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ehidna [41]
3 years ago
14

400 students are assembled in a square form.How many students are assembled in each now?​

Mathematics
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

20

400 is 20 square= 20*20

so  20 students in each row

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Subtract 32 – 15.5<br> help
NikAS [45]

Answer:

16.5

Step-by-step explanation:

32 - 15.5 = 16.5

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3 years ago
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During the Christmas season, a store offers a sale. You buy one item for the regular price then you get 60% discount on a second
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Answer:

24%

Step-by-step explanation:

Given: Regular price of two item is $30 and $20.

           60% discount provided on second item.

Now, lets find out discount provided on second item.

∴Discount provided = \frac{60}{100} \times 20 = \$ 12

Amount paid for second item = Regular price - discount provided

∴Amount paid on second item= 20 - 12= $8

Total amount paid for two item= Regular price of first item + discounted price of second item.

∴Total amount paid for two item= 30+8= \$ 38

Next, lets find out the percent of reduced total cost of two item.

\frac{reduced\ amount}{total\ regular\ price} \times 100

\frac{12}{50} \times 100 = 24\%

∴ By 24\% the total cost of two item reduce during sale.

8 0
4 years ago
there are 16 boxes of crackers there are 10 crackers in each box. how many crackers are in the boxes
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160 in all
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calcola la misura della diagonale di un rettangolo sapendo che il perimetro è 388 cm e la base misura 168.
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3 years ago
The route used by a certain motorist in commuting to workcontains two intersections with traffic signals. The probabilitythat he
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Answer:

a) P(A∩B) = 0.29

b) P1 = 0.1

c) P = 0.35

Step-by-step explanation:

Let's call A the event that the motorist stop at the first signal, and B the event that the motorist stop at the second signal.

From the question we know:

P(A) = 0.39

P(B) = 0.54

P(A∪B) = 0.64

Where P(A∪B) is the probability that he stop in the first, the second or both signals. Additionally, P(A∪B) can be calculated as:

P(A∪B)  = P(A) + P(B) - P(A∩B)

Where P(A∩B) is the probability that he stops at both signals.

So, replacing the values and solving for P(A∩B), we get:

0.64 = 0.39 + 0.54 - P(A∩B)

P(A∩B) = 0.29

Then, the probability P1 that he just stop at the first signal can be calculated as:

P1 = P(A) - P(A∩B) = 0.39 - 0.29 = 0.1

At the same way, the probability P2 that he just stop at the second signal can be calculated as:

P2 = P(B) - P(A∩B) = 0.54 - 0.29 = 0.25

Finally, the probability P that he stops at exactly one signal is:

P = P1 + P2 = 0.1 + 0.25 = 0.35

6 0
4 years ago
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