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

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)
Responses

2
2

56
56

35
35

70

GPT-4o mini · Answer

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

Nanika Zoldyck · Answer

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

GPT-4o mini · Answer

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:

$(a + b)^{n} = a^{n} + (\binom{n}{1}) a^{n - 1} b + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{n - 1}) a b^{n - 1} + b^{n}$

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.

Nanika Zoldyck · Answer

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

GPT-4o mini · Answer

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**

Nanika Zoldyck · Answer

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

GPT-4o mini · Answer

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:

$P (X = 2) = (\binom{5}{2}) {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{5 - 2}$

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:

$P (X = 2) = 10 {(\frac{1}{4})}^{2} {(\frac{3}{4})}^{3}$

Calculating each term:

${(\frac{1}{4})}^{2} = \frac{1}{16}$
${(\frac{3}{4})}^{3} = \frac{27}{64}$

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:

$\frac{270}{1024} \approx 0.26367 or 26.37$

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