First divide the coefficients:-
-18 / -12 = 3/2
a^-2 / a^-4 = a^2
b^-5 / b^-6 = b
so the expression simplifies to 3a^2b / 2
Some fractions that are equivalent to 2/3 are 4/6 and 6/9
to find equivalent fractions, you just have to reduce the fraction or multiply both the top and bottom numbers by the same number to increase the fraction.
2/3 is as reduced as it can be, though.
Answer:
the answer may be this: 8 √3
Answer:
(0, -3)
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
- Subtraction Property of Equality<u>
</u>
<u>Algebra I</u>
- Terms/Coefficients
- Coordinates (x, y)
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
6x - 5y = 15
x = y + 3
<u>Step 2: Solve for </u><em><u>y</u></em>
<em>Substitution</em>
- Substitute in <em>x</em>: 6(y + 3) - 5y = 15
- Distribute 6: 6y + 18 - 5y = 15
- Combine like terms: y + 18 = 15
- [Subtraction Property of Equality] Subtract 18 on both sides: y = -3
<u>Step 3: Solve for </u><em><u>x</u></em>
- Define original equation: x = y + 3
- Substitute in <em>y</em>: x = -3 + 3
- Add: x = 0
Answer:
Probability of having at least 4 Girls
= 0.6875
Step-by-step explanation:
Probability of having at least 4 Girls is 1-probability of having exactly 3 girls
Total number of children= 5 = N
Probability of having a girl p = 0.5
Probability of not having a girl q= 0.5
X= 3
Probability of at least 4 girls is given by
Probability= NCX(p)^x(q)^(N-x)
Probability = 5C3(0.5)^3(0.5)^(5-3)
Probability = 5C3(0.5)^3(0.5)^2
Probability= 5!/3!2!(0.5)^3(0.5)^2
Probability= 10(0.125)(0.25)
Probability= 0.3125
Probability of having at least 4 Girls
= 1- 0.3125
= 0.6875
