Answer:
2y+1=0
Step-by-step explanation:
a number one greater than y is y+1
so sum of y and y+1= 2y+1
Answer:
a) 
b) 
Step-by-step explanation:
Given Data:
Interest rate=
per year
No. of years=
Rate of continuous money flow is given by the function
a) to find the present value of money

Put f(t)=2000 and n=10 years and r=0.08

Now integrate







(b) to find the accumulated amount of money at t=10

Where P is the present worth already calculated in part a




At the start, you have 10 marbles in total and 6 red marbles. The probability of the first event is 6/10.
Since you do not place the marble back in, you now have 9 marbles in total and 3 blue marbles. This probability of the second event is 3/9.
To find the probability of both events occurring, we multiply the two probabilities together. (6•3)/(10•9)=18/90=1/5
Thus, there is a 20% chance of drawing a red marble then a blue marble without placing the red marble back in the bag.
Blue triangle is 1/4, the green triangle is 3/2, and red is 2/8