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3 years ago

PPPPPPlzzz helpppppppp

Mathematics

1 answer:

avanturin [10]3 years ago

7 0

the only 2 possible answers are 2 and -2 .But the answer is y=2 because x is positive

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Answer these two please!!

Gala2k [10]

Answer:

#1: x/4 - 3/4

#2: x/14 + 9/2

Step-by-step explanation:

because i said so, you should mark me brainliest

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3 years ago

Can someone help me I don't understand

MAVERICK [17]

Answer: C 75

Step-by-step explanation: time to distance is ratio 4:5 = 60:X

In 1min He walks 5/4

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2 years ago

Rudy is 6 feet 6 inches tall, weighs 210 pounds, and is very muscular. if you think that rudy is more likely to be a basketball

madam [21]

The answer is representativeness heuristic. This is used when making decisions about the likelihood of an occasion under doubt. It is unique of a group of heuristics (simple rules leading decision or decision-making) suggested by psychologists Amos Tversky and Daniel Kahneman in the 1970s.

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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)

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