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Ede4ka [16]
3 years ago
5

Please help :( [math]

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0

Hey!

------------------------------------------------

First Equation:

2x + 3 = -7

2x + 3 - 3 = -7 - 3 (Subtract 7 to both sides)

2x = -10

2x/2 = -10/2 (Divide 2 to both sides)

x = -5

------------------------------------------------

Second Equation:

4.5x - 7 = 20

4.5x + (-7) + 7 = 20 + 7 (Add 7 to both sides)

4.5x = 27

4.5x/4.5 = 27/4.5 (Divide 4.5 to both sides)

x = 6

------------------------------------------------

Third Equation:

-3x + 7 = 28

-3x + 7 - 7 = 28 - 7 (Subtract 7 to both sides)

-3x/-3 = 21/-3 (Divide -3 to both sides)

x = -7

------------------------------------------------

Answers:

2x + 3 = -7 → x = -5

4.5x - 7 = 20 → x = 6

-3x + 7 = 28 → x = -7

------------------------------------------------

Hope This Helped! Good Luck!

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Answer:

10 cm

Step-by-step explanation:

we know that the perimeter of a rectangle= 2(l+w)

given that the perimeter is 46 cm and length is 13 cm, so by putting the values we get,

2(13+w)=46

= 26+2w=46 (opening the brackets)

= 2w= 46-26

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Cheers.

7 0
3 years ago
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

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Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

5 0
3 years ago
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Answer:

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