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Rudiy27
3 years ago
11

Select all of the given points in the coordinate plane that lie on the graph of the linear equation 4x-y=3

Mathematics
1 answer:
galben [10]3 years ago
4 0

Answer:

(x, y)=(3/4, 3)

Step-by-step explanation:


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Answer:

it would be on the 6th tick

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Does anyone know how to do this?
Strike441 [17]

Answer:

196 m^{2}

Step-by-step explanation:

A square is a  <u>shape that which all sides are equal</u>

<em>14 is one side of the square and to find the area of a square/rectangle, you do  </em><u><em>length*width </em></u>

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2 years ago
PLEASE I NEED HELP! I want to understand this question.
jasenka [17]

Answer:

TP = 5. greater than

Step-by-step explanation:

1. Out of the 4 options, HH, HT, TH, TT, there is one option we want (HT) which is one our of the four options. that means that the theoretical probability is 1/4 = 25%. Since there were 20 flips, the theoretical probability is 25% of 20 which is 5.

2. For experimental probability, its what actually happenned. Out of the 20 flips, 6 were HT so comapared to the Theoretical probability, the experimental probability was higher.

7 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.

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3 years ago
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Ugo [173]

Answer:

12.25 i think sorry if im wrong

Step-by-step explanation:

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3 years ago
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