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Anon25 [30]
3 years ago
13

An average American will throw away 90,000lb of trash over his or her lifetime. Convert this weight to kilograms using the fact

that 1kg=2.2lb. Round to the nearest hundred.
Mathematics
1 answer:
yawa3891 [41]3 years ago
3 0

Answer:

The weight of the trash is 40,900 kilograms

Step-by-step explanation:

∵ 1 kilogram = 2.2 Ib

That means if you want to convert quantity from Ib to kilogram divide the quantity by 2.2 or you can use the ratio method to do that

→  kg    :    Ib

→  1       :    2.2

→  x       :    90,000

Now by using cross multiplication

∵ x × 2.2 = 1 × 90,000

∴ 2.2 x = 90,000

- Divide both sides by 2.2

∴ x = 90,000 ÷ 2.2

∴ x = 40909.09091

- Round it to the nearest hundred

∵ The number before the hundred place is 0

∴ The number in the hundred digit will not change

- Change the unit digit and ten digit to 0

∴ x ≅ 40900 to the nearest hundred

∵ x represents the quantity of the trash in kilogram

∴ The weight of the trash is 40900 kilograms

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Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
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Answer:

(a) 0.343

(b) 0.657

(c) 0.189

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

                        p = \frac{70}{100} = 0.7  

Let P(a vehicle not passing the test) = q

                         q = 1 - p

                         q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

                           = 0.7 × 0.7 × 0.7

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(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

      = (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

      = 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

      = 0.657

(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

                 = 0.063 + 0.063 + 0.063

                 = 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

                 = 0.063 + 0.063 + 0.063 + 0.027

                 = 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

=  (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

                  = 0.973  

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

                         = \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}

                         = \frac{0.343}{0.973}

                         = 0.353

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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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