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Bond [772]
4 years ago
14

What is an important difference between statistics andâ parameters?

Mathematics
1 answer:
Ilia_Sergeevich [38]4 years ago
7 0

Answer:

A  characteristic , usually unknown of the population.

Characteristic of the population , such as its mean ,variance and standard deviations are population <u>Parameters .</u>

An observed characteristic of a sample are known as statistics

Estimate the use of a sample  characteristic (statistic) to guess or approximate a population characteristic (parameter)

<u></u>

Step-by-step explanation:

<u>Introduction</u>:-

The outcome of a statistical experiment may be recorded either as a numerical value or as a descriptive representation.

Example:-1

When a pair of dice are tossed and the sum of the numbers on the faces is the outcome of interest, we record a numerical value.

Example :-2

If the students of a certain school are given blood tests and the type of blood is of interest, then a descriptive representation.

<u>Population:-</u>

Population is consists of the total observation's with which we are concerned

The number of observations in the population is defined to be the size of the population

<u>Sample</u>:- A sample is a subset of population.

Samples are classified in two ways

Large sample : if the size of the sample n≥ 30 is called the large sample.

small sample: if the size of the sample n<30 is called the small sample.

Statistics uses different names and symbols to distinguish characteristics

of the population from those of a sample.

<u>Parameters </u>:-

A  characteristic , usually unknown of the population.

Characteristic of the population , such as its mean ,variance and standard deviations are population <u>Parameters .</u>

In symbolically mean of the population parameter is denoted by<u> μ </u>

In symbolically variance of the population parameter is denoted by<u> σ²</u>

In symbolically standard deviation of the population parameter is denoted by <u> σ</u>

<u>Statistic :-</u>

An observed characteristic of a sample are known as statistics

Estimate the use of a sample  characteristic (statistic) to guess or approximate a population characteristic (parameter)

In symbolically mean of the sample statistic is denoted by<u> x⁻</u>

In symbolically variance of the sample statistic is denoted by<u> S²</u>

In symbolically standard deviation of the statistic is denoted by <u> S</u>

<u></u>

<u></u>

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Answer:

{x = 3, y = 9 thus Answer A is correct

Step-by-step explanation:

{5 y - 7 x = 24 | (equation 1)

{5 y - 9 x = 18 | (equation 2)

Swap equation 1 with equation 2:

{-(9 x) + 5 y = 18 | (equation 1)

{-(7 x) + 5 y = 24 | (equation 2)

Subtract 7/9 × (equation 1) from equation 2:

{-(9 x) + 5 y = 18 | (equation 1)

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Multiply equation 2 by 9/10:

{-(9 x) + 5 y = 18 | (equation 1)

{0 x+y = 9 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(9 x)+0 y = -27 | (equation 1)

{0 x+y = 9 | (equation 2)

Divide equation 1 by -9:

{x+0 y = 3 | (equation 1)

{0 x+y = 9 | (equation 2)

Collect results:

Answer: {x = 3, y = 9

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A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif
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This question is incomplete, the complete question is;

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel consumed at rate r, and the exhaust gases are ejected with constant velocity v(sub e) [relative to the rocket]. A model for the velocity of the rocket at time t is given by the equation:

v(t) = -gt - v(sub e) ln [(m-rt)/m]

where g is the acceleration due to gravity, and t is not too large. If g = 9.8 m/s^2, m = 30,000 kg, r = 160 kg/s, and v(sub e)= 3000 m/s, find the height of the rocket one minute after liftoff.

Answer:

the height of the rocket one minute after liftoff is 14.8441 km

Step-by-step explanation:

Given the data in the question;

Velocity of the rocket at time t is;

v(t) = -gt - V_{e\\} ln( \frac{m-rt}{m})

\int\limits^t_0 {v(t)} \, dt = ₀∫⁶⁰(  -gt - V_{e\\} ln( \frac{m-rt}{m}) )dt

= -g([\frac{t^{2} }{2}]^{60}_0  - V_{e\\} ₀∫⁶⁰In( \frac{m-rt}{m}) dt

= -g([\frac{t^{2} }{2}]^{60}_0  - V_{e\\} [ t In( \frac{m-rt}{m}) + \frac{m}{r} - t - mr\frac{1}{r^{2} } In |m-rt| ]^{60}_0

= -g([\frac{60^{2} }{2}]  - V_{e\\} [ 60 (In(1 -  \frac{60r}{m})-1) + \frac{m(In(m^{2})-In((m-60r)^2))) }{2r} ]

we input our given values; g=9.8 m/s², m=30000 kg, Ve = 3000 m/s, r = 160 kg)

= -9.8([\frac{60^{2} }{2}]  - 3000 [ 60 (In(1 - \frac{60(160)}{m})-1) + \frac{30,000(In(30,000^{2})-In((30,000-60(160))^2))) }{2(160)} ]

\int\limits^t_0 {v(t)} \, dt =  14844.10 m

we convert to kilometer

\int\limits^t_0 {v(t)} \, dt =  14844.10 ÷ 1000

\int\limits^t_0 {v(t)} \, dt =  14.8441 km

Therefore, the height of the rocket one minute after liftoff is 14.8441 km

   

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3 years ago
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