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3 years ago

PLEASE I HAVE 10 min TO FINISH THIS I NEED HELP

Mathematics

2 answers:

Allisa [31]3 years ago

8 0

An acute angle is 90 degrees

Brut [27]3 years ago

5 0

Answer:

Is an acute angle. Acute angle should be 90 degrees

Step-by-step explanation:

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And this is the last one i need done

quester [9]

Answer:

x=-2.5,9

Step-by-step explanation:

Set them equal to each other to say 2x^2-9x=4x+45. Get them all on one side to get 2x^2-13x-45. Factor that out to get (x+2.5)(x-9). Set that equal to 0 to get x=-2.5,9

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2 years ago

What is 0.44×(−534) ??

DanielleElmas [232]

Answer:

−234.96

Step-by-step explanation:

Hope its the right answer that you're looking for :)

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2 years ago

P = A + B + C Find P if A = 23 B = 17 C = 29 P=

madam [21]

Answer:

P = 69!

Step-by-step explanation:

23+17+29 = 69 :)

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2 years ago

Read 2 more answers

X+3=15 what’s x because I keep on getting 18 and it says I’m wrong on Pearson

icang [17]

Answer:

x = 12

Step-by-step explanation:

Subtract 3 from both sides

p.s. it would be 18 if it was x-3 = 15

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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

2 years ago