In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that a student has a cat given that they do not have a dog? ( NO LINKS )
Answers
Given:
Number of students who has a cat and a dog = 5
Number of students who has a cat but do not have a dog = 11
Number of students who has a dog but do not have a cat = 3
Number of students who neither have a cat nor a dog = 2
To find:
The probability that a student has a cat given that they do not have a dog.
Solution:
Let the following events:
A = Student has a cat
B = Do not have a dog
Total number of outcomes is:
[tex]5+3+11+2=21[/tex]
The probability that a student has a cat but do not have a dog is:
[tex]P(A\cap B)=\dfrac{11}{21}[/tex]
The probability that a student do not have a dog is:
[tex]P(B)=\dfrac{11+2}{21}[/tex]
[tex]P(B)=\dfrac{13}{21}[/tex]
The conditional probability is:
[tex]P\left(\dfrac{A}{B}\right)=\dfrac{P(A\cap B)}{P(B)}[/tex]
[tex]P\left(\dfrac{A}{B}\right)=\dfrac{\dfrac{11}{21}}{\dfrac{13}{21}}[/tex]
[tex]P\left(\dfrac{A}{B}\right)=\dfrac{11}{13}[/tex]
Therefore, the probability that a student has a cat given that they do not have a dog is [tex]\dfrac{11}{13}[/tex].
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Answers
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Step-by-step explanation:
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(b) Find the 29th term of the sequence.
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Answers
Answer:
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b. 29th term of the sequence= -35
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Step-by-step explanation:
The first term of the sequence is 77, the 29th term of the sequence is -35 and the sum of the first 40 terms of the sequence is -40.
Given that, [tex]a_{15} =21[/tex] and d=-4.
What is the nth term of the arithmetic sequence?The nth term of the arithmetic sequence is [tex]a_{n} =a+(n-1)d[/tex].
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Answer:
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According to the given question.
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The equivalent expression of the above expression can be given by
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