F ` ( x ) = ( x² )` · e^(5x) + x² · ( e^(5x) )` =
= 2 x · e^(5x) + 5 e^(5x) · x² =
= x e^(5x) ( 2 + 5 x )
f `` ( x ) = ( 2 x e^(5x) + 5 x² e^(5x) ) ` =
= ( 2 x ) ˙e^(5x) + 2 x ( e^(5x) )` + ( 5 x² ) ` · e^(5x) + ( e^(5x)) ` · 5 x² =
= 2 · e^(5x) + 10 x · e^(5x) + 10 x · e^(5x) + 25 x² · e^(5x) =
= e^(5x) · ( 2 + 20 x + 25 x² )
Answer:
a) The probability of selling less than 100 gallons (x≤1) is P=0.16.
b) The mean number of gallons is M=80 gallons.
Step-by-step explanation:
The probability of selling x, in hundred of gallons, on any day during the summer is y(x)=0.32x, in a range for x from [0;2.5].
The probability of selling less than 100 gallons (x≤1) is then:
The mean number of gallons can be calculated as:
Answer:
800 flyersOn Earth day, a local community wants to distribute 800 flyers, 300 banners, and 250 badges to volunteers in packets containing the three items. What is the greatest number of packets that can be made using all these items if each type of item is equally distributed among the packets?
Answer: Peter gets £3
Step-by-step explanation:
Let Angad get x: equation will be 2(3x) + 3x + x = 10
X = 1 = Angad
Peter get 3x = 3*1 = 3
