find the area and perimeter of a rectangle with a width of 2 1/3inch and a length of 5 2/5inches. show your work please.

2 answers:

5 0

A=LxW, but first you should convert the mixed fractions to improper fractions:
7/3 x 27/5 = 189/15 = 12 6/10 = 12 3/5 square inches

P = L+L+W+W here I would convert to improper fractions, but then you need to find a common denominator.
P = 7/3 + 7/3 + 27/5 + 27/5 = 14/3 + 54/5 = 70/15 + 162/15 = 232/15 = 15 7/15 inches

aleksandrvk [35]2 years ago

4 0

Area: If you multiply 2 1/3 by 5 2/5, you get an answer of about: 12.6, which means the area is about: 12.6in. squared.
Perimeter: If you add 2 1/3+ 2 1/3+ 5 2/5+ 5 2/5 you get an answer of about: 15.47. So, the perimeter is about 15.47in.

Note: If you aren't allowed to use a calculator on this test, you should probably multiply the numbers yourself so you get show the work and get the answer in a fraction, which is probably what they want. If you are allowed to use calculators, all the information you need is right here.

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hichkok12 [17]

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Suppose A and B are independent events if P(A) = 0.4 And P(B) = 0.1, what is P(A'uB)? APEX

Aloiza [94]

For Independent Events, P(A) × P(B) = P(A∩B)

so we have, P(A∩B) = 0.4×0.1 = 0.04

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This information can be represented on a Venn diagram as shown below

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P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64