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Alik [6]
3 years ago
15

The length of a rectangle is four times its width. If the perimeter is at most 106 centimeters, what is the greatest possible va

lue for the width? Which inequality models this problem?
A. 2w+2 x (46) _< 106

B. 2w+ 2 x (4) _> 106

C. 2w+2 x (4w) < 106

D. 2w+2 x (46) > 106
Mathematics
2 answers:
andrezito [222]3 years ago
8 0

Answer:

Greatest possible value for the width is 10.6 centimeters.

D. 2w+2 × (46) > 106

Step-by-step explanation:

Perimeter of a rectangle = 2(l + w)

From the given question, P = 106 and l =4w

So that,

  106 = 2(4w + w)

   106 = 2(5w)

   106 = 10 w

      w = 10.6 centimeters

So, the greatest possible value for the width is 10.6 centimeters.

Model;  2w+2 × (46) > 106

      = 2 × 10.6 + 2 × (46)

       = 113.2

113.2 is greater than 106

RideAnS [48]3 years ago
5 0

Answer: A

Step-by-step explanation: The answer is w≤ 13, which means that the greatest value possible is 13.

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3 years ago
P(A)=25,P(B|A)=920,P(A∩B)=?
larisa [96]
Hi! Let me help you!
This problem is a Venn Diagram question. It is much easier to answer this question with the visual it usually comes with. However, we can employ logic to give this one a smart and logical answer.

What we have so far:
P(A) =  25. This means that the shaded area of A is equal to 25.
P(B/A) = 920. This means that area A subtracted 25 to the value of area B which resulted to 920.
P(B) = 945. How did I arrive with 945? Simple, I just added 25 to 920. Go and be a smart lad and find why I did that.

Solution:
Solving P(A∩B) is quite challenging because like I said, a visual is needed for us to give it a pefect answer. Employing logic and using what we have so far:

If P(A∩B) means A and B intersection, we can assume that it is simply as A + B. 

If that is the case, then we can simply pull out our P(A) and P(B), 25 and 945 respectively and add them:

P(A) + P(B) = P(A∩B)
P(A∩B) =  25 + 945
P(A∩B) = 970 <----- What we are looking for

Therefore, P(A∩B) = 970!
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Answer:

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