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eduard
3 years ago
12

Solve the system 5x+3y=-9 &- 2x+y=8

Mathematics
2 answers:
kiruha [24]3 years ago
6 0
I got the same answer as you which is
X=(-3)
Y=(2)
lianna [129]3 years ago
3 0
5x+3y=-9
-2x+y=8

The first step is the trickery (and quite frankly, hardest to explain) part. You need to multiply one equation by a number that will create one variable to be the negative version of the other...(Confusing, right? Don't worry, I'll show you)

We are going to solve for X first.

We will multiply the second equation by -3
5x+3y=-9
-3(-2x+y)=(-3)8

Now simplify:
5x+3y= -9
6x-3y= -24

Now combine the two equations (add them together)
11x= -33
Divide by 11
X=-3

Now plug X in to one of the first equations:
5(-3)+3y= -9
-15+3y =-9
3Y= 6
Y= 2


So your final answer is:
x= -3
Y= 2

Hope this helps!
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A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The
TiliK225 [7]

Correct question:

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The company ships them to restaurants in boxes of 9 ​salmon, to grocery stores in cartons of 16 ​salmon, and to discount outlet stores in pallets of 64 salmon. To forecast​ costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts​ (a) and​ (b) below.

a. Find the standard deviations of the mean weight of the salmon in each type of shipment.

b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets?

Answer:

Given:

Mean, u = 44

Sd = 3

The company ships in boxes of 9, cartons of 16 and pallets of 64.

a) For the standard deviations of the mean weight of the salmon in each type of shipment, lets use the formula: \frac{s.d}{\sqrt{u}}

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\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{9}}

= \frac{3}{3} = 1

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ii) For the standard deviation of the mean weight of salmon in cartons of 16, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{16}}

= \frac{3}{4} = 0.75

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iii) For the standard deviation of the mean weight of salmon in pellets of 64, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{64}}

= \frac{3}{8} = 0.375

Standard deviation = 0.375

b) The distribution of shipping weights would be better characterized by a Normal model for the pallets, because regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample increases.

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3 years ago
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