Answer:
0.918 is the probability that the sample average sediment density is at most 3.00
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 2.80
Standard Deviation, σ = 0.85
Sample size,n = 35
We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling:
![=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cdfrac%7B0.85%7D%7B%5Csqrt%7B35%7D%7D%20%3D%200.1437)
P(sample average sediment density is at most 3.00)
Calculation the value from standard normal z table, we have,
0.918 is the probability that the sample average sediment density is at most 3.00
Answer:
C
Step-by-step explanation:
1st year-> 30000
2nd year-> 31800
3rd year ->33708
4th year->35730.48
5th year -> 37874.3088
37874.3088+35730.48+33708+31800+30000= 169112.7888
≈ 169113
49 because he can play all the cards at the same time if he can, also he can play his cards between 1 and 7 times so you would multiply 7 by its self.