Ask question

Ask question

All categories

2 years ago

11

Historically, the average diameter of the holes drilled has been 0.25 cm and the average range has been 0.10 cm. Samples of size

4 will be collected. For the R-chart, what is the value of upper control limit

1 answer:

Sever21 [200]2 years ago

5 0

Solution :

Given the average range of the holes drilled = 0.10 cm

The average diameter of the holes drilled = 0.25 cm

The sample size = 4

For the sample size of 4, the factor for control limit for the range ( $D_4$ ) = 2.28 (From the table of constants for X-bar and R-chart)

The upper control limit for the R-chart = R-bar x $D_4$

= 0.10 x 2.28

= 0.228

Therefore, the answer is greater than 0.202 but less than or equal to 0.302

You might be interested in

sattari [20]

Answer:

600

Step-by-step explanation:

2000/10=200. 200x3=600

4 0

2 years ago

A system of equations is shown below:

Stolb23 [73]

(-2.-1). You can just plug in the numbers. -1=-6+5 is -1=-1. -1-2(-2)=3 is -1--4=3 is 3=3.

8 0

2 years ago

Read 2 more answers

What is the perpendicular slope intercept form?

In-s [12.5K]

Answer:

y = -1/2x + 2.5

Step-by-step explanation:

- slope of a perpendicular line is the negative reciprocal of the original slope (2 in this case) so y = -1/2 x + c

- to find c we substitute the values from the given point into the equation

(0) = -1/2*(5) + c

- rearrange and calculate and you'll get c = 5/2 or 2.5

- replace the value of c and now you've got y = -1/2x + 2.5

6 0

2 years ago

In your biology class, your final grade is based on several things: a lab score, scores on two major tests, and your score on th

WARRIOR [948]

Answer:

$Weighted\ Average = 73$

Step-by-step explanation:

Given

$Lab = 21\%$

$Tests = 25\%$

$Exam = 29\%$

$Lab\ Score = 60$

$First\ Test = 81$

$Second\ Test = 69$

$Exam = 79$

Required

The weighted average

To do this, we simply multiply each score by the corresponding worth.

i.e.

$Weighted\ Average = Lab\ worth * Lab\ score + Tests\ worth * Tests\ score.....$

So, we have:

$Weighted\ Average = 21\% * 60 + 25\% * 81 + 25\% * 69 + 29\% * 79$

Using a calculator, we have:

$Weighted\ Average = 73.01$

$Weighted\ Average = 73$ --- approximated

4 0

2 years ago

A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The u

OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

$\displaystyle\frac{8000\pi\sqrt{2}}{3}$

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

$x^2+y^2=15^2\to y=\sqrt{225-x^2}$

Then we set the integral for the volume by using shell method:

$\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx$

That can be solved by substitution:

$u=225-x^2\to du=-2xdx$

The limits of integration also change:

For x=5: $u=225-5^2=200$

For x=15: $u=225-15^2=0$

So the integral becomes:

$\displaystyle -\int_{200}^{0}\pi \sqrt{u}du$

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

$\displaystyle \int_{0}^{200}\pi u^{1/2}du$

Then applying the basic rule we get:

$\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}$

Since that is just half of the solid, we multiply by 2 to get the complete volume:

$\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}$

$=\displaystyle\frac{8000\pi\sqrt{2}}{3}$

5 0

3 years ago