a. A simple random sample of size 80 has a mean of 7.31. The population standard deviation is 6.26. Construct a 99% confidence i
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Let
x---------> <span>quantity (ton) of vegetables sold the first day
</span>y---------> quantity (ton) of vegetables sold the second day
z---------> quantity (ton) of vegetables sold the third day
we know that
x=y-3---------> equation 1
z=(5/9)*(x+y)------> equation 2
x+y+z=98-----> equation 3
substitute equation 2 in equation 3
x+y+[(5/9)*(x+y)]=98----> multiply by 9----> 9x+9y+5x+5y=882
14x+14y=882-----> equation 4
to resolve the equations system formed by
x=y-3
14x+14y=882
using a graph tool
see the attached figure
the solution is
x=30
y=33
x+y+z=98----> z=98-(x+y)----> z=98-63---> 35
the answer is
vegetables sold the first day----> 30 ton
vegetables sold the second day----> 33 ton
vegetables sold the third day------> 35 ton
x---------> <span>quantity (ton) of vegetables sold the first day
</span>y---------> quantity (ton) of vegetables sold the second day
z---------> quantity (ton) of vegetables sold the third day
we know that
x=y-3---------> equation 1
z=(5/9)*(x+y)------> equation 2
x+y+z=98-----> equation 3
substitute equation 2 in equation 3
x+y+[(5/9)*(x+y)]=98----> multiply by 9----> 9x+9y+5x+5y=882
14x+14y=882-----> equation 4
to resolve the equations system formed by
x=y-3
14x+14y=882
using a graph tool
see the attached figure
the solution is
x=30
y=33
x+y+z=98----> z=98-(x+y)----> z=98-63---> 35
the answer is
vegetables sold the first day----> 30 ton
vegetables sold the second day----> 33 ton
vegetables sold the third day------> 35 ton