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Pani-rosa [81]
3 years ago
8

(See The Picture To Know The Problem, also it's REALLY bad)

Mathematics
1 answer:
Sophie [7]3 years ago
4 0
The answer would be the second option , B
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Solve pls brainliest
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Have a great day!!!!

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The combined weight of the people in the elevator is 49% of the 3,025 pound capacity. Which would be The best way to estimate th
qaws [65]

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Around 1,500 pounds

Step-by-step explanation:

Because you said "estimate," I will be estimating this.

49% rounds up to 50%

3,025 rounds down to 3,000

50% of 3,000 is 1,500 pounds.

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What two numbers multiply to 6 but add up to -1?
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-3 and 2

Step-by-step explanation:

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During the holiday sale, the store discounted all items 1/3. What is the discount on a $19.95 item? plz help me
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Discount Final Price Each

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3 years ago
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Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card an
ziro4ka [17]

Answer:

1) is not possible

2) P(A∪B) = 0.7

3) 1- P(A∪B) =0.3

4) a) C=A∩B' and P(C)= 0.3

b)  P(D)= 0.4

Step-by-step explanation:

1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4  . Thus the maximum possible value of P(A∩B) is 0.4

2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by

P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7

P(A∪B) = 0.7

3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3

4) the event C that the selected student has a visa card but not a MasterCard is given by  C=A∩B'  , where B' is the complement of B. Then

P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3

the probability for the event D=a student has exactly one of the cards is

P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4

3 0
3 years ago
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