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3 years ago

Use the Trapezoidal Rule with n = 4 to approximate 0.3523 0.7048 0.2456 0.4913

Mathematics

1 answer:

lilavasa [31]3 years ago

8 0

Answer:

0.7048

Step-by-step explanation:

∫₂⁴ 1 / (x − 1)² dx

n = 4, so we divide the interval into 4 regions. The width of the regions is:

(4 − 2) / 4 = 0.5

So the intervals are (2, 2.5), (2.5, 3), (3, 3.5) and (3.5, 4).

Evaluate f(x) at the ends of each interval.

f(2) = 1

f(2.5) = 4/9

f(3) = 1/4

f(3.5) = 4/25

f(4) = 1/9

Find the area of each trapezoid:

A₁ = ½ (2.5 − 2) (1 + 4/9) = 13/36

A₂ = ½ (3 − 2.5) (4/9 + 1/4) = 25/144

A₃ = ½ (3.5 − 3) (1/4 + 4/25) = 41/400

A₄ = ½ (4 − 3.5) (4/25 + 1/9) = 61/900

So the total approximate area is:

A = 0.705

Closest answer is 0.7048.

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4 years ago

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he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

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a) 0.0708

b) 0.9291

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Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

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Using the normal distribution table:

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b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

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P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

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NORMSDIST (-3.92)

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