If 3t − 7 = 5t , then 6t =

There are four steps in the manufacturing process of a stuffed toy: cutting, stuffing, sealing, and packaging. There are two emp

Anit [1.1K]

Answer:

(a) 2min/15units

(b) 1min/20units

(c) The bottleneck stage is cutting. The process capacity is 450units/hour

(d) The capacity of the new employee should be 1unit/2sec. The process capacity will increase four times

Step-by-step explanation:

(a) Cycle time for stage cutting for 2 employees = 16sec/2units × 1min/60sec = 2min/15units

(b) Cycle time for stage sealing for 1 employee = 3sec/1unit × 1min/60sec = 1min/20units

(c) Process capacity of stage cutting = 15units/2min × 60min/1hour = 450units/hour

(d) If the capacity of the new employee is 1unit/2sec, the process capacity = 1unit/2sec × 3600sec/1hour = 1800units/hour. The process capacity increases 4 times (1800/450 = 4)

7 0

4 years ago

1 and 1/2 divided by 4 and 2/8

Bumek [7]

Answer:

1.375

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

T-SHIRTS You can buy 3 T-shirts for $24. Write a proportion that gives the cost

GrogVix [38]

7:56

Figure out what 7/3 is so you can multiply 24 by that number. Is is 2 1/3 so when you multiply 2 1/3 by 24 it gets you to 56 so the proportion is 7:56

8 0

2 years ago

Find the point P on the graph of the function y=√x closest to the point (9,0)

Sphinxa [80]

Answer:

$\displaystyle \frac{17}{2}$ .

Step-by-step explanation:

Let the $x$ -coordinate of $P$ be $t$ . For $P\!$ to be on the graph of the function $y = \sqrt{x}$ , the $y$ -coordinate of $\! P$ would need to be $\sqrt{t}$ . Therefore, the coordinate of $P \!$ would be $\left(t,\, \sqrt{t}\right)$ .

The Euclidean Distance between $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ is:

$\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}$ .

The goal is to find the a $t$ that minimizes this distance. However, $\sqrt{t^2 - 17 \, t + 81}$ is non-negative for all real $t\!$ . Hence, the $\! t$ that minimizes the square of this expression, $\left(t^2 - 17 \, t + 81\right)$ , would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ .

Differentiate $\left(t^2 - 17 \, t + 81\right)$ with respect to $t$ :

$\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17$ .

$\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2$ .

Set the first derivative, $(2\, t - 17)$ , to $0$ and solve for $t$ :

$2\, t - 17 = 0$ .

$\displaystyle t = \frac{17}{2}$ .

Notice that the second derivative is greater than $0$ for this $t$ . Hence, $\displaystyle t = \frac{17}{2}$ would indeed minimize $\left(t^2 - 17 \, t + 81\right)$ . This $t\!$ value would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ , the distance between $P$ $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ .

Therefore, the point $P$ would be closest to $(9,\, 0)$ when the $x$ -coordinate of $P\!$ is $\displaystyle \frac{17}{2}$ .

8 0

3 years ago

5.06 divided by 5.60

allsm [11]

Answer:

0.90357142857

Step-by-step explanation:

6 0

3 years ago