THIS IS THE COMPLETE QUESTION BELOW
The demand equation for a product is p=90000/400+3x where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤ 50.
Answer
$168.27
Step by step Explanation
Given p=90000/400+3x
With the limits of 40 to 50
Then we need the integral in the form below to find the average price
1/(g-d)∫ⁿₐf(x)dx
Where n= 40 and a= 50, then if we substitute p and the limits then we integrate
1/(50-40)∫⁵⁰₄₀(90000/400+3x)
1/10∫⁵⁰₄₀(90000/400+3x)
If we perform some factorization we have
90000/(10)(3)∫3dx/(400+3x)
3000[ln400+3x]₄₀⁵⁰
Then let substitute the upper and lower limits we have
3000[ln400+3(50)]-ln[400+3(40]
30000[ln550-ln520]
3000[6.3099×6.254]
3000[0.056]
=168.27
the average price p on the interval 40 ≤ x ≤ 50 is
=$168.27
Answer:A
Step-by-step explanation: A has everthing correct the median is between the numbers 92 98 making it 95 these are the numbers is order 79,86,88,88,92,98,100,106,115,136
There are 3 coefficents. There is one in 6a, which is 6, one in 8z, which is 8, and one in 1/2n, which is 1/2.
Hope this helps, Ngam123!
Yea it is.4\5 is greater than 2
Start with
Expand both parentheses by multiplying both terms by the number outside:
Sum like terms:
Simplify the "+3" on both sides:
Subtract 2x from both sides:
Divide both sides by 2:
