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vredina [299]
3 years ago
15

A hostel having strength of 300 students requires on an average 36000 L of water per day. It has a tank measuring 10 m × 8 m × 9

m. For how many days will the water in the tank filled to capacity last?
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

The water in the tank filled to capacity lasts 20 days.

Step-by-step explanation:

Total number of students in the hostel = 300

The average need of water  per day = 36,000 L

The dimensions of the tank =   10 m  ×   8 m   ×   9 m

Now, the capacity of the tank = LENGTH X WIGHT X HEIGHT

     = 10 m   ×   8 m   ×   9 m =  720 cubic meters

Now, 1m^{3}   = 1,000L

⇒720 m^{3}   = 720  \times  1,000L  = 720,000L

So, the capacity of the tank = 720,000 L

and the average usage of water by 300 students = = \frac{720,000 L}{36,000 L}  = 20

⇒\textrm{The number of days water lasts}  = \frac{\textrm{Total capacity of the tank}}{\textrm{Average need of water per day}}

or, number of days = 20

Hence,  the water in the tank filled to capacity lasts 20 days.

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Answer:

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Step-by-step explanation:

We are given that five nanotubules are inserted in a single cancer cell. Independently of each other, they become exposed to near-infrared light with probabilities 0.2, 0.4, 0.3, 0.6, and 0.5.

Let the event that a cell is killed be 'A' and the event where the ith nanotubule kill the cell be '\text{B}_i'.

This means that the cell will get killed if \text{B}_1 \bigcup \text{B}_2 \bigcup \text{B}_3 \bigcup \text{B}_4 \bigcup \text{B}_5 happens. This represents that the cell is killed if nanotubule 1 kills the cell, or nanotubule 2 kills the cell, and so on.

Here, P(\text{B}_1) = 0.2, P(\text{B}_2) = 0.4, P(\text{B}_3) = 0.3, P(\text{B}_4) = 0.6, P(\text{B}_5) = 0.5.

So, the probability that the cell will be killed is given by;

P(A)= 1 - [(1 - P(\text{B}_1)) \times (1 - P(\text{B}_2)) \times (1 - P(\text{B}_3)) \times (1 - P(\text{B}_4)) \times (1 - P(\text{B}_5))]

P(A) = 1 - [(1 - 0.2) \times (1 - 0.4) \times (1 - 0.3) \times (1 - 0.6) \times (1 - 0.5)]

P(A) = 1 - (0.8 \times 0.6 \times 0.7 \times 0.4 \times 0.5)

P(A) = 1 - 0.0672 = 0.9328

Hence, the probability that the cell will be killed is 0.9328.

4 0
4 years ago
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