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3 years ago

Find the distance between (1, 3) and (-2, -6) and put it in simplest radical form

Mathematics

2 answers:

julia-pushkina [17]3 years ago

5 0

Answer:

$D=\sqrt{90}\ or\ D=9.49$

Step-by-step explanation:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\D=\sqrt{(-2-1)^2+(-6-3)^2}\\\\D=\sqrt{(-2-1)^2+(-6-3)^2} \\\\D=\sqrt{(-3)^2+(-9)^2} \\\\D=\sqrt{9+81} \\\\D=\sqrt{90}\\\\D=9.49$

Hope this helps!

GarryVolchara [31]3 years ago

3 0

Answer:

$\tt 9.48$

Step-by-step explanation:

(1,3) and (-2,-6)

$\tt Distance= \sqrt{(-6-3)^2+(-2-1)^2}$

$\tt =\sqrt{9^2+3^2}$

$\tt =\sqrt{81+9} =\sqrt{90}$

$\tt Distance=\sqrt{90} =9.48$

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A 90% confidence interval is (35 45). What is the margin of error?

MA_775_DIABLO [31]

Answer:

option A 5

I hope it's correct

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7 0

3 years ago

The time T required to repair a machine is an exponentially distributed random variable with mean 10 hours.

Firdavs [7]

It can be expected about 36.79% of chance that repair time exceeds

The probability that a repair time exceeds 15 hours is 0.3679

What is the exponential distribution?

It explains about the time between events or the distance between two random events is termed the exponential distribution. Here, the occurrence of the events is continuous and also independent. Moreover, the average rate is constant.

The cumulative distribution function of T is obtained below:

From the information given, let the random variable T be the required time to repair a machine follows exponential distribution with parameter λ
with mean. 1/2 hours

That is, E(x) = 1/2 hours.
The parameter of the random variable T is,
E(x) = 1/λ
λ = 1/E(x)
= 1/(1/2)
= 2

The probability density function of T is,
$f(t) = \left \{ {{2e^{-2t} \ \ \ t > 0} \ \atop {0} \ \ \ elsewhere} \ \right.$
The cumulative density function of T is,
FT(t) = P(T <= t)

$= 1 - e^{- \lambda t}$

$= 1 - e^{- 2t}$
The CDF of T is,
P(T <= t) $= 1 - e^{- 2t}$ 0 <= T <= ∞
= 0 otherwise
to obtain the probability that a repair time exceeds

1/2 hours.
(a) The probability that a repair time exceeds 1/2 hours.
From the given information, the CDF of T is,
P(T <= t) $= 1 - e^{- 2t}$ 0 <= T <= ∞
= 0 otherwise

The required probability is,
P(T <= 1/2) = 1 - P(T <= 1/2)
= 1 - [ $= 1 - e^{- 2(1/2)}$ ]
$= e^{-1}$
= 0.3679

om total probability. It can be expected about 36.79% of chance that repair time exceeds

P(X => x) = 1 - P(X < x)
to obtain the probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours.

(b), The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained below:
From the given information, the CDF of T is,
P(T <= t) $= 1 - e^{- 2t}$ 0 <= T <= ∞
= 0 otherwise
The required probability is,
P = P(T => 12.5∩T>12) / P(T>12)
$= e^{- 25 + 24}$

$= e^{- 1}$
= 0.3679
The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained by dividing the
P = P(T => 12.5∩T>12) / P(T>12)
with
P(T>12).