Question:
The question is incomplete. The display technology was not given. Find below the complete question and the answers.
Display from technology:
Hypothesis Test Results
μ : Mean of variable
H₀ : μ=2.7 miles
HA : μ >2.7 miles
Variable Sample Mean Std. Err. DF T-Stat P-value
Length 3.23601 0.285185 499 2.230166 0.0131
Step-by-step explanation:
From the result obtained, we have;
Null hypothesis: H₀ : μ=2.7 miles
Alternative hypothesis: HA : μ >2.7 miles
Test statistics = 2.230166
P-value = 0.0131
Significant level: α = 0.05
Since the P value is less than the significance level, we can reject the null hypothesis.
There is no sufficient evidence to support the claim that the mean tornado length is greater than 2.7 miles
Answer:30/4
Step-by-step explanation:180-11x+5, add all up=180, you get 30/4
unleess i messed up somewhere
Answer:
The overview of the given problem is outlined in the following segment on the explanation.
Step-by-step explanation:
The proportion of slots or positions that have been missed due to numerous concurrent transmission incidents can be estimated as follows:
Checking a probability of transmitting becomes "p".
After considering two or even more attempts, we get
Slot fraction wasted,
= ![[1-no \ attempt \ probability-first \ attempt \ probability-second \ attempt \ probability+...]](https://tex.z-dn.net/?f=%5B1-no%20%5C%20attempt%20%5C%20probability-first%20%5C%20attempt%20%5C%20probability-second%20%5C%20attempt%20%5C%20probability%2B...%5D)
On putting the values, we get
= ![1-no \ attempt \ probability-[N\times P\times probability \ of \ attempts]](https://tex.z-dn.net/?f=1-no%20%5C%20attempt%20%5C%20probability-%5BN%5Ctimes%20P%5Ctimes%20probability%20%5C%20of%20%5C%20attempts%5D)
= ![1-(1-P)^{N}-N[P(1-P)^{N}]](https://tex.z-dn.net/?f=1-%281-P%29%5E%7BN%7D-N%5BP%281-P%29%5E%7BN%7D%5D)
So that the above seems to be the right answer.
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