Data set is normally distributed with a mean of 27 and a standard deviation of 3.5. Find the Z-score for a value of 25, to the n
2 answers:
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Answer:
A
Step-by-step explanation:
Newton's third law states that for every force, there is an equal and opposite reaction. For example, if you collide two objects together, all of the forces in the system would cancel out because energy conserved with the opposite and equal reactions.
Between two examples (2 identical billiard balls and 2 different toy cars), the easiest way to demonstrate Newton's third law is using the 2 balls, since they are identical in shape and mass. When pushing the two balls in the exact same way with the SAME force, the net force of the equation is basically 0;
Therefore, A is correct.
If you pushed a toy car and a toy truck towards each other using the same force, the outcome would not be completely opposite since F = ma and masses are different. Instead, to have a completely opposite and equal reaction, the resulting direction and forces are different. The lighter car would accelerate more in the opposite direction to compensate for the lighter mass. Therefore, B is wrong because even though they are different sizes, newton's third law still happens in the resulting direction and forces, but are simply harder to explain due to so many differing factors.
C is wrong, because diameter of two objects has no correlation to the net force equations. Even if two objects have the same diameter, they could have different masses, shapes, elasticity, etc.
D is wrong, because it goes against Newton's 3rd law. Newton's 3rd law is a LAW. Meaning, it is ALWAYS true (unless in some distant future it's disproven). In any and every collision, energy is conserved in some way that net forces equal 0; either in friction, gravitational force, or resulting force, the net force is always 0.
population mean, μ =135
population standard deviation, σ = 15
sample size, n = 19
Assume a large population, say > 100,
we can reasonably assume a normal distribution, and a relatively small sample.
The use of the generally simpler formula is justified.
Estimate of sample mean
Estimate of sample standard deviation
Thus, using the normal probability table,
Therefore
The probability that the mean weight is between 125 and 130 lbs
P(125<X<130)=0.0731166-0.0018308
=0.0712858
Exponential laws
![x^{\frac{m}{n}}=\sqrt[n]{x^{m}}](https://tex.z-dn.net/?f=%20x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20)
so
(m^(2/3))^(1/2)=m^(2/3 times 1/2)=m^(2/6)=m^(1/3)=![\sqrt[3]{m^{1}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bm%5E%7B1%7D%7D%20)
= ![\sqrt[3]{m}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bm%7D%20)
so
(m^(2/3))^(1/2)=m^(2/3 times 1/2)=m^(2/6)=m^(1/3)=