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2 years ago

What is 2/9 ÷ 1/2 ???

Mathematics

2 answers:

blagie [28]2 years ago

6 0

Answer:

i think its 4/9

Step-by-step explanation:

use the method KCF

-keep the first fraction same

-reverse the sign e.g mutiplyings opposite is dividing

- change over the fraction so if 1/2 is original then flip it over and make it 2/1 so basically top heavy fraction

working out should look like this:

4 2

_ × _

9 9

now multiply as normal

Julli [10]2 years ago

5 0

Answer:

2/9 divided by 1/2 is equal to 2/9 times the reciprocal of 1/2, which is equal to 2.

So, 2/9 divided by 1/2 is equal to 2/9 times 2.

2/9 (2) = 4/9

Let me know if this helps!

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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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