Please help I have a lot of other due assignments and I don’t have Time and paper thx:)

Median: 69

First Quartile: 65.5

Third Quartile: 73

Interquartile Range: 7.5

Step-by-step explanation:

First, put all of the numbers in order.

63, 64, 67, 68, 70, 72, 74, 76.

Median

For the median, it is the center of the data set.

To find the median of this data set, add 68, and 70, then divide by 2.

The Median is 69.

First Quartile

To find the first quartile, find the range below the median (63-68).

Find the median of that range.

Add 64 & 67 together, then divide by 2.

The first quartile = 65.5

Third Quartile

Do the same as the first quartile, but do it with the range above the median (70-76).

Add 72 & 74 together, then divide by 2.

The third quartile = 73.

Interquartile Range

Now, take the first and third quartile, and subtract.

The interquartile range is 7.5.

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204