The answer is 0.3 cause when you subtract all of it you can get it
Hi!
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

To calculate the degrees of freedom you need to use the following equation:

≈34
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one,
we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.
Answer:
Uneducated people.
Step-by-step explanation:
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Answer:
The probability that the selected adult has liver problems is 0.08
Step-by-step explanation:
In this question, from the data given, we want to calculate the probability that an adult selected at random has liver problems.
Let E(L) be the event that an adult has liver problems.
The probability is directly obtainable from the question and it is given as 8%
Thus, the probability that the selected adult has liver problems; P(L) = 8% = 8/100 = 0.08