For there to be 1 car, we consider two possible outcomes:
The first door opened has a car or the second door opened has a car.
P(1 car) = 2/6 x 4/5 + 4/6 x 2/5
P(1 car) = 8/15
For there to be no car in either door
P(no car) = 4/6 x 3/5
P(no car) = 2/5
Probability of at least one car is the sum of the probability of one car and probability of two cars:
P(2 cars) = 2/6 x 1/5
= 1/15
P(1 car) + P(2 cars) = 8/15 + 1/15
= 3/5
Answer:
Step-by-step explanation:
I am joyous to assist you anytime.
He earns $500 every day, so 500 times 14 is 7000
Answer: $7000
Answer:
(2, 4)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = 2x
x = -y + 6
<u>Step 2: Solve for </u><em><u>y</u></em>
<em>Substitution</em>
- Substitute in <em>x</em>: y = 2(-y + 6)
- Distribute 2: y = -2y + 12
- Isolate <em>y</em> terms: 3y = 12
- Isolate <em>y</em>: y = 4
<u>Step 3: Solve for </u><em><u>x</u></em>
- Define equation: x = -y + 6
- Substitute in <em>y</em>: x = -4 + 6
- Add: x = 2
7/15 will be the most you can simplify
