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Answer:
The cost of one taco is $0.87 and the cost of one enchilada is $1.16
Step-by-step explanation:
Let the cost of tacos be X
and cost of enchiladas be y
So, by using given data, we have following equations
3X + 2Y = 4.93 (1)
2X + 4Y = 6.38 (2)
So, first multiply 1st equation by 2 and 2nd equation by 3, then subtracting 1st equation by 2nd.
= 2 × ( 3X + 2Y = 4.93) (1)
= 3 × (2X + 4Y = 6.38) (2)
= - (6X + 4Y = 9.86) (1)
= 6X + 12y = 19.14 (2)
= 8Y = 9.28
Y = 9.28 ÷ 8 = 1.16
By putting the value of Y in equation 1, we get
3X + 2(1.16) = 4.93
3X + 2.32 = 4.93
X = 2.61 ÷ 3 = 0.87
Hence, the cost of one taco is $0.87 and the cost of one enchilada is $1.16.

A reflection about the x axis, about y=0, is the mapping (x',y')=(x,-y) so

A dilation of 2 is the mapping (x'',y'')=(2x', 2y')
So




We can rewrite that without the primes and combine the powers of 2.

Let's graph these and see if we're close,
Plot y= (1/4)^x, y= - (1/4)^{x}, y = - 2^{1-x}
The required probability is 
<u>Solution:</u>
Given, a shipment of 11 printers contains 2 that are defective.
We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.
Now, we know that, 
Probability for first draw to be non-defective 
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective 
(printers after first slot = 10; total defective printers = 2)
Then, total probability 
Answer:10
Step-by-step explanation: