Answer:
u drag the green dot thing over to the coordinate (6,8)
Step-by-step explanation:
i put a black dot where i think ur supposed to drag the green dot to
2x=5 so 5÷2 = x (2.5)
3y = 4 so 4÷3 = y (1.3 recurring)
4z = 3 so 3÷4 = z (0.75)
implement: 24x2.5x1.3•x0.75
hope i answered right
Answer:
The interquartile range is the difference between the highest and lowest values in the middle of a data set.
Step-by-step explanation:
The range is the difference between the maximum and minimum value, hence, it cannot be greater than the maximum value, which is the greatest value in a dataset, the highest value a range could have being equal to the maximum value when the minimum vlaue of the dataset is equal to 0.
The mean is the average value of a dataset, hence, it cannot be greater than the maximum value.
The interquartile range is the middle 50% or half of a dataset and not the difference between the highest and lowest middle values in the middle. It is obtained by taking the difference of the upper and lower QUARTILE.
Answer:
Height: 3/2 inches
Length: 12 inches
Width: 4 inches
Step-by-step explanation:
Let x is the side length of the square
The height of the box by cutting squares off :x
- The new length of the cardboard = 15 -2x (because we cut from 4 corners)
- The new width of the cardboard = 7 -2x (because we cut from 4 corners)
The new volume of it is:
V = (15 -2x) (7 -2x) x
<=> V =
To maximum volume, we use the first derivative of the volume
<=> 
<=> 
<=> 2x -3 = 0 or 6x -35 = 0
<=> x = 3/2 or x = 35/6
To determine which value of x gives a maximum, we evaluate
= 24x -88
= 24(3/2) -88 = -52
= 24(35/6) -88 = 52
We choose x = 3/2 to have the maximum volume because the value of x that gives a negative value is maximum.
So the dimensions (in inches) of the box is:
Height: 3/2 inches
Length: 15-2(3/2) = 12 inches
Width: 7 - 2(3/2) = 4 inches
First split your model into 10 equal sections then you shade/color in 6 of them.
There it goes that is 6/10 bench mark!
~JZ
Hope it helps