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Dvinal [7]
3 years ago
13

What conclusion can you make from the information below?

Mathematics
1 answer:
Alina [70]3 years ago
3 0
C and D i think

I hope this helps
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10 points! Please help!!!!! I dont understand :(
kifflom [539]

Answer:

Z(integer)

Step-by-step explanation:

N is subset of W

W is subset of Z

Z is subset of Q

Q and Q* are subset of R

7 0
3 years ago
Which of the following is an example of the associative property of multiplication?
Zarrin [17]
The second one is the correct answer!
7 0
3 years ago
Read 2 more answers
A tank is filled with water at a rate of 2.4 liters per minute. At this rate, how
TEA [102]

Answer: 35 minutes

Step-by-step explanation:

84 liters divided by 2.4 liters per minute

84/2.4=35 minutes

4 0
3 years ago
Two candidates face each other in an election. The Democratic candidate is supported by 46% of the population, and the Republica
7nadin3 [17]

Answer:

27.98% probability that less than half of them (3 or fewer) would support the Republican candidate

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they support the Republican candidate, or they do not. The people are chosen at random, which means that the probability of them supporting the republican candidate is independent from other people. So 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}.p^{x}.(1-p)^{n-x}

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

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

And p is the probability of X happening.

In this problem we have that:

The Republican candidate is supported by 54%. This means that p = 0.54

Suppose you run a poll of 8 people (randomly choose 8 people). What is the probability that less than half of them (3 or fewer) would support the Republican candidate?

This is P(X \leq 3) when n = 8.

So

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{8,0}.(0.54)^{0}.(0.46)^{8} = 0.0020

P(X = 1) = C_{8,1}.(0.54)^{1}.(0.46)^{7} = 0.0188

P(X = 2) = C_{8,2}.(0.54)^{2}.(0.46)^{6} = 0.0774

P(X = 3) = C_{8,3}.(0.54)^{3}.(0.46)^{5} = 0.1816

So

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0020 + 0.0188 + 0.0774 + 0.1816 = 0.2798

27.98% probability that less than half of them (3 or fewer) would support the Republican candidate

3 0
3 years ago
1. (10pts] Let A = {1, 2, 3, 4, 5}, let B = {1,4,5,7,8,9}, and let C = {2, 4, 6, 7,9}. Determine each of the following (a) An Bn
alisha [4.7K]

Answer and explanation:

Given : Let A = {1, 2, 3, 4, 5}, let B = {1,4,5,7,8,9}, and let C = {2, 4, 6, 7,9}.

To find : Determine each of the following,

a) (A\cap B)\cap C

b) (A\cup B)\cap (A\cup C)

c) A - (B\cup C)

d) (A-C) (C-A)

Solution :

The union of two sets is a new set that contains all of the elements that are in at least one of the two sets.

The intersection of two sets is a new set that contains all of the elements that are in both sets.

a) (A\cap B)\cap C

A\cap B=\{1,4,5\}

Then, (A\cap B)\cap C=\{4\}

b) (A\cup B)\cap (A\cup C)

A\cup B=\{1,2,3,4,5,7,8,9\}

A\cup C=\{1,2,3,4,5,7,9\}

(A\cup B)\cap (A\cup C)=\{1,2,3,4,5,7,9\}

c) A - (B\cup C)

B\cup C=\{1,2,4,5,6,7,8,9\}

A - (B\cup C)=\{3\}

d) (A-C) (C-A)

A-C=\{1,3,5\}

C-A=\{6,7,9\}

(A-C) (C-A)=\{1,3,5\}\times\{6,7,9\}

(A-C) (C-A)=\{(1,6),(1,7),(1,9),(3,6),(3,7),(3,9),(5,6),(5,7),(5,9)\}

5 0
3 years ago
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