Linear regression line y=2.1x+130 predicts sales based on the money spent on advertising.
Linear regression represents the relationship between two variables. the value of y depends on the value of x.
x represents the dollars spent in advertising and y represents the company sales in dollars.
We need to find out sales y when $150 spends on advertising.
Plug in 150 for x and find out y
y = 2.1 x + 130
y = 2.1 (150) + 130
y= 445
The company expects $445 in sales
Answer:
25%
Step-by-step explanation:
let the Market Price of the Watch (M.P.) be x
CP for Rahim=x-[(20/100)*x]
CP = (1 *x) /5 = 4x/5
SP for Rahim = MP = x
Difference (Profit) = SP-CP = x - (4x/5) = (5x - 4x)/5 = x/5
Profit % = (Profit/CP) * 100
Profit % = (x/5)/(4x/5) * 100
Profit % = 25 %
11 teams because
where we find x to be 11.
Answer:
x₂ = 7.9156
Step-by-step explanation:
Given the function ln(x)=10-x with initial value x₀ = 9, we are to find the second approximation value x₂ using the Newton's method. According to Newtons method xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)
If f(x) = ln(x)+x-10
f'(x) = 1/x + 1
f(9) = ln9+9-10
f(9) = ln9- 1
f(9) = 2.1972 - 1
f(9) = 1.1972
f'(9) = 1/9 + 1
f'(9) = 10/9
f'(9) = 1.1111
x₁ = x₀ - f(x₀)/f'(x₀)
x₁ = 9 - 1.1972/1.1111
x₁ = 9 - 1.0775
x₁ = 7.9225
x₂ = x₁ - f(x₁)/f'(x₁)
x₂ = 7.9225 - f(7.9225)/f'(7.9225)
f(7.9225) = ln7.9225 + 7.9225 -10
f(7.9225) = 2.0697 + 7.9225 -10
f(7.9225) = 0.0078
f'(7.9225) = 1/7.9225 + 1
f'(7.9225) = 0.1262+1
f'(7.9225) = 1.1262
x₂ = 7.9225 - 0.0078/1.1262
x₂ = 7.9225 - 0.006926
x₂ = 7.9156
<em>Hence the approximate value of x₂ is 7.9156</em>
