811 <  p <  0.9529

Generally the interval above can interpreted as

    There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval  

b

  Generally  99% is outside the interval obtained in a  above then the claim of Wal-mart is not believable  

c

 n =  125  \  stores  

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  60  

    The number of stores that had more than 2 items price incorrectly is  k =  52  

   

Generally the sample proportion is mathematically represented as  

             \^ p  =  \frac{ k }{ n }

=>          \^ p  =  \frac{ 52 }{ 60 }

=>          \^ p  =  0.867

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

=>   E =  1.96  * \sqrt{\frac{ 0.867  (1- 0.867)}{60} }

=>   E =  0.0859

Generally 95% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.867  - 0.0859  <  p <  0.867  +  0.0859

=>    0.7811 <  p <  0.9529

Generally the interval above can interpreted as

    There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval  

Considering question b

Generally  99% is outside the interval obtained in a  above then the claim of Wal-mart is not believable  

   

Considering question c

From the question we are told that

    The margin of error is  E = 0.05

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the sample size is mathematically represented as  

    n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=> n=  [\frac{1.645 }}{0.05} ]^2 * 0.867  (1 - 0.867 )

=>   n =  125  \  stores  

4 0
3 years ago
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