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ASHA 777 [7]
3 years ago
13

Solve for each variable in the equations that follow 3x+y=25

Mathematics
2 answers:
Karo-lina-s [1.5K]3 years ago
8 0
If you subtract 3x from both sides you get y=25-3x. If you subtract y from both sides you then you divide both sides by three you get x=25-y/3. You can't get a specific answer
torisob [31]3 years ago
3 0

      <u> 3x + y = 25</u>

To solve for  'y' ...

Subtract  3x  from each side:  <em>  y = 25 - 3x</em>
===========================================

<u>3x + y = 25</u>

To solve for  'x' ...

Subtract 'y'  from each side:    3x = 25 - y

Divide each side by  3 :            <em>x = (25 - y) / 3  </em>


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A catering company loses $110 as a result of a delay.
faust18 [17]

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Step-by-step explanation: It would be positive because the question is asking how much money they lost. The lost a positive amount of money.

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1 year 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:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

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
2 years ago
An arc of length 15ft subtends a central angle 0 in a circle of radios 9ft. Find measure of 0 in degrees
Kisachek [45]

Answer:

95.49°

Step-by-step explanation:

The arc length formula is s = rФ, where r is the radius and Ф is the central angle in radians.  Here, r = 9 ft and s = 15 ft.  Thus, the central angle Ф is

Ф = (15 ft) / (9 ft) = 5/3 radians, or

5 rad       180°

------- * ---------- = 95.49°

      3       π rad

6 0
3 years ago
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