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2 years ago

13

You are working in a small, student-run company that sends out merchandise with university branding to alumni around the world.

Every day, you take a sample of 50 shipments that are ready to be shipped to the alumni and inspect them for correctness. Across all days, the average percentage of incorrect shipments is 5 percent. What would be the upper control limit for a p-chart

1 answer:

topjm [15]2 years ago

7 0

Answer:

The answer is "0.142466".

Step-by-step explanation:

Using the p formula for the proportion of nonconforming units through the subgrouping which can vary in sizes:

$p =\frac{np}{n}\\\\$

$\bar{p}=\frac{\Sigma np}{\Sigma n}\\\\$

Defects $= \frac{5}{100} \times 50 \\\\$

$p = \frac{5}{100} \times \frac{50}{50}=0.05\\\\$

It calculates the controls limits through the p-chart that is:

$UCL_{p},LCL_{p}=\bar{p} \pm \sqrt{\frac{\bar{p}(1-\bar{p})}{\bar{n}}}\\\\$

So, the upper control limits:

$= 0.05 + 3 \times \sqrt{(\frac{0.05\times(1-0.05)}{50})} \\\\= 0.142466$

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strojnjashka [21]

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2 years ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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1 year ago

Find the area of the figure

uranmaximum [27]

Answer:

30 units

Step-by-step explanation:

The formula for finding the area of a triangle is 1/2b x h

So b = 12

and h = 5

1/2(12) = 6

6 x 5 = 30

I could be wrong here so

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8 0

2 years ago

Solve the formula A=12bh for h when A=90 and b=15. Write the answer in decimal form.

ohaa [14]

Answer:

Given,

A=90

b=15

A=12bh

Now,

Step-by-step explanation:

Put the given value in the formula

A=12bh

90=12*15*h

90=180h

90/180=h

Hence, h=0.5

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3 years ago

Can someone please help me with this

Mama L [17]

Answer:

27.

<em>Equation of 2 tangent lines at the given curve going through point P is:</em>

<em>y = -7x + 1</em>

<em>y = x + 1</em>

28.

<u>Part 1:</u> 400 feet

<u>Part 2:</u> Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

<u>Part 3:</u> acceleration at any time t is -32 feet/second squared

<u>Part 4:</u> t = 10 seconds

29.

<u>Part 1:</u> Average Rate of Change = -15

<u>Part 2:</u> The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

<u />

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

<em>So equation of 2 tangent lines at the given curve going through point P is:</em>

<em>y = -7x + 1</em>

<em>y = x + 1</em>

<em></em>

28.

<u>Part 1:</u>

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

<u>Part 2:</u>

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

<u>Part 3:</u>

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

<u>Part 4:</u>

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$

We disregard t = 0 because that's basically starting. So we take t = 10 seconds as our answer and we know rock hits the ground at t = 10 seconds.

29.

<u>Part 1:</u>

The average rate of change is basically the slope, which is

Slope = Change in y/ Change in x

The x values are given, from 2 to 3, and we need to find corresponding y values by plugging in the x values in the function. So,

When x = 2, $y=f(2)=-(2)^3 + 4(2) + 2=2$

When x = 3, $y=f(3)=-(3)^3 + 4(3) + 2=-13$

Hence,

Average Rate of Change = $\frac{-13-2}{3-2}=-15$

<u>Part 2:</u>

The instantaneous rate of change is got by differentiating the function and plugging the 2 points and finding the difference.

First, let's differentiate:

$f(x)=-x^3+4x+2\\f'(x)=-3x^2+4$

Now, find the derivative at 3,

$f'(x)=-3x^2+4\\f'(3)=-3(3)^2+4=-23$

finding derivative at 2,

$f'(x)=-3x^2+4\\f'(2)=-3(2)^2+4=-8$

The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

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3 years ago