Answer:
See explanation
Explanation:
Solution:-
- Three students measure the volume of a liquid sample which is 6.321 L.
- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:
Students
Trial A B C
1 6.35 6.31 6.38
2 6.32 6.31 6.32
3 6.33 6.32 6.36
4 6.36 6.35 6.36
- We will define the two terms stated in the question " precision " and "accuracy"
- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.
- The mean measurement taken by each student would be as follows:
![E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\](https://tex.z-dn.net/?f=E%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.35%20%2B6.32%2B6.33%2B6.36%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20A%20%29%20%3D%206.34%5C%5C%5C%5CE%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.31%20%2B6.31%2B6.32%2B6.35%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20B%20%29%20%3D%206.3225%5C%5C%5C%5CE%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.38%20%2B6.32%2B6.36%2B6.36%7D%7B4%7D%20%5C%5C%5C%5CE%20%28%20C%20%29%20%3D%206.355%5C%5C)
- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:
![Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\](https://tex.z-dn.net/?f=Var%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.35%5E2%2B6.32%5E2%2B6.33%5E2%2B6.36%5E2%7D%7B4%7D%20-%206.34%5E2%5C%5C%5C%5CVar%20%28%20A%20%29%20%3D%200.00025%5C%5C%5C%5CVar%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.31%5E2%2B6.31%5E2%2B6.32%5E2%2B6.35%5E2%7D%7B4%7D%20-%206.3225%5E2%5C%5C%5C%5CVar%20%28%20B%20%29%20%3D%200.00026875%5C%5C%5C%5CVar%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.38%5E2%2B6.32%5E2%2B6.36%5E2%2B6.36%5E2%7D%7B4%7D%20-%206.355%5E2%5C%5C%5C%5CVar%20%28%20C%20%29%20%3D%200.000475%5C%5C)
- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:
Var ( A ) < Var ( B ) < Var ( C )
most precise Least precise
- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .
![P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\](https://tex.z-dn.net/?f=P%20%28%20A%20%29%20%3D%20%5Cfrac%7B6.34-6.321%7D%7B6.321%7D%2A100%3D%200.30058%5C%5C%5C%5CP%20%28%20B%20%29%20%3D%20%5Cfrac%7B6.3225-6.321%7D%7B6.321%7D%2A100%3D%200.02373%5C%5C%5C%5CP%20%28%20C%20%29%20%3D%20%5Cfrac%7B6.355-6.321%7D%7B6.321%7D%2A100%3D%200.53788%5C%5C)
- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:
P ( B ) < P ( A ) < P ( C )
most accurate least accurate
Answer:
Not seeing any other information, the best answer I can give is 2m.
Explanation:
M = magnitude
You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.
Answer:
Do you mean 4m^3 and 3.0 tones?
Explanation:
solution:
Mass = m = 3.0 tones
- 1 ton = 1,000 kg
= 3.0 × 1,000
= 3,000 kg
volume = v = 4m^3
Required:
Mass density of oil = p = ?
We know that;
![p = \frac{mass}{volume} = \frac{m}{v} = \frac{3000}{4} = 750kg |m^{3} ans](https://tex.z-dn.net/?f=p%20%3D%20%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%3D%20%20%5Cfrac%7Bm%7D%7Bv%7D%20%3D%20%20%5Cfrac%7B3000%7D%7B4%7D%20%3D%20750kg%20%7Cm%5E%7B3%7D%20ans%20%20)
The answer is:
750kg / m^3
Answer:
see vous se to pe a he ko off a nack u
Answer:
f=1.59 Hz
Explanation:
Given that
Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.
Velocity = 100 mm/s
Maximum amplitude = 10 mm
We know that for a simple undamped system spring mass system
![V_{max}=\omega A](https://tex.z-dn.net/?f=V_%7Bmax%7D%3D%5Comega%20A)
now by putting the values
![V_{max}=\omega A](https://tex.z-dn.net/?f=V_%7Bmax%7D%3D%5Comega%20A)
100 = ω x 10
ω = 10 rad/s
We also know that
ω=2π f
10 = 2 x π x f
f=1.59 Hz
So the natural frequency will be f=1.59 Hz.