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3 years ago

Collection of_______ is called Statistics.

Mathematics

1 answer:

siniylev [52]3 years ago

5 0

Answer:

Collection of numerical information is called Statistics

Step-by-step explanation:

From the Oxford dictionary, the definition of statistics (science) is: "the practice or science of collecting and analyzing numerical data in large quantities, especially to infer proportions in a whole from those in a representative sample".

While the definition of statistic (data) is: "a fact or piece of data obtained from a study of a large quantity of numerical data".

Statistics is a science, not a method, therefore answer can´t be a).

Not all sample data is necessarily a statistic, therefore answer can´t be b).

All statistics are numerical information.

Not all statistics are population data, therefore answer can´t be d).

Cleaned data is information filtered with a specific criterion, but not necessarily used in a statistic.

The most probable answer is c.

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Stephen says that the numbers 38 and 40 are relatively prime. Explain why he is incorrect in making this statement. (make a good

Paha777 [63]

Answer:

He is incorrect in this statement because 38 and 40 both have factors, other than 1 x itself. Prime numbers are numbers the only have 1 factor pair, and that is 1 x itself. Factors are the numbers you multiply to get the end result, and an example is 10x4=40. The factors of 40 are 8x5, 4x10, 2x20, and of course, 1x40. The factors of 38 are 2x19, and 1x38.

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3 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$

Hence, probability of the stronger team winning overall is:

$=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103$

#Define Y as the random variable for winning 2/3 games.:

$P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480$

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

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3 years ago

A board game that normally costs $30 is on sale for 25 percent off. What is the sale price of the game?

Roman55 [17]

25% off translates into 0.25($30) = $7.50 off.

A faster way to determine the sales price is to subtract 0.25 from 1.00 and then mult. the regular price by the result: sales price = 0.75($30) = $22.50 (answer)

5 0

3 years ago

In a casino in Blackpool there are two slot machines (machine 1 and machine 2): one that pays out 10 % of the time, and one that

otez555 [7]

Answer:

Given that;

2 machines, 1 and 2

Machine 1

pays 10% of times when the machine is generous

So the probability that the machine 1 pays given that its generous is

P (Pays/machine 1 is generous) = 10% = 0.10

Machine 2

pays 20% of times when its generous

i.e the probability that the machine pays given that its generous is;

P (Pays / machine 2 is generous) = 20% = 0.20

Also we assume there is equal chance of being generous

i.e

P(machine 1 is generous) = P(machine 2 is generous ) = 0.5

this is to help obtain the probability that machine 1 is generous given that the player loose in the first bet,

i.e P(machine 1 is generous / lost)

Since, the probability that the machine is generous is 0.5,

it can be said that there is 50% chance that machine 1 is generous when the player loses the first bet

therefore The required probability is calculated as;

P(machine 1 is generous/lost) = p(machine 1 is generous ∩ lost) / p (lost)

= [p(lost/machine 1 is generous) × p(machine 1 is generous)] / [{(1-p(pays/machine 1 is generous)) × p(machine 1 is generous)} + {((1-p(pays/machine 2 is generous)) × p(machine 2 is generous)}]

= [(1 - p(pays/machine 1 is hereous)) × p(machine 1 is generous)] × [(( 1- 0.10) × 0.5) + ((1 - 0.20) × 0.5)]

= ( 1- 0.10) × 0.5) / 0.85

= 0.5294

so the probability that the player loses the first bet given that the machine is generous is 0.5294

Since the gotten probability that the player loses the first bet given that machine 1 is generous is close to 0.50 then it can be said that the probability is consistent with the expectations.

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3 years ago

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murzikaleks [220]

Answer:

The third answer

Step-by-step explanation:

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3 years ago