To rationalize the denominator of StartFraction 5 minus StartRoot 7 EndRoot Over 9 minus StartRoot 14 EndRoot EndFraction , you
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Let 
denote the number of minutes spent brushing his teeth, and let 
be the mean and 
the standard deviation for this distribution.
The z-score corresponding to this probability is approximately
, which means
Next, (note the sign change)
The corresponding z-score is approximately
, so you have
Solving the two equations for and , you'll find that the mean is approximately 
and the standard deviation is approximately 
.
The z-score corresponding to this probability is approximately
Next, (note the sign change)
The corresponding z-score is approximately
Solving the two equations for
The vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.
Given an equation showing profits of A Christmas vendor as
P=-0.1+30g-1200.
We have to find the number of gingerbread houses that the vendor needs to sell in order to earn profit of $665.60 and $1500.
To find the number of gingerbread houses we have to put P=665.60 in the equation given which shows the profit earned by vendor.
665.60=-0.1+30g-1200
0.1-30g+1200+665.60=0
0.1-30g+1865.60=0
Divide the above equation by 0.1.
-300g+18656=0
Solving for g we get,
g=[300±]/2*1
g=[300±
g=[300±]/2
g=(300±124)/2
g=(300+124)/2 , g=(300-124)/2
g=424/2, g=176/2
g=212,88
Because 212 is much greater than 88 so vendor prefers to choose selling of 88 gingerbread houses.
Put the value of P=1500 in equation P=-0.1+30g-1200.
-0.1+30g-1200=1500
0.1-30g+1500+1200=0
0.1-30g+2700=0
Dividing equation by 0.1.
-300g+27000=0
Solving the equation for finding value of g.
g=[300±]/2*1
=[300±
=[300±]/2
Because comes out with an imaginary number so it cannot be solved for the number of gingerbread houses.
Hence the vendor has to sell 88 gingerbread houses to earn a profit of $665.60 and there is no chance that the vendor will earn $1500.
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