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faltersainse [42]
4 years ago
5

Below two situations are described in words. In each case, decide

Mathematics
1 answer:
irina [24]4 years ago
4 0

(a)

If by "age in years" you mean that you only consider whole years, then this is not a one-to-one function, because you will have the same output for 365 inputs:

  • If you're 0 to 364 days old, your age in years is 0
  • If you're 365 to 729 days old, your age in years is 1
  • ....

So, you can see that it is not true that every different input yields a different output.

(b)

This is a one-to-one function: since one mile is 1.609 kilometers, you can convert miles per hour into kilometers per hour by multiplying by 1.609:

x mph = 1.609x km/h

The inverse function would be

x km/h = x/1.609 mph.

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Answer:

x=-1

Step-by-step explanation:

-2x - 5 = -3

add 5 to both sides

-2x = 2

divide both sides by -2

x = -1

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4 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:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = 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 <em>A</em>, <em>B</em> and <em>C</em> 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.

7 0
3 years ago
Complete the square. Fill in the number that makes the polynomial a perfect-square
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Answer:

h² - 10h + <u>25</u>

Step-by-step explanation:

h² - 10h + _ → (h - 5)² = h² - 10h + 25

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3 years ago
Which represents the dimensions of the rectangular park?
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