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2 years ago

Janice needs to re-shingle the roof of her house. One bundle of shingles costs $35.99 and covers 2.25 m2.

Mathematics

1 answer:

larisa [96]2 years ago

7 0

To calculate the number of shingles, we first calculate for the total area of the roof of Janice's house. From the figure, we can calculate the area of the two shapes: pyramid and a rectangular prism.

A(pyramid) = lw+l√[(w/2)²+h²]+w√[(l/2)²+h²]
A = 208 m²
A(prism) = 2(wl+hl+hw)
A = 293 m²

Total area = 501 m²

Number of shingles she needs = 501 m² / 2.25 m² = 223

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Alright, lets get started.

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Differences between two samples are least likely to be statistically significant if the samples are _______ and the standard dev

neonofarm [45]

Differences between two samples are least likely to be statistically significant ifthe samples are small and the standard deviations of the samples are large

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2 years ago

For the diagram below figure b is a reflection of y=2x of the line y =x

Scilla [17]

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2 years ago

A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur

11111nata11111 [884]

(a) 0.059582148 probability of exactly 3 defective out of 20

(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

= 0.05^3 * 0.95^17 * 20! / (3!17!)

= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

= 0.05^3 * 0.95^17 * 20*19*3

= 0.000125* 0.418120335 * 1140

= 0.059582148

(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

= 0.05*0.95^3*24/(1!3!)

= 0.05*0.857375*24/6

= 0.171475

So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.

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2 years ago