Find the slope of the line that passes through these two points. (-2,1);(3,2) simplify completely

Anyone can help me to solve this?

marishachu [46]

Answer:

$x^3+3x^2+2x$

Step-by-step explanation:

Definitions

Integer: A whole number that can be positive, negative, or zero.

Consecutive Integers: Integers that follow each other continuously in the order from smallest to largest.

Product: The result of multiplying two of more values together.

Let x be the first integer.

Therefore, to find the consecutive integers, add 1 to each term:

x
x + 1
(x + 1) + 1 = x + 2

Therefore, the product of the first three consecutive integers is:

$\begin{aligned}x(x+1)(x+2) & = (x \cdot x+x \cdot 1)(x+2)\\& = (x^2+x)(x+2)\\ & = x^2 \cdot x+x^2 \cdot 2+x \cdot x+x \cdot 2\\& = x^3+2x^2+x^2+2x\\& = x^3+3x^2+2x\end{aligned}$

6 0

1 year ago

Read 2 more answers

The four-member math team at Pecanridge Middle School is chosen from the math club, which has three girls and five boys. How man

cestrela7 [59]

Answer:

$Total\ Selection = 30\ ways$

Step-by-step explanation:

Given

Girls = 3

Boys = 5

Required

How many ways can 2 boys and girls be chosen?

The keyword in the question is chosen;

This implies combination and will be calculated as thus;

$Selection =\ ^nC_r = \frac{n!}{(n-r)!r!}$

For Boys;

n = 5 and r = 2

$Selection =\ ^5C_2$

$Selection = \frac{5!}{(5-2)!2!}$

$Selection = \frac{5!}{3!2!}$

$Selection = \frac{5 * 4 * 3!}{3!*2 * 1}$

$Selection = \frac{20}{2}$

$Selection = 10$

For Girls;

n = 3 and r = 2

$Selection =\ ^3C_2$

$Selection = \frac{3!}{(3-2)!2!}$

$Selection = \frac{3!}{1!2!}$

$Selection = \frac{3 * 2!}{1 *2!}$

$Selection = \frac{3}{1}$

$Selection = 3$

Total Selection is calculated as thus;

$Total\ Selection = Boys\ Selection * Girls\ Selection$

$Total\ Selection = 10 * 3$

$Total\ Selection = 30\ ways$

5 0

4 years ago

Leon has a bag of 4 red balls, 3 green balls and 6 yellow balls. He takes a ball out and replaces it 50 times. Out of those 50

faltersainse [42]

Well if he gets one of hes green balls out 12 times the next time he has a chance of 18%

6 0

3 years ago

The Lewis family went out to an expensive restaurant for dinner. The bill was $108.00. Mr.Lewis wanted to leave a 15% tip. How m

Luden [163]

So you find 15% of the $108 which is $16.20 you do this by 15/100*108. Then you add $16.20 to 108 and you get $124.20.

So the answer is $124.20

3 0

3 years ago

A coin is tossed 13 times.

Serga [27]

Answer:

a) 2¹³

b) 0.0873

c) 0.9983

d) 0.9538

Step-by-step explanation:

a) How many different outcomes are possible?

For every toss, there are two possible outcomes (heads or tails), so if we toss the coin 13 times the total of outcomes possible is 2¹³

b) What is the probability of getting exactly 4 heads?

The definition of probability is an event is: number of favourable events/ number of total events.

When we have two possible outcomes p, q and we repeat the experiment multiple times, we apply the Binomial distribution, in this distribution the probability of getting exactly k successes in n trials is given by

$P(X=k) =nCk (p^{k} )(1-p)^{n-k}$ where p represents the probability of success. nCk refers to the combinations of k elements out of n elements.

So we have to know how many different ways we can get exactly 4 heads.

let's name p = probability of getting heads = 0.5

1 -p = probability of getting tails = 0.5

We are going to toss the coin 13 different times and we want to get heads 4 times and tails 9 times.

So, by the Binomial Distribution we would have:

P(X=4) = 13C4 (0.5)⁴(0.5)⁹ = 715 (0.5)¹³ = 0.0873

Thus, the probability of getting exactly 4 heads is 0.0873.

c) What is the probability of getting at least 2 heads?

We are going to use the same binomial distribution but now we need at least to heads, this can be written as 1 - (P(X=0) + P(X=1))

P(X=0) + P(X=1) = 13C0 (0.5)⁰(0.5)¹³ + 13C1 (0.5)¹(0.5)¹² = 0.0001220703125 + 0.0015869140625 = 0.001708984375

1 - 0.001708984375 = 0.998291015625

The probability of getting at least 2 heads is 0.9983

d) What is the probability of getting at most 9 heads?

We are going to use the same binomial distribution but now we need at most 9 heads. This can be written as 1 - (P(X=10) + P(X = 11) + P(X=12) + P(X=13))

1 - (P(X=10) + P(X = 11) + P(X=12) + P(X=13)) =1 -(13C10(0.5)¹⁰(0.5)³ + 13C11(0.5)¹¹(0.5)²+ 13C12(0.5)¹²(0.5)¹ + 13C13(0.5)¹³(0.5)⁰) = 0.95385742188

The probability of getting at most 9 heads is 0.9538

6 0

3 years ago