The probability of selecting two non-defective DVD players from a group of six is 2/5. This is based on the assumption that the selection is done without replacement.
We can use the formula for calculating probabilities of combinations:
P(not defective) = number of ways to choose 2 non-defective DVD players / total number of ways to choose 2 DVD players
Total number of ways to choose 2 DVD players out of 6 is:
C(6,2) = 6! / ([2!] [4!]) = 15
Number of ways to choose 2 non-defective DVD players out of 4 is:
C(4,2) = 4! / ([2!] [2!]) = 6
Therefore, the probability that Cecil chooses 2 non-defective DVD players is:
P(not defective) = 6/15 = 2/5
So the value of P(not defective) is 2/5.
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in a congressional district, 55% of the registered voters are democrats. which of the following is equivalent to the probability of getting less than 50% democrats in a random sample of size 100?
A. P( z< 50 — 55/ 100 )
B. P( z< 50 — 55/ √55(45)/100)
C. P( z< 55 — 5 / √55(45)/100)
D. P( z< 50 — 55/√100(55) (45))
The correct answer to the question, "Which of the following is equivalent to the probability of getting less than 50% democrats in a random sample of size 100?" is: B. P( z < 50 — 55/ √55(45)/100).
To find the probability, we first calculate the z-score using the formula:
z = (x - μ) / σ
where x is the value (50%), μ is the mean (55%), and σ is the standard deviation.
The standard deviation can be calculated as:
σ = √(np(1-p))
where n is the sample size (100) and p is the proportion of democrats (0.55).
Now, plug in the values into the z-score formula:
z = (50 - 55) / √(100 * 0.55 * 0.45)
The probability is then found as P(z < z-score), which is represented by the option B.
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mental ability
hardest queston for grade 7
you are god if you did and explained properly
i will mark you as brainliest
optinons are-:
173
153
182
142
Answer:
153
Step-by-step explanation:
The relationship in the row is below
4³ +2³ +1³ =64 + 8 +1 = 72
1³ + 2³ +6³= 1 + 8 + 216
3³ + 1³ + 5³ = 27 +1 +125 = 153
All numbers below the first level are raised to power 3 and added together
after a (not very successful) trick or treating round, candice has 12 tootsie rolls and 10 twizzlers in her pillow case. her mother asks her to share the loot with her three younger brothers. (a) how many different ways can she do this?
Using the stars and bars technique, Candice can distribute her 24 pieces of candy among her four siblings in 2,925 different ways. If she must give each sibling at least one of each type of candy, there are 67,200 ways to distribute the candy among the four siblings.
(A) To solve this problem, we can use the technique of stars and bars. We have a total of 24 pieces of candy to share among four children. We can represent this using 24 stars, with 3 bars to separate the stars into four groups, one for each child. For example, the following arrangement represents giving 6 pieces of candy to the first child, 10 pieces to the second child, 3 pieces to the third child, and 5 pieces to the fourth child:
*****|**********|***|****
The number of ways to arrange the stars and bars is equal to the number of ways to choose 3 positions out of the 27 possible positions for the stars and bars. Therefore, the number of different ways that Candice can share her candy with her three younger brothers is:
C(27, 3) = 27! / (3! * 24!) = 2925
(B) Now, we need to ensure that each child receives at least one Tootsie roll and one Twizzler. We can give each child one of each candy to start, and then distribute the remaining 13 Tootsie rolls and 7 Twizzlers using the stars and bars technique. We have 13 Tootsie rolls and 7 Twizzlers to distribute among four children, which can be represented using 13 stars and 3 bars for the Tootsie rolls, and 7 stars and 3 bars for the Twizzlers. The number of ways to arrange the stars and bars for each type of candy is:
C(16, 3) = 560 for the Tootsie rolls
C(10, 3) = 120 for the Twizzlers
To find the total number of ways to distribute the candy, we can multiply the number of ways for each type of candy:
560 * 120 = 67200
Therefore, there are 67,200 different ways for Candice to share her candy with her three younger brothers after her mother asks her to give at least one of each type of candies to each of her brothers.
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Complete question:
After a (not very successful) trick or treating round, Candice has 15 Tootsie rolls and 9 Twizzlers in her pillow case. Her mother asks her to share some of the loot with her three younger brothers.
(A) How many different ways can she do this?
(B) How many different ways can she do this after her Mother asks her to give at least one of each type of candies to each of her brothers?
What percentage of people would exed to score higher than a 2.5, but lower than 3.5? The mean: X=3.00 The SDis= + 0.500 18% 999 o 50% 03%
Therefore, approximately 68.26% of people are expected to score higher than 2.5 but lower than 3.5.
Based on the information provided, the mean (X) is 3.00 and the standard deviation (SD) is 0.50. To find the percentage of people expected to score higher than 2.5 but lower than 3.5, we will use the standard normal distribution (z-score) table.
First, we need to calculate the z-scores for both 2.5 and 3.5:
z1 =[tex] (2.5 - 3.00) / 0.50 = -1.0[/tex]
z2 = [tex](3.5 - 3.00) / 0.50 = 1.0[/tex]
Now, we can use the standard normal distribution table to find the probability of the z-scores. For z1 = -1.0, the probability is 0.1587 (15.87%). For z2 = 1.0, the probability is 0.8413 (84.13%).
To find the percentage of people expected to score between 2.5 and 3.5, subtract the probability of z1 from the probability of z2:
Percentage = [tex](0.8413 - 0.1587) x 100 = 68.26%[/tex]
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