Drag the values from the problem and fill in the blanks to create and solve a system of equations.

LiRa [457]

They are the same because if you add a zero it doesnt change it

4 0

3 years ago

1) Lisa took a trip to Kuwait. upon leaving she decided to convert all of her dinars back into dollars . how many dollars did sh

Yuliya22 [10]

Lisa took a trip to Kuwait upon leaving she decided to convert all of her dinars back into dollars how many

3 0

3 years ago

Factor 2x^2-128 completely

Alina [70]

2x² - 128

2(x² - 64), but (x² - 64) = x² - 8² (difference of 2 squares),[a²-b²=(a-b)(a+b)]

2(x² - 64) = 2(x² - 8²) = 2(x-8)(x+8)

8 0

4 years ago

Order these numbers from least to greatest.<br> 3.3661, 3.5, 3.36, 3.036

anyanavicka [17]

Answer:

3.036, 3.36, 3.3661 , 3.5

Step-by-step explanation:

Look at the first 2 numbers

3.3661, 3.5, 3.36, 3.036

We can order them as 3.0 , 3.3 , 3.3 . 3.5

So 3.036 is first one

Now we have 2 of the 3.3's

One is 3.3661 and the other one is 3.3600

Because if you add a number to the end it will always be a zero and it wont change the answer

So 3.36 is second and 3.3661 is third one

3.5 is bigger than 3.3

So 3.5 is last

5 0

2 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:

$P(X > x) = 0.08$