Evaluate y = 3x for x = 2.

The production manager is going to present this information to the company's board of directors. Which graph should the manager

viktelen [127]

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.

3 0

4 years ago

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The temperature at 6pm was 0 degrees Fahrenheit at 10 pm the temperature was -11.2 degrees Fahrenheit write an expression that y

lana66690 [7]

The answer is the temperature goes down 2.8° every hour and the expression is x=-11.2/4 because 4 is how many hours between 6pm and 10pm and its -11.2 because that was how much the temperature dropped from 6pm to 10pm

8 0

3 years ago

Find the circumference a circle with the diameter AB A (-8,4) and B (4,-1)

Simora [160]

Answer:

C = 13π ≈ 40.82 units

Step-by-step explanation:

C = 2 r π = d π

AB = $\sqrt{(-8-4)^2 +(4+1)^2}$ = 13 units

C = 13π ≈ 40.82 units ( π ≈ 3.14 )

7 0

3 years ago

Erica makes necklaces from beads. Each necklace consists of blue, white and green beads in the ratio of 6 : 4: 1 One necklace ne

Assoli18 [71]

Answer:

For one Necklace you would need...

24 blue beads.

16 white beads.

4 green beads.

Ratio: 24: 16: 4

Step-by-step explanation:

If you quadruple the amount of beads from the original amount (6 : 4: 1), the total beads for that necklace will be 44.

Step 1: 6 x 4 = 24 (amount of blue beads in necklace).

Step 2: 4 x 4 = 16 (amount of white beads in necklace).

Step 3: 1 x 4 = 4 (amount of green beads in necklace).

Now check to see if the total of all types of beads sum up to 44:

Step 4: 24 + 16 + 4 = 44

______________-

There is your answer:

You need 24 blue beads, 16 white beads, and 4 green beads to have a total sum of 44 beads in that one necklace.

7 0

3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

4 years ago

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