To best emphasize the number of defects. Manager should use graph 3 (refer the image shown):
If we talk about graph 1, it can also be used but usually we put the time line on the horizontal axis, for the convenience and the quantity to be measured on the y-axis. In the graph 1, the time is placed on the vertical axis (x-axis) so it would not be a good pick for the manager.
Same is the case with graph 2 again we have time on the vertical axis. So it is not a good idea to with graph 2.
Graph 3 could be the best to emphasize the number of defects because first of all time is placed on the horizontal axis and the quantity to be shown is on the vertical axis. Secondly, the range of the vertical axis is less so it is easy to observe the data set on the graph quite distinctively. Therefore, graph 3 is the best pick.
Graph 4 is placed correctly in terms of vertical and horizontal axes but the range of vertical axis is quite high due to which the dispersion or the display of the data is quite compressed and it gets hard to visualize.
<span>If i started a business making braclets for $2 and necklaces for $3 with a budget >200 what would be the inequality equation?
How would I graph
Solution:
let x= number of bracelets
y= number of necklaces
Thus, the equation would be:
</span>$2x+$3y>200
In graphing this equation, you just have to substitute any number for the values of x and y as long as when substituted, their sum is greater than 200
Answer:
17
Step-by-step explanation:
Let us suppose two years ago my brother's age was x years
Then, my age was 3x
Three years from now, my brother's age will be (x +2+3) = (x+5) years
And my age will be (3x+2+3) = (3x+5) years
But it is given that i will be twice as old as my brother.
So, 2(x+5)= (3x+5)
or, x= 5 years
So my brother's present age is 5+2= 7 years
And my age is 5*3+2= 17 years
Answer:
Y= -6x + 1
Step-by-step explanation:
Y=mx + b
M is your slope
B is your y intercept
Answer:
- the domain is [0,129]
- 112 mm
- 1942
Step-by-step explanation:
1. The function is good for years 1880 to 2009, 0 to 129 years after 1880. Values of x can be anything in the domain [0, 129].
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2. 1950 -1880 = 70. The year 1950 corresponds to x=70, so the function tells us the water level rose ...
f(70) = 1.6·70 = 112 . . . . . mm
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3. We want to find x when f(x) = 100. That will be the solution to ...
100 = 1.6x
100/1.6 = x = 62.5
Then 62.5 years after 1880 is year 1942.5. The water level was 100 mm higher than in 1880 in the year 1942.
