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pishuonlain [190]
3 years ago
5

(20 points) find a fraction equal to 1/3 by multiplying 1/3 by 2/2 write the fraction, then add it to 3/6 what is the sum?

Mathematics
1 answer:
kenny6666 [7]3 years ago
5 0

Answer:

it is 5/6

Step-by-step explanation:

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A youth ice hockey game has 3 periods that are each 20 minutes long. Colin plays 12 minutes each period. Which ratio shows Colin
labwork [276]
The answer is 12:20.......................

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Is the point (-3,2) part of the solution set to the system y < -4x - 3, x + 8y > 7
gregori [183]

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Yes

Step-by-step explanation:

If you replace each x with -3 and each y with 2 you get:

1) 2<-4*(-3)

2<12

True

2) -3+8*2>7

13>7

True

Therefore the point is part of the solution set

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Help please !!!! I need this I don’t need a detailed explanation
natta225 [31]
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The answer is C, -6/5
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Can someone please help me, Thanks - Alyssa T.
Oxana [17]
A=LW
192=(13+x)((9+x)
192=117+22x+x^2 (subtract 192 from both sides)
x^2+22x-75+0 (factor)
(x+25)(x-3)=0
so x=-25 or 3
Check work
since we are adding x it can only be the positive solution so x=3 so the new dimensions are:
L=13+3=16m
W=9+3=12
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7 0
3 years ago
Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Vis
Mice21 [21]

Answer:

Step-by-step explanation:

Given that,

Visa card is represented by P(A)

MasterCard is represented by P(B)

P(A)= 0.6

P(A')=0.4

P(B)=0.5

P(B')=0.5

P(A∩B)=0.35

1. P(A U B) =?

P(A U B)= P(A)+P(B)-P(A ∩ B)

P(A U B)=0.6+0.5-0.35

P(A U B)= 0.75

The probability of student that has least one of the cards is 0.75

2. Probability of the neither of the student have the card is given as

P(A U B)'=1-P(A U B)

P(A U B)= 1-0.75

P(A U B)= 0.25

3. Probability of Visa card only,

P(A)= 0.6

P(A) only means students who has visa card but not MasterCard.

P(A) only= P(A) - P(A ∩ B)

P(A) only=0.6-0.35

P(A) only=0.25.

4. Compute the following

a. A ∩ B'

b. A ∪ B'

c. A' ∪ B'

d. A' ∩ B'

e. A' ∩ B

a. A ∩ B'

P(A∩ B') implies that the probability of A without B i.e probability of A only and it has been obtain in question 3.

P(A ∩ B')= P(A-B)=P(A)-P(A∩ B)

P(A∩ B')= 0.6-0.35

P(A∩ B')= 0.25

b. P(A ∪ B')

P(A ∪ B')= P(A)+P(B')-P(A ∩ B')

P(A ∪ B')= 0.6+0.5-0.25

P(A ∪ B')= 0.85

c. P(A' ∪ B')= P(A')+P(B')-P(A' ∩ B')

But using Demorgan theorem

P(A∩B)'=P(A' ∪ B')

P(A∩B)'=1-P(A∩B)

P(A∩B)'=1-0.35

P(A∩B)'=0.65

Then, P(A∩B)'=P(A' ∪ B')= 0.65

d. P( A' ∩ B' )

Using demorgan theorem

P(A U B)'= P(A' ∩ B')

P(A U B)'= 1-P(A U B)

P(A' ∩ B')= 1-0.75

P(A' ∩ B')= 0.25

P(A U B)'= P(A' ∩ B')=0.25

e. P(A' ∩ B)= P(B ∩ A') commutative law

Then, P(B ∩ A') = P(B) only

P(B ∩ A') = P(B) -P(A ∩ B)

P(B ∩ A') =0.5 -0.35

P(B ∩ A') =0.15

P(A' ∩ B)= P(B ∩ A') =0.15

5 0
3 years ago
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