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Viefleur [7K]
1 year ago
9

Use the points (2, 2485.6) and (6, 1172.5) to write an equation of the line of fit in slope-intercept form. Let x be the years s

ince 2010 and let y be the number of CDs in millions sold.
Mathematics
1 answer:
mojhsa [17]1 year ago
6 0

Answer:

y = -328.275x + 1829.05

Step-by-step explanation:

In this case, we need to write an equation of this form:

y = ax + b  (1)      a: slope    b: intercept

In order to calculate the slope, we need to use the following expression:

a = y₂ - y₁ / x₂ - x₁     (2)

So, we have two points here, so, from these points we can identify each of the elements so we can determine the slope.

Point 1: (2, 2485.6)

Point 2: (6, 1172.5)

The slope of this equation will be:

a = 1172.5 - 2485.6 / 6 - 2

a = -328.275

Now that we have the slope, to get the intercept "b", we just use any of the given points, and replace in the general equation (1) to solve for b:

2485.6 = -328.275(2) + b

b = 2485.6 - 656.55

b = 1829.05

With the value of a and b, we can write the general equation of the line for this:

<h2>y = -328.275x + 1829.05</h2>

Hope this helps

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asambeis [7]

Answer:

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Step-by-step explanation:

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n\geq (\frac{z\sigma}{ME})^2

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We substitute our given values to calculate the sample size:

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1 year ago
For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a
aleksley [76]

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a. -1.60377

b. 0.25451

c. 0.344

d. Option b) 78th

Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

z = (x-μ)/σ

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z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

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z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

For 39 Skittles

z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

For 42 Skittles

z = (42 - 38.4)/2.12

= 1.69811

Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

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z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

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P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

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