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professor190 [17]
2 years ago
8

Pa answer po thank you ​

Mathematics
1 answer:
Arisa [49]2 years ago
4 0

Answer:

hope it helps you

Step-by-step explanation:

To make such a frequency distribution table, first, write the class intervals in one column. Next, tally the numbers in each category based on the number of times it appears. Finally, write the frequency in the final column. A frequency distribution table drawn above is called a grouped frequency distribution table.

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serg [7]

Answer:

35 by 40

Step-by-step explanation:

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3 years ago
Which statement describes a disadvantage of Marnie paying a bill by telephone with a credit card or a debit card? Marnie will no
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Answer:

The Answer is B Marnie will have to share her card account information.

Plus I took this test so

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Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe
nirvana33 [79]

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.

P(A|D^c) = probability that a person is tested positive given he don't have a disease = 0.004

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

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

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

(B)

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

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

P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}

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

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

8 0
3 years ago
Which pair of expressions represents inverse functions
Natali5045456 [20]

Answer: C. \dfrac{x+3}{4x-2} and \dfrac{2x+3}{4x-1}


Step-by-step explanation: if a function f(x) has g(x) as its inverse then it satisfies fog(x)=x and gof(x)=x

C. f(x)=\dfrac{x+3}{4x-2} and g(x)=\dfrac{2x+3}{4x-1}

   fog(x)=f(\dfrac{2x+3}{4x-1})

              =\dfrac{\dfrac{2x+3}{4x-1}+3 }{\dfrac{4(2x+3)}{4x-1}-2 }

               =x

   gof(x)=g(\dfrac{x+3}{4x-2})

              =\dfrac{\dfrac{2(x+3)}{4x-2} +3}{\dfrac{4(x+3)}{4x-2}-1 }

               =x

hence C. is the pair of inverse functions


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