If the MAD for a forecast is 3.2, we can say that a. The forecast is 3.2 units too high on the average b. The forecast is 3.2 un
1 answer:
You might be interested in
Answer:
Colour | Probability
YELLOW | 0.20
WHITE | 0.18
GREEN | 0.12
RED | 0.18
BLUE | 0.32
Step-by-step explanation:
The complete question is presented on the attached image to this solution.
From the image of the spinner and the given probabilities, the combination of YELLOW, WHITE and GREEN make up exactly half of the total space on the spinner.
0.2 + 0.18 + 0.12 = 0.50
Then, BLUE and RED make up the other half of the spinner.
Probability of landing on either BLUE or RED = 1 - 0.50 = 0.50
Let P(BLUE) = x and P(RED) = y
x + y = 0.50
Two of the lines shown on the diagram are diameters of the circle.
The first line that is a diameter splits the circle into two parts; (WHITE, YELLOW, GREEN) and (RED, BLUE).
From here, we have established that x + y = 0.50
The other line that is a diameter splits The circle into two part; (WHITE, BLUE) and (YELLOW, GREEN, RED)
From this other information,
P(WHITE) + P(BLUE) = 0.50
0.18 + x = 0.50
x = 0.50 - 0.18 = 0.32
P(YELLOW) + P(GREEN) + P(RED) = 0.50
0.20 + 0.12 + y = 0.50
y = 0.50 - 0.32 = 0.18
To check, x + y = 0.32 + 0.18 = 0.50 (Correct!!)
Hope this Helps!!!
Answer:
- a)
- b)
- c)
- d)
Step-by-step explanation:
We will use the product rule from combinatorics.
- a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are
strings.
- b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are
strings.
- c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are
strings.
- d) Remember that the binomial coefficient
is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in
ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are
strings.