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Elodia [21]
2 years ago
15

As you were in a hurry to catch a flight, you picked two out of your four credit cards at random, without even looking. One of t

he four cards has reached its balance limit and can no longer be used for purchases. What is the probability that you picked one good (=usable) and one bad card? (Hint: Give the cards names, list all possible pairs picked, and then count pairs that have one good and one bad card.) Group of answer choices
Mathematics
1 answer:
agasfer [191]2 years ago
3 0

Answer:

P = 0.5

Step-by-step explanation:

There are four cards.

Lets say,

Card 1 = A

Card 2 = B

Card 3 = C

Card 4 = D

Lets suppose that A has reached its limit. Listing all possibilities of picking 2 random cards.

1st possibility = A and B

2nd possibility = A and C

3rd possibility = A and D

4th possibility = B and C

5th possibility = B and D

6th possibility = C and D

There are three possibilities that the card which has reached its limit will be picked with the usable card. Calculation probability.

P = total number of wanted outcomes / total number of possible outcomes

P = 3/6

P = 0.5

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Identify the constant term in this expression.
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Answer:

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Plsss help me graph this!!! Attached is the question
e-lub [12.9K]

The graph is a straight line graph with a slope of -1/2

Equation of a line in point-slope form

The equation of a line in point slope form is expressed as:

y - y₁ = m(x - x₁)

Given the following parameters

Point = (-2, 3)

Slope = -1/2

Substitute the given parameters into the formula

y-3 = -1/2(x-(-2))

y-3 = -1/2(x+2)

Write in slope-intercept form

2(y-3) = -(x+2)

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2y - 6 = -x -2

2y = -x + 4

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The graph of the equation is as shown below

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1 year ago
Please help! it’s urgent
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Step-by-step explanation:

{9}^{ \frac{3}{2} }  \times  {27}^{ \frac{1}{2} }  =  {(3)}^{ \frac{3}{2} \times 2 }  \times  {(3)}^{ \frac{1}{2}  \times  3}  \\  =  {3}^{3}  \times  {3}^{ \frac{3}{2}  }  \\  =  {3}^{3 +  \frac{3}{2} }  \\  =  {3}^{4\frac{1}{2} }  \\  =  {3}^{ \frac{9}{2} }

8 0
2 years ago
Susie and Amelia solved the same equation using two separate methods. Their work is shown in the table below:
kvv77 [185]
<span>Stephen and Aaron solved the same equation using two separate methods. Their work is shown in the table below:

Stephen Aaron:
3x - 2 = 5x - 6 3x - 2 = 5x - 6 
3x - 2 + 2 = 5x - 6 + 2 3x - 3x - 2 = 5x - 3x - 6
3x = 5x - 4 -2 = 2x - 6
3x - 5x = 5x - 5x - 4 -2 - 6 = 2x 
-2x = -4 -8 = 2x
x = 2 -4 = x

Identify who made the error and what he did wrong.
Aaron made the error when he subtracted 6.

Aaron made the error when he subtracted 3x.

Stephen made the error when he added 2.

Stephen made the error when he subtracted 5x.

answer:

</span>In the Aaron`s work:
- 2 = - 2 x - 6
and after that:
- 2 - 6 = 2 x
It should be:
- 2 + 6 = 2 x
or: - 2 + 6 = 2 x - 6 + 6
Answer:
A ) Aaron made the error when he subtracted 6.
5 0
3 years ago
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