Answer:
4000
Step-by-step explanation:
A=P(2)^(t/d)
A=future amount
P=starting amount
t=time elapsed
h=doubling time
given
P=10^3
t=6
d=2
A=10^3(2)^(6/2)
A=10^3(2)^2
A=10^3(4)
A=4*10^3
A=4000
4000 bacteria
Answer:
A
Step-by-step explanation:
Replace the point (0,0) in each inequality
A. y - 4 < 3x - 1
B. y - 1 < 3x - 4
C. y + 4 < 3x - 1
D. y + 4 < 3x + 1
A. 0 - 4 < 3(0) - 1
- 4 < - 1
True
B. 0 - 1 < 3(0) - 4
- 1 < - 4
False
C. y + 4 < 3x - 1
0 + 4 < 3(0) - 1
4 < - 1
False
D. y + 4 < 3x + 1
0 + 4 < 3(0) + 1
4 < 1
False
1) the rule for #1 is is to subtract the small number from the big number so (15-4) (get rid of all the signs when doing this see how i took away the sign from 15) the answer is then 11 but you pit the sign of the bigger number since 15 is the bigger number and it has the negative sign you put the negative sign on 11 so the answer for 1 is -11
2) to solve -3*-5 you have to realize that they are both negative and the rule of multiplying negative integers is that a negative times a negative = a postive (weird but those are the rules) so -3*-5=15
Answer:
The probability is 0.3576
Step-by-step explanation:
The probability for the ball to fall into the green ball in one roll is 2/1919+2 = 2/40 = 1/20. The probability for the ball to roll into other color is, therefore, 19/20.
For 25 rolls, the probability for the ball to never fall into the green color is obteined by powering 19/20 25 times, hence it is 19/20^25 = 0.2773
To obtain the probability of the ball to fall once into the green color, we need to multiply 1/20 by 19/20 powered 24 times, and then multiply by 25 (this corresponds on the total possible positions for the green roll). The result is 1/20* (19/20)^24 *25 = 0.3649
The exercise is asking us the probability for the ball to fall into the green color at least twice. We can calculate it by substracting from 1 the probability of the complementary event: the event in which the ball falls only once or 0 times. That probability is obtained from summing the disjoint events: the probability for the ball falling once and the probability of the ball never falling. We alredy computed those probabilities.
As a result. The probability that the ball falls into the green slot at least twice is 1- 0.2773-0.3629 = 0.3576
