Given fx (x) = λe^λx
Fx (x) = 1 – e^-λx x…0
To find distribution of Min (X1,….Xn)
By applying the equation
fx1 (x) = [n! / (n – j)! (j – 1)!][F(x)]^j-1[1-F(x)]^n-j f(x)
For minimum j = 1
[Min (X1,…Xn)] = [n!/(n-1)!0!][F(x)]^0[1-(1-e^-λx)]^n-1λe^-λx
= ne^[(n-1) λx] λe^(λx)
= nλe^(-λx[1+n-1])
= nλe^(-nλx)
Answer:
<h2>
6.4 cm</h2>
Step-by-step explanation:
Volume of cartoon = 5 cm × 8 cm × 15 cm
= 600 cm³
Volume of milk = 5 cm × 8 cm × 12 cm
= 480 cm³
Space = 600 cm³ - 480 cm³ = 120 cm³
Let depth be ' h '
15 × 5 × h = 480
Multiply the numbers
75h = 480
Divide both sides of the equation by 75
75 h / 75 = 480 / 75
Calculate
h = 6.4 cm
therefore, The depth of the milk is 6.4 cm
Hope this helps...
Best regards!!
Answer:
Step-by-step explanation:
4x² - 7x + 3 = 4x² - 4x - 3x + (-3)*(-1)
= 4x*(x - 1) - 3(x - 1)
= (x-1) (4x - 3)
Answer:
5
Step-by-step explanation:
