The answer is 7.3. Hope that helps.
Let
x ----------> the height of the whole poster
<span>y ----------> the </span>width<span> of the whole poster
</span>
We need
to minimize the area A=x*y
we know that
(x-4)*(y-2)=722
(y-2)=722/(x-4)
(y)=[722/(x-4)]+2
so
A(x)=x*y--------->A(x)=x*{[722/(x-4)]+2}
Need to minimize this function over x > 4
find the derivative------> A1 (x)
A1(x)=2*[8x²-8x-1428]/[(x-4)²]
for A1(x)=0
8x²-8x-1428=0
using a graph tool
gives x=13.87 in
(y)=[722/(x-4)]+2
y=[2x+714]/[x-4]-----> y=[2*13.87+714]/[13.87-4]-----> y=75.15 in
the answer is
<span>the dimensions of the poster will be
</span>the height of the whole poster is 13.87 in
the width of the whole poster is 75.15 in
Answer:
x equals 2
Step-by-step explanation:
.3-.2+.3 is .4
.4 divided by .2 is 2
then you are left with 2 = x
Answer:
Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .
Step-by-step explanation:
We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;
Mean, 
= 1.3 cm and Standard deviation, 
= 0.01 cm
Also, since distribution is normal;
Z = 
~ N(0,1)
Let X = range of diameters
So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)
P(X < 1.33) = P( < 
) = P(Z < 3) = 0.99865
P(X <= 1.27) = P( < 
) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865
= 0.00135
P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .
Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .
You cant answer the Question with out the picture <span />
