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Mathematics 9 Online
OpenStudy (anonymous):

WILL GIVE MEDAL! Find the value of x such that the data set has the given mean. 3.8, 4.2, 5.3, x; mean 4.8

OpenStudy (amoodarya):

\[4.8=\frac{3.8+4.2+5.3+x}{4}\]you can find x

OpenStudy (anonymous):

@amoodarya so would it be 3.325?

OpenStudy (anonymous):

@amoodarya x= 3.325?

OpenStudy (amoodarya):

i think so

OpenStudy (amoodarya):

no it is not x=4(4.8)-5.3-4.2-3.8=5.9

OpenStudy (welshfella):

5.9 is correct

OpenStudy (anonymous):

@welshfella could you explain to me how you get 5.9? I'll give you a medal

OpenStudy (amoodarya):

\[4.8=\frac{3.8+4.2+5.3+x}{4}\\\frac{4.8}{1}=\frac{3.8+4.2+5.3+x}{4}\\4.8 *1=3.8+4.2+5.3+x\\\]

OpenStudy (welshfella):

first you cross multiply 4 * 4.8 = 19.2 so 3.8 + 4.2 + 5.3 + x = 19.2 adding the numbers on the left 13.3 + x = 192 now subtract 13.3 from both sides (this will leave x on its own on left side) 13.3 + x = 19.2 - 13.3 - 13.3 x = 5.9

OpenStudy (welshfella):

cross multiply:- |dw:1433875265692:dw| -

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