Answer:
The equation is y= 0,65 x
If x is the price of the ticket without the coupon, and the theater offers a discount if you have a coupon, then having a coupon means that the price a person ultimately pays (y) is the original price (x) minus a 35% of this price: y= x -0.35 x . By association: y= (1-0.35) x and then y= 0.65 x.
The line should be in the first quadrant because the first quadrant allows you to represent a situation in which the dependent variable (y) and the independent variable (x) are both positive. This is the case in this exercise, because both prices, the one without discount (x) and the one with discount (y) are necessary positive (you can not pay a negative price!).
Step-by-step explanation:
- The price without discount (or without the coupon) is x.
- The price with discount (or with coupon) is y.
- y and x are both related: y is a percentage of x, specifically, y is 35% smaller than x. This means that y =0.65 x.
4(2t+6)
8t+24 is the answer
Answer:
0.45% probability that they are both queens.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes
The combinations formula is important in this problem:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
Desired outcomes
You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.
The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So
![D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6](https://tex.z-dn.net/?f=D%20%3D%20C_%7B4%2C2%7D%20%3D%20%5Cfrac%7B4%21%7D%7B2%21%284-2%29%21%7D%20%3D%206)
Total outcomes
Combinations of 2 from a set of 52(number of playing cards). So
![T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326](https://tex.z-dn.net/?f=T%20%3D%20C_%7B52%2C2%7D%20%3D%20%5Cfrac%7B52%21%7D%7B2%21%2852-2%29%21%7D%20%3D%201326)
What is the probability that they are both queens?
![P = \frac{D}{T} = \frac{6}{1326} = 0.0045](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BD%7D%7BT%7D%20%3D%20%5Cfrac%7B6%7D%7B1326%7D%20%3D%200.0045)
0.45% probability that they are both queens.
Answer:
-3/5
Step-by-step explanation:
First, make the mixed fraction into an improper fraction:
-1 2/5 = -(5/5 + 2/5) = -(7/5)
Note that two negatives = one positive:
- (-4/5) = + 4/5
Combine the terms. Subtract the numerators:
-(7/5) + (4/5) = (-7 + 4)/5 = -3/5
-3/5 is your answer.
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