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2 years ago

Which of the following is the measure of side a? 23.3 144 8 16

Mathematics

1 answer:

jenyasd209 [6]2 years ago

4 0

Step-by-step explanation:

it is a right-angled triangle, so the side lengths need to fulfill the Pythagoras rule :

c² = a² + b²

where c is the Hypotenuse, the baseline opposite of the 90 degree angle. and a and b (often called legs) are the sides "enclosing" the 90 degree angle.

so, we have

20² = a² + 12²

400 = a² + 144

a² = 256

a = 16

FYI - we could have picked the right answer also by eliminating the wrong given answer options.

as this is a right-angled triangle :

a cannot be longer than the baseline (20).

that eliminates already the first 2 options.

and then look at the picture : a cannot be shorter than b (the other side line (12)).

that eliminates the third option.

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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:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

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

$-0.01x = \ln{0.08}$

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

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

3 years ago

Transform x2 + 6x + 8 = 0 into the form (x − p)2 = q?

liraira [26]

So hmmm x²+6x+8=0

alrite.. let's do some grouping now
( x² + 6x + [?]²) + 8 = 0

notice above, we have a missing fellow in order to get a perfect square trinomial... hmm who would that be?

let's take a peek at the middle guy of the trinomial.. 6x.. hmmm let's factor it, 2*3*x, wait a minute! 2 * 3 * x... we already have x² on the left-side, since the middle term is just 2 * the square root of the other two terms, that means that the guy on the right, our missing guy must be "3"

alrite, let's add 3² then, however, bear in mind that, all we're doing is borrowing from our very good friend Mr Zero, 0

so if we add 3², we also have to subtract 3², let's do so

(x² + 6x +3² - 3²) + 8 = 0

(x² + 6x +3²) + 8 - 3² = 0

(x+3)²=3² - 8

(x+3)² = 1

5 0

3 years ago

Read 2 more answers