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Paladinen [302]
4 years ago
12

The normal time to perform a repetitive manual assembly task is 4.25 min. In addition, an irregular work element whose normal ti

me is 1.75 min must be performed every 8 cycles. Two work units are produced each cycle. The PFD allowance factor is 16%. Determine the anticipated amount of time worked per 8-hour shift that corresponds to the PFD allowance factor of 16%.
Mathematics
1 answer:
34kurt4 years ago
8 0

Answer:

76.8 minutes

Step-by-step explanation:

Every 8 cycles we have 6 regular work elements plus 1 irregular work element.  

So, the normal time for 8 cycles is  

3*4.25+1.75 = 14.5 minutes.

Since the PDF allowance factor is 16%, the standard time for 8 cycles is 14.5 increased 16%

14.5*1.16 = 16.82 minutes.

So, the anticipated amount of time worked every 14.5 minutes

is 2.32 minutes.

An 8-hour shift has 480 minutes = 33.1034*14.5, so  the anticipated amount of time per 8-hour shift is

33.1034*2.32 = 76.8 minutes

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katovenus [111]

Answer:

  all real numbers

Step-by-step explanation:

The function f(x) = 3|x +4| +1 is defined for all values of x. Its domain is <em>all real numbers</em>.

7 0
4 years ago
Read 2 more answers
the perimeter of a triangular plot of land is 2400 ft. the longest feet is 200 ft less than twice the shortest. the middle side
noname [10]
P = a + b + c
p = 2400
a = longest = 2c - 200
b = middle = a - 200 = 2c - 200 - 200 = 2c - 400

2400 = (2c - 200) + (2c - 400) + c
2400 = 5c - 600
2400 + 600 = 5c
3000 = 5c
3000/5 = c
600 = c <== shortest side

a = 2c - 200.....= 2(600) - 200 = 1200 - 200 = 1000<== longest side
b = 2c - 400...= 2(600) - 400 = 1200 - 400 = 800 <== middle side
3 0
3 years ago
Let A be a 3×3 matrix and suppose we know that −2a1+3a2−5a3=0 where a1,a2 and a3 are the columns of A. Write a non-trivial solut
belka [17]

Answer:

One of the obvious non-trivial solutions is  (x_1, x_2, x_3)=(-2, 3, -5).

Step-by-step explanation:

Suppose the matrix A is as follows:

A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&3_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right]

The observed system Ax=0 after multiplying looks like this

Ax=0 \iff \left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \cdot \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] =0 \iff \\ \\a_{11}x_1+a_{12}x_2+a_{13}x_3=0\\a_{21}x_1+a_{22}x_2+a_{23}x_3=0\\a_{31}x_1+a_{32}x_2+a_{33}x_3=0\\\\

Since we now that -2A_1+3A_2-5A_3=0, where A_i\ ,\  i=1, 2, 3 are the columns of the matrix A, we actually know this:

-2\cdot \left[\begin{array}{ccc}a_{11}\\a_{21}\\a_{31}\end{array}\right] +3\cdot \left[\begin{array}{ccc}a_{12}\\a_{22}\\a_{32}\end{array}\right] -5\cdot \left[\begin{array}{ccc}a_{13}\\a_{23}\\a_{33}\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right]

Once we multiply and sum up these 3 by 1 matrices, we get that these equations hold:

-2a_{11}+3a_{12}-5a_{13}=0\\-2a_{21}+3a_{22}-5a_{23}=0\\-2a_{31}+3a_{32}-5a_{33}=0

This actually means that the solution to the previously observed system of equations (or equivalently, our system Ax=0) has a non-trivial solution (x_1, x_2, x_3)=(-2, 3, -5).

8 0
3 years ago
Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou
monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

              B) 0.6796

              C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )

                   = P ( Z > 2.33 ) = 0.0099

                   = 0.9%

B) percentage of oranges weighing less than 8.7 ounces

  P( x < 8.7 ) = P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )

                    = P ( Z < 0.4667 ) = 0.6796

                    = 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = P (\frac{6.2-8}{1.5} <  \frac{x - u}{ std} < \frac{7-8}{1.5} )

                          = P ( -1.2 < Z < -0.66 )

                          = Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0
4 years ago
BRAINLIEST if right! 
Alisiya [41]
The formula for the area of a circle is pi times the radius squared. Here are the steps to solving this problem.

1. First, find the radius. Since the radius is allows half of the diameter, then we divide the diameter by two. 50/2= 25

2. Next, we square the radius. 25^2= 625

3. After that we multiply 625 ft by pi(3.14).
625 times 3.14= 1962.5

4. The last step is to round the the answer to the nearest square foot. Since 5 is at the end, we round up to 1,963.

So the answer is 1,963 ft^2.
4 0
3 years ago
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