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2 years ago

4x+7-6x = 10 + 3x + 12

Mathematics

2 answers:

Nuetrik [128]2 years ago

7 0

Step-by-step explanation:

???

solve it for x ?

if that's the case :

4x + 7 - 6x = 10 + 3x + 12

-2x + 7 = 22 + 3x

-15 = 5x

x = -3

ra1l [238]2 years ago

6 0

Answer:

Step-by-step explanation:

4x+7-6x = 10 + 3x + 12

4x + 7 - 6x - 10 - 3x - 12 = 0

-5x - 15

x = -3

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

check

4 * (-3) + 7 - 6 * (-3) - 10 - 3 * (-3) - 12 = 0

the answer is good

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Help me pls math genius

Alecsey [184]

Answer:

x = -3

y = -1

Step-by-step explanation:

in the given question :-

=》x - 4y = 1

=》x = 4y + 1

now replacing the value of x as ( 4y + 1 ) in equation (2)

=》3x + 2y = -11

=》3 (4y + 1) + 2y = -11

=》12y + 3 + 2y = -11

=》14y = -11 - 3

=》y = -14 ÷ 14

=》y = -1

now, putting the value of y in equation (1)

=》x - 4y = 1

=》x - ( 4 × -1 ) = 1

=》x + 4 = 1

=》x = 1 - 4

=》x = -3

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3 years ago

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aniked [119]

Answer:

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Step-by-step explanation:

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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:

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

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:

$P(X > x) = 0.08$