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Anastaziya [24]
2 years ago
13

The playing time X of classical CDs has the normal distribution with mean 54 and standard

Mathematics
1 answer:
pickupchik [31]2 years ago
7 0

Using the normal distribution, it is found that the relative frequency of classical CDs with playing time X between 49 and 69 minutes is 0.8403 = 84.03%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 54, \sigma = 5

The relative frequency of classical CDs with playing time X between 49 and 69 minutes is the <u>p-value of Z when X = 69 subtracted by the p-value of Z when X = 49</u>, hence:

X = 69:

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

Z = \frac{69 - 54}{5}

Z = 3

Z = 3 has a p-value of 0.9987.

X = 49:

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

Z = \frac{49 - 54}{5}

Z = -1

Z = -1 has a p-value of 0.1584.

0.9987 - 0.1584 = 0.8403.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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Answer:

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Step-by-step explanation:

y = x + b

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6 = (0) + b

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A construction crew is lengthening a road. Let L be the total length of the road (in miles). Let D be the number of days the cre
djverab [1.8K]

Answer:

Set of Values for Domain = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80}

Set of Values for Range = {402, 404, 406, 408, 410, 412, 414, 416, 418, 420, 422, 424, 426, 428, 430, 432, 434, 436, 438, 440, 442, 444, 446, 448, 450, 452, 454, 456, 458, 460, 462, 464, 466, 468, 470, 472, 474, 476, 478, 480, 482, 484, 486, 488, 490, 492, 494, 496, 498, 500, 502, 504, 506, 508, 510, 512, 514, 516, 518, 520, 522, 524, 526, 528, 530, 532, 534, 536, 538, 540 542, 544, 546, 548 550, 552, 554, 556, 558, 560}.

Step-by-step explanation:

Solution...

Assume that there is a link for the road between your house and the market the road maintenance team is meant to repair the road.

now the system of work is that the length of the road they will work upon is equal to 2 times the amount of days + 400.

This is to say that originally there is 400 miles of the road, and once they work in a day they add 2 more miles to the original 400. if they work in 2 days they add 4 more miles to the original 400. If they work in 3 days they add 6 more miles to the original 400.

So L = 2(D) + 400.

Answer 1.

description of the values in both the domain and range of a function.

the domain is the set of all values for which the function is defined.

this means that the domain is all numbers between 0 to 80.

Why do we begin from 0, because when you put 0 into the equation you have, L = 2(0) + 400, and this is equal to 2 x 0 + 400 or 400 + 0 (since all numbers multiplied by 0 gives 0).

Why is 80 the maximum, this is because they work for at most 80 days.

So the domain is

D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80}

And the range is the set of all values that L (which is the function) takes, to get this you have to calculate the days from day 0 to day 80.

The range for the function is...

L = {402, 404, 406, 408, 410, 412, 414, 416, 418, 420, 422, 424, 426, 428, 430, 432, 434, 436, 438, 440, 442, 444, 446, 448, 450, 452, 454, 456, 458, 460, 462, 464, 466, 468, 470, 472, 474, 476, 478, 480, 482, 484, 486, 488, 490, 492, 494, 496, 498, 500, 502, 504, 506, 508, 510, 512, 514, 516, 518, 520, 522, 524, 526, 528, 530, 532, 534, 536, 538, 540 542, 544, 546, 548 550, 552, 554, 556, 558, 560}.

So how did we get all these. Using the figures registered as D to get the L.

Before solving, remember that, (HD is H•D or H × D,

2D is 2•D or 2 × D,

29F is 29•F or 29 × F... When numbers are used to multiply letters which stand as a varying value like X, Y, D etc, then they are written as 2x or 2•x or 2(x). They all mean multiply 2 by x)

So, i.e.

L = 2(D) + 400.

When it's day 1, D is 1

L = 2(1) + 400 which equals 2 + 400 = 402. So for 1 day, the crew has worked 402 miles.

On Day 2, D is 2

L = 2(2) + 400 which equals 4 + 400 = 404. So for 2 days, the crew has worked 404 miles.

On Day 3, D is 3

L = 2(3) + 400 which equals 6 + 400 = 406. So for 3 days, the crew has worked 406 miles.

On Day 4, D is 4

L = 2(4) + 400 which equals 8 + 400 = 408. So for 4 days, the crew has worked 408 miles.

On Day 5, D is 5

L = 2(5) + 400 which equals 10 + 400 = 410. So for 5 days, the crew has worked 410 miles.

On Day 6, D is 6

L = 2(6) + 400 which equals 12 + 400 = 412. So for 6 days, the crew has worked 412 miles.

...

...

...

For 7 days, the crew has worked 414 miles.

For 8 days, the crew has worked 416 miles.

For 9 days, the crew has worked 418 miles.

For 10 days, the crew has worked 420 miles.

For 20 days, the crew has worked 440 miles.

For 30 days, the crew has worked 460 miles.

For 40 days, the crew has worked 480 miles.

For 50 days, the crew has worked 500 miles.

For 60 days, the crew has worked 520 miles.

For 70 days, the crew has worked 540 miles.

For 75 days, the crew has worked 550 miles.

...

...

...

On Day 79, D is 79

L = 2(79) + 400 which equals 158 + 400 = 558. So for 79 days, the crew has worked 558 miles..

On Day 80, D is 80

L = 2(80) + 400 which equals 160 + 400 = 560. So for 80 days, the crew has worked 560 miles.

That is how the domain and range is determined.

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