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zubka84 [21]
3 years ago
15

A store manager timed Janette to see how long it would take her to fold and put away a sweater, a shirt, a pair of pants, and a

scarf. It took her 26.1 seconds for the shirt, 24.3 seconds for the sweater, 32.8 seconds for the pants, and 18.2 seconds for the scarf. What was the average time it took Janette to fold and put away all four items?
Mathematics
2 answers:
Tasya [4]3 years ago
7 0
Add them all up and divide by 4
101.4/4 = 25.35 or 25.4 seconds
ArbitrLikvidat [17]3 years ago
7 0

Answer:

25.4

Step-by-step explanation:

thats the answer... your welcome

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Ainat [17]

Answer:

(i) The area of the rabbit cage when the width is 5.2 m is 81.5 m²

(ii) The area of the rabbit cage if Wilson has 40 meters of wire mesh is 75 m²

Step-by-step explanation:

(i) The given relation of the area, A to the width P of the rabbit cage is A = 3·p²

The graph of the function between the values of 0 and 6 inclusive is found as follows;

A,              3·p²

0,               0

1,                1

2,               12

3,               27

4,               48

5,               75

6,               108

Please find attached the graph of A to 3·p²

From the graph, we have when the the width, p, of the rabbit cage = 5.2, the area, A ≈ 81.5 m²

The area of the rabbit cage when the width is 5.2 m = 81.5 m²

(ii) Also from the graph given that the total wire mess with Wilson = 40 meters, we have;

The formula for the perimeter of the cage = The formula for the perimeter of a rectangle = 2×length + 2×width

The formula for the perimeter of the cage = 2×3×p + 2× p = 8·p

Where the total length of the wire mesh available = 40 meters for the cage

The 40 meters of wire mesh will be used round the perimeter of the cage

∴ 40 m. = 8·p

p = 40/8 = 5 m.

At p = 5 m. the area is given as A = 75 m².

Therefore, the area of the rabbit cage if Wilson has 40 meters of wire mesh = 75 m².

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