Answer:
Variance 0²
standard deviation 0×
mean α
median x-bar
Step-by-step explanation:
these are the four different symbols that have been used to represent specific statistical measures like for example, mean which is used to calculate the average value of a sample or population. Variance is the what is expected from the squared standard deviation of a random variable from is average vale/ mean. Median is the value that lies in the middle in a random sample and lastly the standard deviation is the square root of the variance and it measures how the values of sample are spread or how far are they from the mean. these are also commonly use in excel a lot to calculate and evaluate data.
Answer:
7/ (3a-1)
Step-by-step explanation:
3a^2 -13a +4 28+7a
------------------ * --------------------
9a^2 -6a+1 a^2 -16
Factor
(3a-1)(a-4) 7(4+a)
------------------ * --------------------
(3a-1) (3a-1) (a-4)(a+4)
Cancel like terms
1 7
------------------ * --------------------
(3a-1) 1
Leaving
7/ (3a-1)
If you’re looking for the probability it’s 1/10
Answer:
A = (0, -8)
Step-by-step explanation:
B = (A+C)/2 . . . . the midpoint is the average of the end points
A = 2B -C = 2(-3, -5) -(-6, -2) . . . solve for A, substitute point values
A = (0, -8)
Answer:
No, the on-time rate of 74% is not correct.
Solution:
As per the question:
Sample size, n = 60
The proportion of the population, P' = 74% = 0.74
q' = 1 - 0.74 = 0.26
We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.
Now,
The proportion of the given sample, p = 
Therefore, the probability is given by:
![P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]](https://tex.z-dn.net/?f=P%28p%5Cleq%200.634%29%20%3D%20%5B%5Cfrac%7Bp%20-%20P%27%7D%7B%5Csqrt%7B%5Cfrac%7BP%27q%27%7D%7Bn%7D%7D%7D%5D%5Cleq%20%5B%5Cfrac%7B0.634%20-%200.74%7D%7B%5Csqrt%7B%5Cfrac%7B0.74%5Ctimes%200.26%7D%7B60%7D%7D%7D%5D)
P![(p\leq 0.634) = P[z\leq -1.87188]](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87188%5D)
P![(p\leq 0.634) = P[z\leq -1.87] = 0.0298](https://tex.z-dn.net/?f=%28p%5Cleq%200.634%29%20%3D%20P%5Bz%5Cleq%20-1.87%5D%20%3D%200.0298)
Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %
Thus the on-time rate of 74% is incorrect.
