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Sauron [17]
3 years ago
15

5

Mathematics
1 answer:
Daniel [21]3 years ago
7 0

Answer:

Option A

..............................................................

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What is 12÷2/5? Ty to everyone helping
Dvinal [7]

Answer:

<u>1.2</u> is the answer.

Step-by-step explanation:

12/2=6

6/5=1.3

Have a nice day!

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2 years ago
Give a recursive formula for this sequence: 1,4,7,10,13,16....<br> PLEASE HELP ME! Tysm! (:
Valentin [98]

Answer:

a_{n+1} = a_{n} + 3 ; a₁ = 1

Step-by-step explanation:

Given

1, 4, 7, 10, 13, 16, ......

Note the difference in consecutive terms is constant, that is

4 - 1 = 7 - 4 = 10 - 7 = 13 - 10 = 16 - 13 = 3

Thus to obtain the next term in the sequence ( recursive formula )

Add 3 to the previous term

a_{n+1} = a_{n} + 3 ( with a₁ = 1 )

7 0
3 years ago
Which property is shown by 6+9=9+6?
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8 0
3 years ago
A new test to detect TB has been designed. It is estimated that 88% of people taking this test have the disease. The test detect
Elodia [21]

Answer:

Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

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

<em>X</em> = the test is positive.

The information provided is:

P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99

Compute the probability that a person does not have the disease as follows:

P(D^{c})=1-P(D)=1-0.88=0.12

The probability of a person not having the disease is 0.12.

Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264

Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188

Compute the probability that a randomly selected person is tested negative  as follows:

P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})

           =0.0264+0.1188\\=0.1452

Thus, the probability of the test indicating that the person does not have the disease is 0.1452.

4 0
3 years ago
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