Answer:
a) -1, 0
b) -5, -4
c) 2, 3
d) -8, -7
Step-by-step explanation:
I think its "Yes it can because -3.5 lies to the left of -1.5"
To find the area of the shaded region you need find the area
of the shaded region and subtract the area of the unshaded region.
Area of a rectangle = width x length
A = (x + 10) x (2x + 5)
Next apply FOIL or
First Outer Inner Last
A = (x * 2x) (x * 5) (10 * 2x) (10 * 5)
A= 2x2 + 5x + 20x + 50
A= 2x2 +25x +50
Area of a square= sides2
A= (x + 1)2
A= (x+1) (x+1)
Next apply FOIL or
First Outer Inner Last
A = (x *x) (1*x) (1*x) (1*1)
A = x2 + 1x + 1x +1
A= x2 + 2x +1
A= 2x2 +25x +50 - 2x2 +25x +50
A= 50x + 100
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.
1532 ÷ 36 = 42 and remain 20.
So there were 20 CDS in the remains box