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Akimi4 [234]
3 years ago
11

A cuboid measures 60cm by 40cm by 20cm.

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

a) The area of the largest face = 2400 cm², The area of the largest face = 0.24 m²

b) The volume of the cuboid = 48000 cm³, The volume of the cuboid = 0.048 m³

Step-by-step explanation:

The volume of any cuboid = L × W × H, where

L is its length

W is its width

H is its height

∵ A cuboid measures 60 cm by 40 cm by 20 cm.

→ Assume that L = 60 cm, W = 40 cm, H = 20 cm

a)

∵ The largest face is the face who has the largest dimensions

∵ L and W are the largest dimensions

∴ The area of the largest face = L × W

∴ The area of the largest face = 60 × 40

∴ The area of the largest face = 2400 cm²

∵ 1 cm = 1/100 m

∴ 1 cm² = (1/100)² m² = 1/10000 m²

∴ The area of the largest face = 2400 × 1/10000

∴ The area of the largest face = 0.24 m²

b)

∵ The volume of the cuboid = L × W × H

∴ The volume of the cuboid = 60 × 40 × 20

∴ The volume of the cuboid = 48000 cm³

∵ 1 cm = 1/100 m

∴ 1 cm3 = (1/100)³ m³ = 1/1000000 m³

∴ The volume of the cuboid = 48000 × 1/1000000

∴ The volume of the cuboid = 0.048 m³

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Step-by-step explanation:

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Tatiana [17]

Answer:

Last option (1/2)^3 = .125

Step-by-step explanation:

1/2 (1.035)^5 =0.579637

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A machine is shut down for repairs if a random sample of 100 items selected from the daily output of the machine reveals at leas
Harlamova29_29 [7]

Answer:

0.7995 = 79.95% probability that the sample will contain at least three defectives.

Step-by-step explanation:

Binomial probability distribution

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The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

Suppose that a random sample of 20 items is selected from the machine.

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The machine produces 20% defectives

This means that p = 0.2

Mean and standard deviation:

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Probability that the sample will contain at least three defectives

Using continuity correction, this is P(X \geq 3 - 0.5) = P(X \geq 2.5), which is 1 subtracted by the pvalue of Z when X = 2.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{2.5 - 4}{1.79}

Z = -0.84

Z = -0.84 has a pvalue of 0.2005

1 - 0.2005 = 0.7995

0.7995 = 79.95% probability that the sample will contain at least three defectives.

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