The answer to the question is y=-4x+23
They get 2/3 because 2 divided by 3=2/3.
Here are the factors of 424,380:
<span>1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 20, 22, 30, 33, 44, 55, 60, 66, 110, 132, 165, 220, 330, 643, 660, 1286, 1929, 2572, 3215, 3858, 6430, 7073, 7716, 9645, 12860, 14146, 19290, 21219, 28292, 35365, 38580, 42438, 70730, 84876, 106095, 141460, 212190, and 424380</span>
Answer:
(a) 0.14%
(b) 2.28%
(c) 48%
(d) 68%
(e) 34%
(f) 50%
Step-by-step explanation:
Let <em>X</em> be a random variable representing the prices paid for a particular model of HD television.
It is provided that <em>X</em> follows a normal distribution with mean, <em>μ</em> = $1600 and standard deviation, <em>σ</em> = $100.
(a)
Compute the probability of buyers who paid more than $1900 as follows:


*Use a <em>z</em>-table.
Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.
(b)
Compute the probability of buyers who paid less than $1400 as follows:


*Use a <em>z</em>-table.
Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.
(c)
Compute the probability of buyers who paid between $1400 and $1600 as follows:


*Use a <em>z</em>-table.
Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.
(d)
Compute the probability of buyers who paid between $1500 and $1700 as follows:


*Use a <em>z</em>-table.
Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.
(e)
Compute the probability of buyers who paid between $1600 and $1700 as follows:


*Use a <em>z</em>-table.
Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.
(f)
Compute the probability of buyers who paid between $1600 and $1900 as follows:


*Use a <em>z</em>-table.
Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.
Is useful<span> in </span>proving<span>various theorems about </span>triangles<span> and other polygons.
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