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Tanzania [10]
3 years ago
10

What is the commutative, associative, and distributive property of 3a plus 5a plus 2b

Mathematics
2 answers:
anyanavicka [17]3 years ago
5 0
Question: What is the commutative, associative, and distributive property of 3a plus 5a plus 2b0.


Answer:Defintion of  Commutative property holds for both addition and multiplication. That is: the operations addition and multiplication are commutative over the set of real numbers. That means, for any two real numbers x and. y, x + y = y + x and xy = yx. Definiton of associative In mathematics, the associative property is a property of some binary operations. In propositional logic, associativity is a valid rule of replacement for expressions in logical proofs and distrive property means The Distributive Property is an algebra property which is used to multiply a single term and two or more terms inside a set of parentheses. I really hope this helped thx for the question!
VLD [36.1K]3 years ago
3 0
Commutative: a+b=b+a
associative: (a+b)+c=a+(b+c)
distributive: ab+ac=a(b+c)

so
3a+5a+2b
we can show distributive with the a terms
3a+5a=a(3+5)=8a
8a+2b
we can distribute again
2(4a)+2b=2(4a+b)

not sure how to show commutative or associative
I guess
3a+5a+2b=2b+3a+5a
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3 years ago
According to a​ survey, 60 ​% of the residents of a city oppose a downtown casino. Of these 60 ​% about 8 out of 10 strongly opp
katrin [286]

Answer:

(a) 0.48

(b) 0.20

(c) it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the​ casino.

Step-by-step explanation:

​(a) Find the probability that a randomly selected resident opposes the casino and strongly opposes the casino. ​

The probability that a radomly selected resident opposes the casino and strongly opposes the cassino is the product of the two probabilities, that a resident opposes the casino and that it strongly opposes the casino (once it is in the first group) as it is shown below.

Use this notation:

  • Probability that a radomly selected resident opposes the casino: P(A)

  • Probability that a resident who opposes the casino strongly opposes it: P(B/A), because it is the probability of event B given the event A

i) Determine the <em>probability that a radomly selected resident opposes the casino</em>, P(A)

Probability = number of favorable outcomes / number of possible outcomes

  • P(A) is <em>given as 60%</em>, which in decimal form is 0.60

ii) Next, determine,the <em>probability that a resident who opposes the casino strongly opposes it</em>, P(B/A):

  • It is given as 8 out of 10 ⇒ P(B/A) = 8/10

iii) You want the probability of both events, which is the joint probability or  intersection: P(A∩B).

So, you can use the definition of conditional probability:

  • P(B/A) = P(A∩B) / P(A)

iv) From which you can solve for P(A∩B)

  • P(A∩B) = P(B/A)×P(A) =  (8/10)×(0.60) = 0.48

(b) Find the probability that a randomly selected resident who opposes the casino does not strongly oppose the casino.

In this case, you just want the complement of the probability that <em>a radomly selected resident who opposes the casino does strongly oppose the casino</em>, which is 1 - P(B/A) = 1 - 8/10 = 1 - 0.8 = 0.2.

​(c) Would it be unusual for a randomly selected resident to oppose the casino and strongly oppose the​ casino?

You are being asked about the joint probability (PA∩B), which you found in the part (a) and it is 0.48.

That is almost 0.50 or half of the population, so you conclude it is not unusual for a radomly selected resident to oppose the casino and strongly oppose the​ casino.

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statuscvo [17]

Answer:

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Step-by-step explanation:

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Hope this helps :)

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*see photo for complete solution*

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