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±![\sqrt{90000-74624}]/2](https://tex.z-dn.net/?f=%5Csqrt%7B90000-74624%7D%5D%2F2)
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±![\sqrt{90000-108000}] /2](https://tex.z-dn.net/?f=%5Csqrt%7B90000-108000%7D%5D%20%2F2)
=[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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Answer:
The width of the photo is 
.
Step-by-step explanation:
From the given figure it is notices that the total width of the frame is
The photo is covered by a frame border and the width of the border is
To find the width of the photo we have to subtract the width of upper frame border and lower frame border from the total width of frame.
Width of the photo is
Therefore the width of the photo is .
Answer:
x=1/8y+9/8
y=<u>y=−8x+9</u>
Step-by-step explanation:
<u>Let's solve for y.</u>
<u>8x+y=9</u>
<u>Step 1: Add -8x to both sides.</u>
<u>8x+y+−8x=9+−8x</u>
<u>y=−8x+9</u>
<u />
<u>Let's solve for x.</u>
<u>8x+y=9</u>
<u>Step 1: Add -y to both sides.</u>
<u>8x+y+−y=9+−y</u>
<u>8x=−y+9</u>
<u>Step 2: Divide both sides by 8.</u>
<u>8x/8=−y+9/8</u>
<u>x=−1/8y+9/8</u>
<u />
<u />
The purpose of the tensor-on-tensor regression, which we examine, is to relate tensor responses to tensor covariates with a low Tucker rank parameter tensor/matrix without being aware of its intrinsic rank beforehand.
By examining the impact of rank over-parameterization, we suggest the Riemannian Gradient Descent (RGD) and Riemannian Gauss-Newton (RGN) methods to address the problem of unknown rank. By demonstrating that RGD and RGN, respectively, converge linearly and quadratically to a statistically optimal estimate in both rank correctly-parameterized and over-parameterized scenarios, we offer the first convergence guarantee for the generic tensor-on-tensor regression. According to our theory, Riemannian optimization techniques automatically adjust to over-parameterization without requiring implementation changes.
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