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andrey2020 [161]
3 years ago
8

PLEASE HELP ASAP

Mathematics
1 answer:
Oliga [24]3 years ago
7 0

The answer is A.)  this is because look at the hints shown in the equation. X^2 is one of them which forms an parabla which is a curve line

IF YOU LIKE MY ANSWER PLS GIVE ME BRAINLIEST!!

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The answer is A

Step-by-step explanation:

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Need help with b and c
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Step-by-step explanation:

the formula is:

a_n = \frac{n+1}{n+2}

[tex]a_100 = \frac{101}{103}[\tex]

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There are 22 paper clips in a draw 12 is gold an 10 are silver what fraction of the paper clip is gold?
yKpoI14uk [10]

Answer:

6/11

Step-by-step explanation:

there are 12 gold clips out of 22, so 12/22 and divide both sides by two to simplify.

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3 years ago
Explain how you would find the perimeter of this triangle.
Lynna [10]

Answer:

<em>15.4 </em>

Step-by-step explanation:

\frac{u}{4} = tan51.3° ⇒ u = 4tan51.3° ≈ 5

\frac{4}{v} = cos51.3° ⇒ v = \frac{4}{cos51.3} ≈ 6.4

<em>P </em>≈ 4 + 5 + 6.4 = <em>15.4</em>

4 0
2 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
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