If the measure of angle A is 40 degrees and the length of side b is 15 inches, which can be the length of side a if it is possib
1 answer:
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Answer:
Step-by-step explanation:
Let
n -----> number of tickets
C ----> represent the cost of buy n tickets online
we have the ordered pairs
(1,16.50) and (2,30.50)
<em>Find out the slope of the linear equation</em>
The formula to calculate the slope between two points is equal to
substitute the values
<em>Find the equation of the line in slope intercept form</em>
we have
substitute
substitute
The domain of the function is all positive integers (whole numbers) including zero
{0,1,2,3,4,...}
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%.