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nikitadnepr [17]
3 years ago
11

The staff takes 1/2 of an hour to assemble 1/7 of the booths for a carnival. How long does it take the staff to assemble all of

the booths?
Mathematics
2 answers:
Nata [24]3 years ago
6 0
It would take the staff 3 and 1/2 hours
masha68 [24]3 years ago
6 0
T= time
b= 1 booth (7b = 7 booths)
x= total time (unknown)

1/2 hr ( 30 minutes) * 7 booths = 30*7= 210 minutes

210 minutes / 60 minutes (an hour) =3 1/2 hours


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TB +/- Adj = ATB

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Question 2(Multiple Choice Worth 1 points)
kramer

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6 0
2 years ago
Find x and y<br><br> helppppp !!
Anni [7]

3x + 6 = 48 (alternate angles are equal)

- 6

3x. = 42

÷3

x = 14 degrees

180-48 - 2y + 5y-9 =180

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Explanation:

To find the last angle on the top straight line, do:

180 - (the 2 given angles).

So, 180 - (3x + 16, which is 48 due to alternate angles being equal). Then, minus the 2y.

(180 - 48 - 2y) & simplify => 132 - 2y

This gives you the equation for the missing angle on our top straight line.

Thus, co-interior angles add to 180. So, we add the new equation (132 - 2y) to 5y - 9.

Simplify

=> 123 + 3y (because - 2+5 =3)

and put it equal to 180. Solve for y

Hope this helps!

7 0
1 year ago
Here are 100 seats on an airplane, and each of 100 passengers has a ticket for a different seat. The passengers line up to board
Kryger [21]

Answer:

There are 1% probability that the last person gets to sit in their assigned seat

Step-by-step explanation:

The probability that the last person gets to sit in their assigned seat, is the same that the probability that not one sit in this seat.

If we use the Combinatorics theory, we know that are 100! possibilities to order the first 99 passenger in the 100 seats.

LIke we one the probability that not one sit in one of the seats, we need the fraction from the total number of possible combinations, of combination that exclude the assigned seat of the last passenger. In other words the amount of combination of 99 passengers in 99 seats: 99!

Now this number of combination of the 99 passenger in the 99 sets, divide for the total number of combination in the 100 setas, is the probability that not one sit in the assigned seat of the last passenger.

P = 99!/100! = 99!/ (100 * 99!) = 1/100

There are 1% probability that the last person gets to sit in their assigned seat

7 0
3 years ago
Un carpintero quiere cortar una plancha de triple y de 1m de largo y 60 cm de ancho, en cuadrados lo más grandes posibles. El ca
Ugo [173]

Answer: 20\ cm

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GCF=2^2*5\\\\GCF=4*5\\\\GCF=20

Therefore, the side lenght of each square must be:

s=20\ cm

7 0
3 years ago
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