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sashaice [31]
2 years ago
9

three positive numbers are in Arithmetic Progression (A.P). the sum of the squares of the three numbers is 155. while the sum of

the numbers is 21. if the common difference is positive. find the values of x and y
Mathematics
1 answer:
Alecsey [184]2 years ago
6 0
Given:
Three numbers in an AP, all positive.
Sum is 21.
Sum of squares is 155.
Common difference is positive.

We do not know what x and y stand for.  Will just solve for the three numbers in the AP.
Let m=middle number, then since sum=21, m=21/3=7
Let d=common difference.
Sum of squares
(7-d)^2+7^2+(7+d)^2=155
Expand left-hand side
3*7^2-2d^2=155
d^2=(155-147)/2=4
d=+2 or -2
=+2  (common difference is positive)

Therefore the three numbers of the AP are
{7-2,7,7+2}, or
{5,7,9}


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2

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Step-by-step explanation:

From the  question we are told that

     The population mean is  \mu =  28.29

      The standard deviation is \sigma  =  33.493

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Generally the standard error for the  sample  mean (\= x ) is mathematically evaluated as

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substituting values  

       \sigma _{\=x} =  \frac{33.493}{\sqrt{56} }

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Apply central limit theorem[CLT] we have  that

        P(\= X < 33) =  [z <  \frac{33 -  \mu }{\sigma_{\= x}} ]

substituting values

       P(\= X < 33) =  [z <  \frac{33 -  28.29 }{4.48} ]

       P(\= X < 33) =  [z <  1.05 ]

From the z-table  we have that  

       P(\= X < 33) = 0.8531

For the second question

    Apply central limit theorem[CLT] we have  that

    P(\= X > 30 ) =  [z >  \frac{30 -  \mu }{\sigma_{\= x}} ]

substituting values

   P(\= X < 33) =  [z >  \frac{30 -  28.29 }{4.48} ]

From the z-table  we have that  

     P(\= X < 30) = 0.6480

Thus  

     P(\= X > 30) = 1- P(\= X < 30) = 1- 0.6480

     P(\= X > 30) = 0.3520

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