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1 year ago

Expand (2x-3y)^4 using Pascal's Triangle. Show work

1 answer:

3 0

(2x+3y)⁴
1) let 2x = a and 3y = b

(a+b)⁴ = a⁴ + a³b + a²b² + ab³ + b⁴
Now let's find the coefficient of each factor using Pascal Triangle

0 | 1
1 | 1 1
2 | 1 2 1
3 | 1 3 3 1
4 | 1 4 6 4 1

0,1,2,3,4,.. represent the exponents of binomials
Since our binomial has a 4th exponents, the coefficients are respectively:

(1)a⁴ + (4)a³b + (6)a²b² + (4)ab³ + (1)b⁴
Now replace a and b by their real values in (1):

2⁴x⁴ +(4)8x³(3y) + (6)(2²x²)(3²y²) + (4)(2x)(3³y³) + (1)(3⁴)(y⁴)

16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴

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What is 1/999 as a decimal

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Step-by-step explanation: I have no idea. just put it in a calculator

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1 year ago

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Write an arithmetic series for which S5 = 10.

Ludmilka [50]

S5=10 means the 5th number is 10.
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2 years ago

What is the graph of the system y = -2x + 3 and 2x + + 4y = 8?

Maslowich

Answer:what are you solving it by

Step-by-step explanation:

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2 years ago

Exams are approaching and Helen is allocating time to studying for exams. She feels that with the appropriate amount of studying

svp [43]

Answer: 0.05

Step-by-step explanation:

Let M = Event of getting an A in Marketing class.

S = Event of getting an A in Spanish class,

i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45

Required probability = P(neither M nor S)

= P(M'∩S')

= P(M∪S)' [∵P(A'∩B')=P(A∪B)']

=1- P(M∪S) [∵P(A')=1-P(A)]

= 1- (P(M)+P(S)- P(M∩S)) [∵P(A∪B)=P(A)+P(B)-P(A∩B)]

= 1- (0.80+0.60-0.45)

= 1- 0.95

= 0.05

hence, the probability that Helen does not get an A in either class= 0.05

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2 years ago

A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if

topjm [15]

Answer:

a) The expected value is $\frac{-1}{15}$

b) The variance is $\frac{49}{45}$

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is $\binom{5}{2}$ . Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is $\binom{5}{1}\cdot \binom{5}{1}$ . So, we define the following probabilities.