Luis could make 0, 1, 2 or 3 touchdowns. We know that in the past he
made 0 touchdowns 4 times, 1 touchdown 7 times, 2 touchdowns 9 times and 3 touchdowns 6 times. To make a probability distribution table, add up 4, 7, 9 and 6; the sum is 26. Thus, based upon past experience, the probability of his making 0 touchdowns is 4/26, of 1 touchdown 7/26, of 2 touchdowns, 9/26 and of 3 touchdowns 6/26: {0.154, 0.269, 0.346, 0.231}. Important: Note that these four decimal probabilities add up to 1, as they must.
The empirical probability of Luis' scoring 2 touchdowns is 0.346.
To find the probability of his scoring more than 1 touchdown, add together the probabilities of his scoring 2 and 3 touchdowns: 0.346+0.231 = 0.577.
The expected value can be calculated as follows:
0(0.154) + 1(0.269) + 2(0.346) + 3(0.231). Note that each term in this expression comes from {0, 1, 2, 3}, and that the fractions all represent the probabilities of each of these four possible outcomes.
The sum is 1.65. This is the "expected value" of the number of touchdowns Luis will likely make. 1.65 is obviously more than 0 or 1 touchdowns, but less than 3 or 4 touchdowns.
Answer:
12 square centimetres
Step-by-step explanation:
6*2=12
Let the number of raspberry bushes in one garden = x
And the number of raspberry bushes in second garden = y
Garden one has 5 times as many raspberry bushes as second garden,
So the equation will be,
x = 5y -------(1)
If 22 bushes were transplanted from garden one to the second, number of bushes in both the garden becomes same,
Therefore, (x - 22) = (y + 22)
x - y = 22 + 22
x - y = 44 ------(2)
Substitute the value of x from equation (1) to equation (2)
5y - y = 44
4y = 44
y = 11
Substitute the value of 'y' in equation (1),
x = 5(11)
x = 55
Therefore, Number of bushes in garden one were 55 and in second garden 11 originally.
Learn more,
brainly.com/question/12422372
no it cant. irrational number is any real number that cannot be expressed as a ratio.
