Check the picture below.
A)
well, we start off by making a table of values for the function, with a few "t" values, as you see in the picture, then graph those points to get the graph.
B)
at t = 0, h(0) = 98, and at t = 1, h(1) = 82, so it went from 98 to 82, for a difference of 16.
C)
at t = 1, h(1) = 82, and at t = 2 h(2) = 34, so it went from 82 to 34, for a difference of 48.
does it fall the same distance on both intervals? well, they're different, so nope.
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
<h2>Hello my friend.</h2>
The Pi value is approximately equal to 3.14.
<h2>I hope I have helped a lot.</h2>
Answer:
8
Step-by-step explanation:
