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3 years ago

Find (f x g)(2)? Thanks

Mathematics

1 answer:

omeli [17]3 years ago

7 0

Let me break this up.

Note: (f o g)(2) = f(g(2)).

Note: g(2) = (2, 2) means when x is 2 y is 2.

In other words, g(2) is 2.

We know that f(g(2)) = f(2).

You can see that f(2) = -3. This is given in the point (2, -3).

So, f(g(2)) = -3.

Done.

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Which order from least to greatest of 2.8, 2.8x10², 2.8x10⁻3, 2.8x10⁴, 2.8x10⁻5

Dominik [7]

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Maurinko [17]

Answer:

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Step-by-step explanation:

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3 years ago

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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3 years ago

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loris [4]

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3 years ago