5(−3)+4=60−1(-5) help

The numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacemen

mamaluj [8]

<h2>Answer:</h2>

(a)

The probability is : 1/2

(b)

The probability is : 1/2

<h2>Step-by-step explanation:</h2>

The numbers 1, 2, 3, 4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacement.

The total combinations that are possible are:

(1,2) (1,3) (1,4) (1,5)

(2,1) (2,3) (2,4) (2,5)

(3,1) (3,2) (3,4) (3,5)

(4,1) (4,2) (4,3) (4,5)

(5,1) (5,2) (5,3) (5,4)

i.e. the total outcomes are : 20

(a)

Let A denote the event that the first number is 4.

and B denote the event that the sum is: 9.

Let P denote the probability of an event.

We are asked to find:

P(A|B)

We know that it could be calculated by using the formula:

$P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}$

Hence, based on the data we have:

$P(A\bigcap B)=\dfrac{1}{20}$

( Since, out of a total of 20 outcomes there is just one outcome which comes in A∩B and it is: (4,5) )

and

$P(B)=\dfrac{2}{20}$

( since, there are just two outcomes such that the sum is: 9

(4,5) and (5,4) )

Hence, we have:

$P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}$

(b)

Let A denote the event that the first number is 3.

and B denote the event that the sum is: 8.

Let P denote the probability of an event.

We are asked to find:

P(A|B)

Hence, based on the data we have:

$P(A\bigcap B)=\dfrac{1}{20}$

( since, the only outcome out of 20 outcomes is: (3,5) )

and

$P(B)=\dfrac{2}{20}$

( since, there are just two outcomes such that the sum is: 8

(3,5) and (5,3) )

Hence, we have:

$P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}$

7 0

3 years ago

Which function is linear?

natali 33 [55]

Answer:

B) f(x)=13x+4 and C) f(x)=|3x| − 4

5 0

2 years ago

Evaluate the integral by interpreting it in terms of areas Draw a picture of the region the integral

denis23 [38]

Answer:

Step-by-step explanation:

The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!