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3 years ago

True or false The coefficient of (x^(k)) (y^(n-k)) in the expansion of (x+y)^n equals (n choose k)

1 answer:

8 0

This is under the binomial theorem. A binomial theorem is an equation that predicts the sequence when a binomial is raised to certain power. The general form of the equation is (a+b)^n. The equation for the binomial theorem would be

nCk a^(n-k) b^k, where k is the kth term of the expanded form, n is the nth power. The coefficient of the term in the binomial expansion is nCk or n!/k!(n-k)!.

You might be interested in

Which phrases can be used to represent the inequality 12 minus 3 x greater-than-or-equal-to 9? Select two options.

AlexFokin [52]

Answer:

Twelve less three times a number is at least nine.

Twelve minus the product of three and a number is no smaller than nine.

Step-by-step explanation:

complete question is:

Twelve less than three times a number is greater than or equal to nine.

Twelve less than three times a number is at least nine.

Twelve less three times a number is at least nine.

Twelve minus the product of three and a number is no more than nine.

Twelve minus the product of three and a number is no smaller than nine.

ANSWER:

<u>Twelve less three times a number is at least nine.</u>

The phrase "twelve less three..." means that the "three times a number" is being subtracted from twelve. So, it has to be at least nine.

<u>Twelve minus the product of three and a number is no smaller than nine.</u>

Even the least number you suppose is 1 we'll get 9

e.g: 12-3(1)=9. Therefore, this statement also represent the inequality

3 0

4 years ago

Read 2 more answers

You have a six-sided die that you roll once. Let Ri denote the event that the roll is i. Let G j denote the event that the roll

Misha Larkins [42]

Answer:

a. $P (R3 | G1)=\frac{1}{5}$

b. $P (R6| G3)= \frac{1}{3}$

c. $P(G3|E)=\frac{2}{3}$

d. $P (E|G3)=\frac{2}{3}$

Step-by-step explanation:

The sample space associated with the random experiment of throwing a dice is is the equiprobable space {R1, R2, R3, R4, R5, R6}. Then,

a. The conditional probability that 3 is rolled given that the roll is greater than 1? $P (R3 | G1) = \frac{P (R3\bigcap G1)}{P(G1)} = \frac{1/6}{5/6} = \frac{1}{5}$

b. What is the conditional probability that 6 is rolled given that the roll is greater than 3? $P (R6| G3) = \frac{P (R6\bigcap G3)}{P(G3)} = \frac{1/6}{3/6} = \frac{1}{3}$

c. What is P [GIE], the conditional probability that the roll is greater than 3 given that the roll is even? $P(G3|E) = \frac{P (G3\bigcap E)}{P(E)} = \frac{2/6}{3/6} = \frac{2}{3}$

d. Given that the roll is greater than 3, what is the conditional probability that the roll is even? $P (E|G3) = \frac{P (E\bigcap G3)}{P(G3)} = \frac{2/6}{3/6} = \frac{2}{3}$

6 0

3 years ago

30 people are running for an office to be chosen randomly. You plan to use a 6-sided dice and a spinner. How many sections need

allsm [11]

The number of sections on the spinner so that each person has heir own shape and number is 5

<h3>How to determine the number of sections?</h3>

The given parameters are:

People, n = 30

Die = 6 sided

Let the sections on the spinner be s.

The equation to calculate s is represented as:

People = Die * s

This gives

30 = 6 * s

Divide both sides by 6

s = 5

Hence, the number of sections on the spinner so that each person has heir own shape and number is 5