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frez [133]
3 years ago
7

Which equation represents a line that passes through (5, 1) and has a slope of StartFraction one-half EndFraction?

Mathematics
2 answers:
MissTica3 years ago
8 0

Answer:

y - 1 = 1/2(x -5)

Just took the test, hope this helps!

natta225 [31]3 years ago
7 0

Answer:

above is correct the letter answer is C on edg2020

Step-by-step explanation:

edg2020

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Write the equation of a line through (3,1), parallel to y = x + 1
Vera_Pavlovna [14]

Answer:

The answer is C ( y= 2/3x - 1 )

Step-by-step explanation:

Equation Formula : y - y1 = m ( x - x1)

y - 1 = 2/3 ( x - 3 )

y - 1 = 2/3x - 2

y = 2/3x - 2 + 1

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A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The
adelina 88 [10]

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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