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: Last Option

Step-by-step explanation:
In this case we have a uniform probability. In the graph the horizontal axis represents the possible values of the variable x and the vertical axis represents the probability P(x).
To calculate the probability that x is between 4.71 and 7.4 we calculate the area under the curve.
The horizontal length between 4.71 and 7.4 is:
.
Then notice that the vertical length in this interval is 0.125.
Then the area of a rectangle is:

Where l is the length and w is the width.
In this case we have to:


So


Answer:
1 D (10, 120)
2 C (20, 40)
Step-by-step explanation:
Answer:
y = 2x+7
Step-by-step explanation:
y = mx + c
m = slope
c is the y intercept
so
y = 2x+ 7
You can do this by finding a number close to it, and then finding out how to do it.
For example with 63(square root sign), find out what is close to it.
7x7=49 while 8x8=64
Because of this, we now know that it is between 7 and 8 but is also really close to 8.
Our number could be 7.9
Hope this helped!!!