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Anvisha [2.4K]
3 years ago
7

the values in the table represent a linear function. what is the common difference of the arithmetic sequence x: 1 2 3 4 5 y: 11

25 39 53 67
Mathematics
2 answers:
Andreyy893 years ago
5 0

Answer:

The common difference of the arithmetic sequence is 14.

Step-by-step explanation:

The arithmetic sequence is defined as

x:   1     2    3    4    5

y:  11   25  39  53   67

The common difference of the arithmetic sequence is

d=\frac{y_2-y_1}{x_2-x_1}

Consider any two points, i.e.,(1,11) and (2,25).

d=\frac{25-11}{2-1}

d=\frac{14}{1}

d=14

Therefore the common difference of the arithmetic sequence is 14.

GenaCL600 [577]3 years ago
3 0

Answer:

common difference: 14

Step-by-step explanation: You have to find out what number it takes to add to the number to get the following number. ex 11+14=25      25+14=39 etc


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Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let D^c be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that  one person in 10,000 people has a rare genetic disease.

So,P(D)=\frac{1}{10000}

Only 0.4% of the people who don't have it test positive.

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P(D^c)=1-\frac{1}{10000}

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P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}

P(D|A)=\frac{2470}{2471}=0.9995

P(D|A)=0.9995

A)The probability that someone who tests positive has the disease is 0.9995

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P(D^c|A^c)=probability that someone does not have disease given that he tests negative

P(A^c|D^c)=probability that a person tests negative given that he does not have disease =1-0.004

=0.996

P(A^c|D)=probability that a person tests negative given that he has a disease =1-0.988=0.012

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P(D^c|A^c)=0.99999

B)The probability that someone who tests negative does not have the disease is 0.99999

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Answer:

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