Does a circular permutation of 5 distinct objects always have fewer arrangements than a linear permutation of 5 distinct

Three dentists wanted to compare the ages of their patients. They each calculated their patients’ mean age and the mean absolute

SCORPION-xisa [38]

<span>B. Dr. Appiah’s patients’ ages vary less than do Dr. Singh’s patients’ ages.

C. Dr. Cantwell’s patients’ ages and Dr. Singh’s patients’ ages vary about the same amount.</span>

6 0

3 years ago

The Statistical Abstract of the United States published by the U.S. Census Bureau reports that the average annual consumption of

ololo11 [35]

Answer:

Hence,

a) The probability that the sample average would be less than 90 pounds is 0.0210.

b) The probability that the sample average would be between 98 and 105 pounds is 0.5045.

c) The probability that the sample average would be less than 112 pounds is 0.9935.

d) The probability that the sample average would be between 93 and 96 pounds is 0.1341.

Step-by-step explanation:

a) $P(X < 90) = P(Z < (90 - 99.9) / (30\sqrt(38)))$

= P(Z < -2.03) = 0.0210

b ) $P(98 < x$

= P(-0.39 < Z < 1.05) = 0.5045

c ) $P(X < 112) = P(Z < (112 - 99.9) / (30\sqrt(38)))$

= P(Z < 2.49) = 0.9935

d ) $P(93 < x < 96) = P((93 -99.9) / (30 \sqrt(38)) < Z < (96 -99.9) / (30 \sqrt(38)))$

= P( -1.42 < Z < -0.8 )

= 0.2119 - 0.0778 = 0.1341

8 0

3 years ago

II'll mark brainliest who is the first to answer both questions correctly.

Strike441 [17]

Answer:

A) The probability that each player gets an ace, a 2 and a 3 with the order unimportant = (72/1925) = 0.0374

B) The probability of winning the jackpot = (1/1200) = 0.0008333

Step-by-step explanation:

A) There are four Aces, four 2's, and four 3's, forming a set of 12 cards

These cards are to be divided at random between 4 players.

What is the probability that each player gets an ace, a 2 and a 3.

We start with the first player, the probability of these cards for the first player, with order not important (because order isn't important, there are 6 different arrangement of the 3 cards)

6 × (4/12) × (4/11) × (4/10) = (16/55)

Then the second player getting that same order of cards

6 × (3/9) × (3/8) × (3/7) = (9/28)

Third player

6 × (2/6) × (2/5) × (2/4) = (2/5)

Fourth player

6 × (1/3) × (1/2) × (1/1) = 1

Probability that each of the players get different cards is then a multiple of the probabilities obtained above

= (16/55) × (9/28) × (2/5) × 1

= (288/7700) = (72/1925) = 0.0374

B) The concluding part of the B question.

To win the jackpot, the numbers on your ticket must match the three white balls and the SuperBall. (You don't need to match the white balls in order). If you buy a ticket, what is your probability of winning the jackpot?

Probability of wimning the jackpot is a product the probability of getting the 3 white balls correctly (order unimportant) and the probability of picking the right red superball

Probability of picking 3 white balls from 10

First slot, any of the 3 lucky numbers can fill this slot, (3/10)

Second slot, only 2 remaining lucky numbers can fill this slot, (2/9)

Third slot, only 1 remaining lucky number can fill this slot, (1/8)

(3/10) × (2/9) × (1/8) = (1/120)

Probability of picking the right red superball

(1/10)

Probability of winning the jackpot = (1/120) × (1/10) = (1/120) × (1/10) = (1/1200) = 0.0008333

Hope this Helps!!!

5 0

3 years ago

Write an expression that is equivalent to thank you:)

Serga [27]

Answer:

18x + 8x^2 + 7 divided by 1 + 2x

4 0

3 years ago

Read 2 more answers

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0

4 years ago