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2 years ago

Please help me find the missing measurement!

Mathematics

2 answers:

STALIN [3.7K]2 years ago

4 0

Answer: 11 in

a = lw

99 = 9x

11 = x

pashok25 [27]2 years ago

4 0

Answer:

81 ITS 81

Step-by-step explanation:

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7 0

3 years ago

Read 2 more answers

1/x-2 graphed <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-2%7D" id="TexFormula1" title="\frac{1}{x-2}" alt="\fra

bearhunter [10]

Answer:

Step-by-step explanation:

The given equation is expressed as

y = 1/(x - 2)

We will assume values for x and determine the corresponding values for y. Therefore,

When x = 0, y = 1/(0 - 2) = - 1/2

When x = 1, y = 1/(1 - 2) = - 1

When x = 2, y = 1/(2 - 2) = 0

When x = 3, y = 1/(3 - 2) = 1

When x = 4, y = 1/(4 - 2) = 1/2

To plot the graph, we would choose a suitable scale for the x and y axis of the graph.

Let 2 cm represent 1 unit on the vertical axis.

Let 2 cm represent 0.5 unit on the x axis.

The graph is shown in the attached photo

4 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:

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

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:

$P(X > x) = 0.08$