Does anyone get this???

Identify whether or not each pair of terms can be combined to so oily the expression (3x + 1) + (2x -5)

eimsori [14]

Answer:

3x and 2x can be combined bc it has ex at the end.

1 and 2x cannot be combined bc 1 doesn't have an x.

1 and -5 can be combined

6 0

2 years ago

Add.<br><br> 0.125 + 0.25 + 0.33

ivann1987 [24]

Answer:

0.705

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the value of 40/z, when z = 10? <br><br>I WILL GIBE BRAINLY POINTS

Mamont248 [21]

Replace x with 10 and simplify the fraction.

40/10 = 4/1 = 4

The answer is 4

5 0

1 year ago

3. Two balls are drawn in succession without replacement from an urn containing 4 red balls and 3 blue balls. Find the probabili

choli [55]

Answer:

X ____ 0 ______ 1 _______ 2

P(x) __ 1/7 _____ 4/7 _____ 2/7

Step-by-step explanation:

Given that :

Number of red balls = 4

Number of Blue balls = 3

Total number of balls = (4 + 3) = 7

Picking without replacement :

Number of picks = 2

1st pick is red :

Probability = (required outcome / Total possible outcomes)

USing combination formula :

nCr = n! / (n-r)! r!

Total possible outcome :

7C2 = 7! / 5!2! = 7*6 / 2 = 42 / 2 = 21

X = number of red

For red = 0;

P(x =0) = required outcome / Total possible outcomes

P(x = 0)

Required outcome : [4C0 * 3C2)]

P(x = 0) = [(4C0 * 3C2)] / 21 = (1*3)/21 = 3/21 = 1/7

P(x = 1)

Required outcome : [4C1 * 3C1)]

P(x = 1) = [(4C1 * 3C1)] / 21 = (4*3)/21 = 12/21 = 4/7

P(x = 2)

Required outcome : [4C2 * 3C0)]

P(x = 2) = [(4C2 * 3C0)] / 21 = (6*1)/21 = 6/21 = 2/7

Hence ;

X ____ 0 ______ 1 _______ 2

P(x) __ 1/7 _____ 4/7 _____ 2/7

8 0

2 years ago

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

So

$P = 0.75(0.25) = 0.1875$