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Alex17521 [72]
3 years ago
6

-9=2x+9y what is the answer

Mathematics
2 answers:
MrMuchimi3 years ago
4 0

Answer:

9y= -9-2x

y=-(2/9)x-1

Step-by-step explanation:

adell [148]3 years ago
3 0
-9-2x-9y=0
9+2x+9y=0
answer: 2x+9y+9=0

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A t shirt maker wants to open his first store.If he chooses the store on Main Street he will pay 650 in rent and will charge 32$
ch4aika [34]

Answer: 35 t shirts

Step-by-step explanation:

Let number of t shirts be x

Let profit made be y

On main street, the store costs $650,

Selling the t shirt at $32 per 1

He would make a revenue of 32x

Profit = revenue - cost accrued

y1 = 32x - 650

On Broad street, the store costs $440,

Selling the t shirt at $26 per 1

He would make a revenue of 26x

Profit = revenue - cost accrued

y2 = 26x - 440

To make same profit on either location

y1 = y2

32x - 650 =26x - 440

32x -26x = -440+650

6x = 210

x = 210/6

= 35 t shirts

5 0
3 years ago
I need help with this asap
Aleksandr [31]
I need point sorry for wasting your time
3 0
3 years ago
Read 2 more answers
Which point is a distance of 5 units from the point (1,9)?
Fantom [35]
You just need to plot them on a graph and see how many spaces are in between each.  So if you plot (1,9) remember it's X, Y axis.  Start from zero on the graph.  Since it's positive go over 1 to the right and up 9.  Then do the same for the rest. Then count how many units are between each point on the graph and find the one that is 5 units apart.
4 0
3 years ago
An Xത chart with three-sigma limits has parameters as follows: UCL = 104; Center Line = 100; LCL = 96; n = 5 Suppose the process
ruslelena [56]

Answer:

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

Step-by-step explanation:

We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.

The probability of one point of being over UCL=104 is:

z=(X-\mu)/\sigma=(104-98)/8=6/8=0.75\\\\P(X>104)=P(z>0.75)=0.227

The probability of one point of being under LCL=96 is:

z=(X-\mu)/\sigma=(96-98)/8=-2/8=-0.25\\\\P(X

Then, the probability of exhibit lack of control is:

P=P(X>104)+P(X

The probability of having at least one point out of control in the first three points is:

P(x\geq1)=1-P(0)\\\\P(x\geq1)=1- 0.052=0.948\\\\\\P(x=0) = \binom{11}{0} p^{0}q^{3}= 1*1*0.0515=0.0515\\\\

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

7 0
3 years ago
What is the domain of the function f(x)=x+1/x^2-6x+8
Andre45 [30]
The answer to the question is all real numbers except 2 and 4.
7 0
3 years ago
Read 2 more answers
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