Cherese started a Green Club at her school and needs to appoint a board of members. There are eight people available for these roles. How many different selections of four out of eight people could she choose? Write Pascal’s Triangle from row 0 to row 8 and use it to solve the problem.(1 point)

Question

Responses

2
2

56
56

35
35

70

Answer 1

To find the number of different selections of four out of eight people, we can use combinations, which is represented as \( C(n, k) \), where \( n \) is the total number of people to choose from, and \( k \) is the number of people to choose. In this case, we want \( C(8, 4) \).

Using Pascal's Triangle, we list rows from 0 to 8. Each entry in Pascal's Triangle can be calculated by adding the two entries directly above it.

Here is Pascal’s Triangle from row 0 to row 8:

Row 0:         1
Row 1:       1   1
Row 2:     1   2   1
Row 3:   1   3   3   1
Row 4: 1   4   6   4   1
Row 5: 1   5  10  10   5   1
Row 6: 1   6  15  20  15   6   1
Row 7: 1   7  21  35  21   7   1
Row 8: 1   8  28  56  70  28   8   1

To find \( C(8, 4) \), we look at row 8 in Pascal's Triangle. The 5th element (0-indexed) in row 8 is 70.

Thus, the number of different selections of four out of eight people is 70.

So the correct response is: 70

Answer 2

Which of the following establishes the Binomial Theorem for the expansion of polynomials in the form (a+b)n ?(1 point)
Responses

(a+b)n=bn+nC1abn−1+nC2a2bn−2+⋯+nCn−1an−1b+an
left parenthesis a plus b right parenthesis superscript n baseline equals b superscript n baseline plus subscript n baseline upper C subscript 1 baseline a b superscript n minus 1 baseline plus subscript n baseline upper C subscript 2 baseline a squared b superscript n minus 2 baseline plus dot dot dot plus subscript n baseline upper C subscript n minus 1 baseline a superscript n minus 1 baseline b plus a superscript n baseline

(a+b)n=an+bn
left parenthesis a plus b right parenthesis superscript n baseline equals a superscript n baseline plus b superscript n baseline

(a+b)n=an+nC1an−1b+nC2an−2b2+⋯+nCn−1abn−1+bn
left parenthesis a plus b right parenthesis superscript n baseline equals a superscript n baseline plus subscript n baseline upper C subscript 1 baseline a superscript n minus 1 baseline b plus subscript n baseline upper C subscript 2 baseline a superscript n minus 2 baseline b squared plus dot dot dot plus subscript n baseline upper C subscript n minus 1 baseline a b superscript n minus 1 baseline plus b superscript n baseline

(a+b)n=an+1Cnan−1b+2Cnan−2b2+⋯+n−1Cnabn−1+bn

Answer 3

The Binomial Theorem gives a formula for the expansion of a binomial expression raised to a power \( n \). The correct form for the expansion is:

\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In expanded form, the theorem provides:

\[ (a+b)^n = a^n + \binom{n}{1} a^{n-1} b + \binom{n}{2} a^{n-2} b^2 + \ldots + \binom{n}{n-1} a b^{n-1} + b^n \]

Now, looking at the provided options, the correct choice is:

(a+b)n=an+nC1an−1b+nC2an−2b2+⋯+nCn−1abn−1+bn

This option correctly follows the Binomial Theorem expansion and captures all the necessary elements.

Answer 4

Jaime applies the Binomial Theorem for the expansion of (x−5)4 . Which of the following is the x2 term?(1 point)
Responses

−30x2
negative 30 x squared

6x2
6 x squared

150x2
150 x squared

−150x2

Answer 5

To find the \( x^2 \) term in the expansion of \( (x - 5)^4 \) using the Binomial Theorem, we can use the formula:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In this case, we have \( a = x \), \( b = -5 \), and \( n = 4 \).

We want to find the coefficient of the \( x^2 \) term. This corresponds to \( k \) for which \( n-k = 2 \), or \( k = 4 - 2 = 2 \).

Using this, the \( x^2 \) term can be calculated as follows:

\[ \text{Term} = \binom{4}{2} x^2 (-5)^2 \]

Calculating each part:

1. Calculate \( \binom{4}{2} \):

\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

2. Calculate \( (-5)^2 \):

\[ (-5)^2 = 25 \]

Now, substituting these values into the term:

\[ \text{Term} = 6 x^2 \cdot 25 = 150 x^2 \]

Thus, the \( x^2 \) term in the expansion of \( (x - 5)^4 \) is 150x².

So the correct answer is:

150x²

Answer 6

The probability of drawing a red ball out of a bag containing one red ball and three yellow balls is 14 . What is the probability of drawing the red ball twice out of five attempts? Apply the formula nCkpkqn−k to solve.(1 point)
Responses

3 percent
3 percent

26 percent
26 percent

9 percent
9 percent

16 percent

Answer 7

To find the probability of drawing the red ball twice out of five attempts, we can use the binomial probability formula:

\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

Where:

• \( n \) is the total number of trials (in this case, 5 attempts),
• \( k \) is the number of successful outcomes (in this case, drawing the red ball 2 times),
• \( p \) is the probability of drawing a red ball,
• \( q \) is the probability of not drawing a red ball (drawing a yellow ball).

Step 1: Calculate \( p \) and \( q \)

Since there is 1 red ball and 3 yellow balls, the total number of balls is:

\[ 1 + 3 = 4 \]

The probability \( p \) of drawing a red ball is:

\[ p = \frac{1}{4} \]

The probability \( q \) of drawing a yellow ball is:

\[ q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \]

Step 2: Apply the formula

Now we want to find the probability of drawing the red ball \( k = 2 \) times in \( n = 5 \) attempts:

\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^{5-2} \]

Step 3: Calculate \( \binom{5}{2} \)

\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

Step 4: Calculate \( P(X = 2) \)

Now we can substitute everything back into the formula:

\[ P(X = 2) = 10 \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^3 \]

Calculating each term:

1. \( \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)
2. \( \left(\frac{3}{4}\right)^3 = \frac{27}{64} \)

Now substitute these values into the equation:

\[ P(X = 2) = 10 \times \frac{1}{16} \times \frac{27}{64} \]

Combine the fractions:

\[ P(X = 2) = 10 \times \frac{27}{1024} = \frac{270}{1024} \]

Step 5: Simplify and calculate the percentage

Now simplify \( \frac{270}{1024} \):

Since \( 270 \) and \( 1024 \) have no common factors (other than 1), we can calculate the percentage:

\[ \frac{270}{1024} \approx 0.26367 \text{ or } 26.37% \]

Conclusion

Rounding to the nearest whole number, the probability of drawing the red ball twice out of five attempts is approximately 26 percent.

So the correct answer is: 26 percent