How do we figure out 6 2/5 times 3 1/6
1 answer:
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Two numbers that add up to -19 and multiply to 48 are -16 and -3:
So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.
x+4=0
x = -4
x-4=0
x = 4
x^2-3=0
x^2 = 3
x = +/- √3
So, the solutions are -4, -√3, √3, and 4.
So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.
x+4=0
x = -4
x-4=0
x = 4
x^2-3=0
x^2 = 3
x = +/- √3
So, the solutions are -4, -√3, √3, and 4.
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%.