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Effectus [21]
3 years ago
10

If it takes 100 minutes to complete a circuit, demand at the next work center is 5 parts per minute, and a kanban container hold

s 20 parts, how many containers are required?
Mathematics
1 answer:
Harrizon [31]3 years ago
4 0
5 times 100 = 500
500 divided by 20 = 25
25 containers
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If the x- and y-values in each pair of a set of ordered pairs are interchanged, the resulting set of ordered pairs is known as t
dem82 [27]

Answer:

Inverse of a function

Step-by-step explanation:

If the x- and y-values in each pair of a set of ordered pairs are interchanged, the resulting set of ordered pairs is known as the inverse of a function.

For example, given the following function:

y = 2x

If x=0 → y= 0

If x=1 → y= 2

If x=2 → y= 4

Now, if we find the inverse of the function:  

y = 2x → x = 2y → y = x/2

Now:

If x=0 → y= 0

If x=2 → y= 1

If x=4 → y= 2

Comparing both cases, you will notice that the ordered pairs are effectively interchanged.

3 0
3 years ago
Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

8 0
3 years ago
Read 2 more answers
How to subtract radians?
Dafna1 [17]
You first need to make sure the denominators are equal, then when they are, you can add/subtract the numerators. Then you can place the result over the common denominator.

I hope this helps! :) 
4 0
2 years ago
Read 2 more answers
A study conducted an experiment with 22,000 women over 10 years to investigate whether a high-stress job leads to heart problems
solniwko [45]
The sample of the group would be the 22,000 participants.
3 0
2 years ago
What is the answer and how can I find it
finlep [7]
The answer is C 
1/2(5x + 12) 

because the perimeter is equal 20x + 6 

4 0
3 years ago
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