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3 years ago

How many leaves on a tree diagram are needed to represent all possible combinations tossing a coin 3 times?

2 answers:

8 0

N = 3 is the number of times the coin is tossed. Since there are 2 outcomes per toss, this means that there are 2^n = 2^3 = 2*2*2 = 8 outcomes total. Therefore, there are 8 leaves total.

If we make
H = heads
T = tails

Then here are the 8 outcomes in the sample space
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT

vladimir1956 [14]3 years ago

4 0

There are 8 combinations. A P E X :)

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Y= - 2x + 4 2x + y = 4

Tomtit [17]

Answer:

Infinite amount of solutions

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Terms/Coefficients
Coordinates (x, y)
Solving systems of equations using substitution/elimination

Step-by-step explanation:

Step 1: Define Systems

y = -2x + 4

2x + y = 4

Step 2: Solve for x

Substitution

Substitute in y: 2x + (-2x + 4) = 4
Combine like terms: 4 = 4

Here we see that 4 does indeed equal 4.

∴ the systems of equations has an infinite amount of solutions.

7 0

3 years ago

Read 2 more answers

Please answer i need it quick :(

Oksi-84 [34.3K]

Not sure but I think its 120
I think 120 times 2 and I got 240. Then I added the number of oak trees which was 120 and got 360

5 0

3 years ago

Read 2 more answers

Expand (2x+2)^6 How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

8 0

2 years ago

Can someone help me? Files attached! What answer choices go to what figure? :D

den301095 [7]

Since the center is considered the corner that they both share, here are the answers figure a goes with the first, figure b goes with the last, figure c goes with the second, and figure d goes with the third. Hope this helps.

3 0

4 years ago

Please help with a math problem :/ Please show steps... x^2 -13x = -36

Marianna [84]

X^2 - 13 x = -36
add 36 to both sides

x^2 - 13x +36 = 0
figure out the factor
(x-4)(x-9) = 0
so (x-4) or (x-9) equals 0
so solution is
x-4=0 add 4 to both sides and get x = 4
x-9=0 add 9 to both sides and get x=9
x={ 4, 9}

4 0

4 years ago