The equation for the first option is Y=10x + 200
The equation for the second option is Y=30x + 100

X = one month
To find when they would be the same about you have to set the equations equal to each other

10x + 200 = 30x + 100
-10x -100 -10x -100

100= 20x

5 = x

After five months she would save the same amount. To find how much is saved you have to plug in 5 for x in one of the equations. You can always double-check 5 by plugging it in both equations and making sure you get the same answer

Y= 10(5) + 200 = 250

Y= 30(5) + 100 = 250

She would have saved $250 after five months by using either method

6 0

3 years ago

Solve for

faust18 [17]

Given equation : n(17+x)=34x−r.

We need to solve it for x.

Distributing n over (17+x) on left side, we get

17n + nx = 34x - r.

Adding r on both sides, we get

17n+r + nx = 34x - r+r.

17n + r + nx = 34x.

Subtracting nx from both sides, we get

17n + r + nx-nx = 34x-nx

17n + r = 34x -nx.

Factoring out gcf x on right side, we get

17x + r = x(34-n).

Dividing both sides by (34-n), we get

$\frac{(17x + r)}{(34-n)} = \frac{x(34-n)}{(34-n)}$

$\frac{(17x + r)}{(34-n)} = x$

<h3>Therefore, final answer is $x=\frac{(17x + r)}{(34-n)}.$ </h3>

7 0

3 years ago

Read 2 more answers

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:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

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

$-0.01x = \ln{0.08}$

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

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

3 years ago