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makkiz [27]
3 years ago
8

Use inverse variation equation to fill the table.

Mathematics
2 answers:
ASHA 777 [7]3 years ago
7 0

The correct answers to the question above includes:

a=.1
b=20.775
c=0.02


mestny [16]3 years ago
4 0

Answer:

Inverse Variation states that a relationship between two variables x and y, represent an inverse variation if it can be expressed in the form of:

i.e y \propto \frac{1}{x}  or y = \frac{k}{x} ;  where k is the constant of variation.

In the given problem :

P=\frac{8.31}{v} where v is the volume(liters) and P is the pressure.

We have the value of the constant K

i.e  K = 8.31

so, Pv = 8.31                  ......[1]

From the given table;

if v = 83.1 liters and P= a ;

Substitute in [1] to solve for a;

83.1a = 8.31

Simplify:

a = 0.1 kilopascals

Similarly,

if v = b liters and P = 0.4 kilopascals

Substitute in [1] to solve for b;

0.4b = 8.31

Simplify:

b = 20.775 liters

and

if v = 415.5 liters and P =c ;

Substitute in [1] to solve for c;

415.5c = 8.31

Simplify:

c = 0.02 kilopascals .

Therefore, value of:

a = 0.1

b = 20.775

c = 0.02


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Answer:

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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This means that \mu = 75, \sigma = 22

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This means that n = 144, s = \frac{22}{\sqrt{144}} = 1.8333

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The probability that it is less than 70 is the pvalue of Z when X = 70. So

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Z = \frac{X - \mu}{s}

Z = \frac{70 - 75}{1.8333}

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0.0064*100% = 0.64%

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