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pogonyaev
2 years ago
8

if a certain number x is doubled, the result is less than or equal to 12. find the range of values of the number​

Mathematics
1 answer:
ki77a [65]2 years ago
4 0

Answer:

2x≤12

2x/2 ≤12/2

x≤6

therefore x=6

x+x

6+6=12

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eimsori [14]
  • i = \sqrt{-1}
  • Product rule of radicals: √ab = √a x √b

Firstly, factor out i: i\sqrt{50}

Next, apply the product rule of radicals here as such:

i\sqrt{50}=i\sqrt{10\times 5}=i\sqrt{5\times 2\times 5}=5i\sqrt{2}

<u>Your answer is 5i√2, or the second option.</u>

8 0
3 years ago
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3(8) + 4

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7 0
2 years ago
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g A population is infected with a certain infectious disease. It is known that 95% of the population has not contracted the dise
trasher [3.6K]

Answer:

There is approximately 17% chance of a person not having a disease if he or she has tested positive.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = a person has contracted the disease.

+ = a person tests positive

- = a person tests negative

The information provided is:

P(D^{c})=0.95\\P(+|D) = 0.98\\P(+|D^{c})=0.01

Compute the missing probabilities as follows:

P(D) = 1- P(D^{c})=1-0.95=0.05\\\\P(-|D)=1-P(+|D)=1-0.98=0.02\\\\P(-|D^{c})=1-P(+|D^{c})=1-0.01=0.99

The Bayes' theorem states that the conditional probability of an event, say <em>A</em> provided that another event <em>B</em> has already occurred is:

P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}

Compute the probability that a random selected person does not have the infection if he or she has tested positive as follows:

P(D^{c}|+)=\frac{P(+|D^{c})P(D^{c})}{P(+|D^{c})P(D^{c})+P(+|D)P(D)}

              =\frac{(0.01\times 0.95)}{(0.01\times 0.95)+(0.98\times 0.05)}\\\\=\frac{0.0095}{0.0095+0.0475}\\\\=0.1666667\\\\\approx 0.1667

So, there is approximately 17% chance of a person not having a disease if he or she has tested positive.

As the false negative rate of the test is 1%, this probability is not unusual considering the huge number of test done.

7 0
3 years ago
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MArishka [77]
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