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Agata [3.3K]
3 years ago
13

The Hubble Space Telescope is an optical imaging telescope with extremely good angular resolution. Someone discovers an object t

hat Hubble cannot see at all. How might they have found it?
Engineering
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

<h2><em><u>Through</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>use</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>Hubble</u></em><em><u> </u></em><em><u>Space</u></em><em><u> </u></em><em><u>Telescope</u></em><em><u> </u></em><em><u>that's</u></em><em><u> </u></em><em><u>why</u></em><em><u> </u></em><em><u>they</u></em><em><u> </u></em><em><u>might</u></em><em><u> </u></em><em><u>find</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>or</u></em><em><u> </u></em><em><u>enlarging</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>resolution</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>telescope</u></em><em><u>.</u></em></h2>

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Eva8 [605]

Answer:

-effective technical skills.

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Explanation:

is this what ur looking for? if so there ya go lol

7 0
2 years ago
Read 2 more answers
A continuous random variable, X, whose probability density function is given by f(x) = ( λe−λx , if x ≥ 0 0, otherwise is said t
Ganezh [65]

Answer:

a) F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

b) P(10 < X

Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

Part a

Let X the random variable of interest. We know on this case that X\sim Exp(\lambda)

And we know the probability denisty function for x given by:

f(x) = \lambda e^{-\lambda x} , x\geq 0

In order to find the cdf we need to do the following integral:

F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

Part b

Assuming that X \sim Exp(\lambda =0.1), then the density function is given by:

f(x) = 0.1 e^{-0.1 x} dx , x\geq 0

And for this case we want this probability:

P(10 < X

And evaluating the integral we got:

P(10 < X

4 0
3 years ago
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Explanation:

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2 years ago
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Answer:

I think true

Explanation:

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