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3 years ago

a water faucet leaked 2 1/6 liters of water every 1/3. hour it leaked at a rate of how many liter per hour?

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

7 0

Answer: The required rate at which the water faucet leaked is 6.5 liters per hour.

Step-by-step explanation: Given that a water faucet leaked $2\frac{1}{6}$ liters of water every $\frac{1}{3}$ hour.

We are to find the rate in liter per hour at which the water faucet leaked.

We will be using the UNITARY method to solve the problem.

We have

Number of liters of water faucet leaked in $\frac{1}{3}$ hour. is

$2\dfrac{1}{6}=\dfrac{13}{6}~\textup{liters}.$

Therefore, the number of liters of water faucet leaked in 1 hour is

$\dfrac{\frac{13}{6}}{\frac{1}{3}}=\dfrac{13}{6}\times3=\dfrac{13}{2}=6.5~\textup{liters}.$

Thus, the required rate at which the water faucet leaked is 6.5 liters per hour.

ladessa [460]3 years ago

4 0

2 1/18. Hope this will help you

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ivolga24 [154]

Answer:

1) We have the system:

5*x - 3*y = 15

4*x + 3*y = 6

To solve this, we first need to isolate one of the variables in one of the equations, let's isolate x in the first equation:

x = 15/5 + (3/5)*y = 3 + (3/5)*y

Now we can replace this in the other equation to get:

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and solve this for y.

12 + (12/5)*y + 3*y = 6

(12/5 + 3)*y = 6 - 12 = -6

(12/5 + 15/5)*y = -6

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Now we can replace this in the equation:

x = 3 + (3/5)*y

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x = 3 + (3/5)*1.11 = 3.67

Then the solution of this system is the point (3.67, 1.11)

2) Now we have the system:

2*x + 5*y = 26

4*x + 3*y = 24

The solution method is the same as before:

x = 26/2 - (5/2)*y = 13 - (5/2)*y

Now we replace this in the other equation:

4*( 13 - (5/2)*y) + 3*y = 24

52 - 10*y + 3*y = 24

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x = 13 - (5/2)*4 = 13 - 10 = 3

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3*x + 3*y = 39

2*x - 3*y = -2

first we isolate x in the first equation:

x = 39/3 - 3*y/3 = 13 - y

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26 - 5*y = -2

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x = 13 - 5.6 = 7.4

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It was so easy xD

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Jobisdone [24]

Answer:

$\frac{3}{5}$ .

Step-by-step explanation:

We have been given two sets as A: {71,73,79,83,87} B:{57,59,61,67}. We are asked to find the probability that both numbers are prime, if one number is selected at random from set A, and one number is selected at random from set B.

We can see that in set A, there is only one non-prime number that is 87 as it is divisible by 3.

So there are 4 prime number in set A and total numbers are 5.

$P(\text{Prime number from A})=\frac{4}{5}$

We can see that in set B, there is only one non-prime number that is 57 as it is divisible by 3.

So there are 3 prime number in set B and total numbers are 4.

$P(\text{Prime number from B})=\frac{3}{4}$

Now, we will multiply both probabilities to find the probability that both numbers are prime. We are multiplying probabilities because both events are independent.

$P(\text{Prime number from A and B})=\frac{4}{5}\times \frac{3}{4}$

$P(\text{Prime number from A and B})=\frac{1}{5}\times \frac{3}{1}$

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Therefore, the probability that both numbers are prime would be $\frac{3}{5}$ .

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