Answer:
The value of the ve = 9m/sec
Step-by-step explanation:
From the given formula, it can be conclude that this is evenly accelerated movement.
First I will rewrite given formula
vi = √ ve∧2 - 2ad First we will square on both sides and get
vi∧2 = ve∧2 - 2ad Now we will add monom (+2ad) to both sides and get
vi∧2 + 2ad = ve∧2 -2ad + 2ad =>
ve∧2 = vi∧2 +2ad Now we will rooted both sides and get
ve = √vi∧2 + 2ad
Now we will replace given data vi=7m/sec, a=8m/s∧2 and d=2m in the last formula
ve= √7∧2 + 2*8*2 = √49+32 = √81 = 9
ve= 9m/sec
Good luck!!!
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
Answer:
Step-by-step explanation:
Answer:
10+ x
Step-by-step explanation:
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