it would be the 3rd one i just took the test
Some friends played a board game. During the game, one unlucky player had to move back 9 spaces
7 turns in a row. Find a number to represent that player's movements for those 7 turns
Answer:
-63
Step-by-step explanation:
Because -9 x 7 = -63. If the signs are different then it will be negative. -9 x 7 = -63 is the same like 9 x 7 = 63. I hope this answer your question!
Answer:
Step-by-step explanation:
Our term is 15+20x
we can divide it to : 15 and 20x
so the factors are : (1,3,5,15)
20 can be divided by : 1 , 2 , 4 ,5 ,10 , 20
so the factors of 20 x are :
1 , 2 , 4 , 10 , 20 , x , 2x , 4x , 5x , 10 x , 20x
Answer:
15: 1, 3, 5, 15
20x: 1, 2, 4, 5, 10, 20, x ^_^
Step-by-step explanation:
Answer:
The answer is invalid
Step-by-step explanation:
Choose Invalid .
You welcome
Answer: Valid
8 * 12 = 96
Answer:
Smallno. =16
Bug no. =36
Step-by-step explanation:
Y=big number
X=small number
X+Y=52
Y=2X+4
Substitute
X+2X+4=52
3X=48
X=16(small no.)
Y=36(large no.)
Checking:
36+16=52(yay correct)
36=(2*16)+4 (coorrectoo!)
Hopethishelps!
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