MATHEMATICS COLLEGE

3/8+ 5/6 divided by 5

Answers

Answer 1

Answer:

Answer:

Answer is 29/120 I think i worked it out so if it inst right truly im sorry but i belive this is the answer.

Answer 2

Answer:

Answer:

= 13

Step-by-step explanation:

3 + 5 / 5

8 6

do the division first as per the law of operation of math requires.

5 / 5 = 5 x 1 = 1

6 6 5 6

3 + 1 = 9 + 4 = 13

8 6 24 24

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4272/6 use partial qutiants to divide

Sec (90 -A) sinA = cot (90-A). tan( 90- A)

Answers

Step-by-step explanation:

sec(90-A) . Sin A = cot (90-A) . tan(90-A)

cosec X sinA = tanA X cotA

1/sinA X sinA = tanA X 1/tanA

1=1

Hence proved

L.H.S=sec(90-A)·sinA

=cosecA·sinA ;[sec(90-A)= cosecA]

=1/sinA·sinA ;[cosecA=1/sinA]

R.H.S=cot(90-A)·tan(90-A)

=tanA·cotA ;[cot(90-A)=tanA, tan(90-A)=cotA]

=tanA·1/tanA ;[cotA=1/tanA]

thus, L.H.S=R.H.S

[Proved]

1. How many one-fifths are there in 17 1/2

Answers

Answer:

Step-by-step explanation:

= 17 1/2 ÷ 1/5

= 35/2 (improper fraction should be used as a dividend) × 5/1 (the divisor will be reciprocal and the operation will be multiplication)

= 175/2

= 87 1/2

Since the question is how many ⅕, the fraction will not be included.

According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function?

Answers

Answer:

Step-by-step explanation:

We play a card game where we receive 13 cards at the beginning out of the deck of 52. we play 50 games one evening. for each of the the following random variable identify the name and parameters of the distribution: a) The number of aces I get in the first game b) The number of games in which I recieve at least one ace during the evening c) The number of games in which all my cards are from the same suit d) The number of spades I receive in 5th game

Answers

The answer & explanation for this question is given in the attachment below.

Final answer:

The number of aces in the first game and the number of spades in the 5th game follow a Hypergeometric Distribution while the number of games receiving at least one ace can be modeled by a Binomial distribution. The event of all cards being from the same suit can be thought of as a Uniform distribution.

Explanation:

a) The number of aces you get in the first game follows a Hypergeometric Distribution. In such a distribution, you are drawing cards without replacement. The parameters are N=52 (the population size), K=4 (the number of success states in the population i.e., the number of aces in a deck), and n=13 (the number of draws).

b) The number of games in which you receive at least one ace can be modeled by a Binomial distribution. Each game you play (out of 50) is a single trial, with the probability of success (getting at least one ace) being the same for every trial. The parameters are n=50 (the number of trials/games) and p (the probability of getting at least one ace).

c) The likelihood of all your cards being from the same suit in a game is heavily reliant on chance, can be modeled as a Uniform distribution given its rare occurrence. Essentially, the parameters would be minimum = 0 and maximum = 1. However, determining the parameters would require calculation of the specific probabilities, which is complex due to the nature of the game.

d) The number of spades you receive in the 5th game also follows a Hypergeometric distribution, similar to the situation in the first game. The parameters in this case are N=52, K=13 (number of spades in a deck), and n=13 (the number of drawn cards).