let the two numbers be x and y
From the first sentence,
xy=24
x+y=10
Then make y in equation 2 the subject of the formular and substitute in equation 1
x+y=10
y=10-x
substituting in equation 2
x(10-x)=24
open the bracket
10x-x^2=24
=-x^2+10x=24
Transfer the constant to the left hand side
=-x^2+10x-24=0
Then factorise completely
Look at the photo above
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%.