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3 years ago

Please help me

Mathematics

1 answer:

Stels [109]3 years ago

4 0

Answer:

Step-by-step explanation:

Rational number between 2 and 3 = $\frac{2+3}{2}=\frac{5}{2}$

$2 < \frac{5}{2} < 3$

Now, find the rational number between 2 and 5/2

Rational number between 2 and 5/2 = $\frac{1}{2}(2 + \frac{5}{2})=\frac{1}{2}*(\frac{2*2}{1*2}+\frac{5}{2})=\frac{1}{2}(\frac{4}{2}+\frac{5}{2})=\frac{1}{2}*\frac{9}{2}=\frac{9}{4}$

$2 < \frac{9}{4} < \frac{5}{2} < 3$

Rational number between 5/2 and 3 is

$= \frac{1}{2}(\frac{5}{2}+3)=\frac{1}{2}(\frac{5}{2}+\frac{3*2}{1*2})=\frac{1}{2}*(\frac{5}{2}+\frac{6}{2})=\frac{1}{2}*\frac{11}{2}=\frac{11}{4}$

Rational numbers between 2 and 3 are

9/4 , 5/2 , 11/4

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Answer:

See explanation

Step-by-step explanation:

Solution:-

The random variable, Y be the temperature of chemical reaction in degree fahrenheit be a linear expression of a random variable X : The temperature in at which a certain chemical reaction takes place.

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- The median of the random variate "X" is given to be equal to "η". We can mathematically express it as:

P ( X ≤ η ) = 0.5

- Then the median of "Y" distribution can be expressed with the help of the relation given:

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- The left hand side of the inequality can be replaced by the linear relation:

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Hence,

- Through conjecture we proved that: (1.8*η + 32) has to be the median of distribution "Y".

- Recall that the definition of proportion (p) of distribution that lie within the 90th percentile. It can be mathematically expressed as the probability of random variate "X" at 90th percentile :

P ( X ≤ p_.9 ) = 0.9 ..... 90th percentile

- Now use the conjecture given as a linear expression random variate "Y",

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Y = aX + b

Where, a : is either a positive or negative constant.

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P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

= P ( a*X ≤ a*p_% )

= P ( X ≤ p_% )

= np_%

- When a is negative,

P ( Y ≤ a*p_% + b ) = P ( a*X + b ≤ a*p_% + b )

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3 years ago

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Answer:

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Answer:

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3) A Type I error happens when we reject a null hypothesis that is true. In this case, that would mean that the conclusion is that there is evidence to support the claim that the greater the incentive, the more puzzles are solved, but that in reality there is no significant difference.

4) A Type II error happens when a false null hypothesis is failed to be rejected. In this case, that would mean that there is no enough evidence to support the claim that the greater the incentive, the more puzzles are solved, but in fact this is true.

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Step-by-step explanation:

As the claim is that the greater the incentive, the more puzzles were solved, the null hypothesis will state that this claim is not true. That is that there is no significant relation between the incentive and the amount of puzzles that are solved. In other words, the mean amount of puzzles solved for the different incentives is equal (or not significantly different):

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The research (or alternative hypothesis) is that the greater the incentive, the more puzzles were solved. That means that the mean puzzles solved for an incentive of 50 cents is significantly higher than the mean mean puzzles solved for an incentive of 25 cents and this is significantly higher than the mean puzzles solved for an incentive of 5 cents.

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A Type II error happens when a false null hypothesis is failed to be rejected. In this case, that would mean that there is no enough evidence to support the claim that the greater the incentive, the more puzzles are solved, but in fact this is true.

The probability of a Type I error is equal to the significance level, as this is the chance of having a sample result that will make the null hypothesis be rejected.

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3 years ago

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Answer:

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Step-by-step explanation:

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