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Elena-2011 [213]
3 years ago
9

The probability that a given urine sample analyzed in a pathology lab contains protein is 0.73 (event A). The probability that t

he urine sample contains the protein albumin is 0.58 (event B).
Event A: The sample contains protein.Event B: The sample contains the protein albumin.
The probability that a sample contains protein and that the protein is albumin is 0.58. Which statement is true?
A.Events A and B are dependent because P(A and B) P(A) + P(B).
B.Events A and B are dependent because P(A and B) = P(A) + P(B).
C.Events A and B are dependent because P(A and B) = P(A)P(B).
D.Events A and B are dependent because P(A and B)     P(A)      P(B).
Mathematics
1 answer:
Sholpan [36]3 years ago
5 0
Event\ A\ and\ event\ B\ are\  independent\\\\ \ \Leftrightarrow\ \ \ \ P(A\ \cap\ B)=P(A)\ \cdot\ P(B)\\---------------------\\\\P(A)=0.78\ \ \ and\ \ \ P(B)=0.58\ \ \ and\ \ \ P(A\ \cap\ B)=0.58\\\\P(A)\cdot P(B)=0.78\cdot0.58\  \neq \ 0.58\\\\\Rightarrow\ \ \ Event\ A\ and\ event\ B\ are\  dependent\ \ \ \Rightarrow\ \ \ Ans.\ (?)
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1/ 8 pounds of dog food

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(a) If the breadth of rectangle in one-third of its length and perimeter is 80 cm, find the length and breadth of rectangle. ​
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Step-by-step explanation:  

Given that :  

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Work out 12+8÷(9-5) 0.018÷0.06 Express as single fraction 5/7÷2/5
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Step-by-step explanation:

I don't know if the first set of numbers is all in one set, but I'll do my best to give you an answer.

Really all you need to do is use PEMDAS for the first question.

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1 2 + 8 \div (9 - 5) \\ 12 + 8 \div 4 \\ 12 + 2 \\ 14

Then to simplify that fraction next to it, notice that 0.018 is 3x 0.06.

that's a 3:1 ratio, so it ends up simplifying to this:

\frac{3}{1}

Lastly, to solve the division of that fraction. If you divide by a fraction, you multiply whatever it's dividing by its inverse.

So...

\frac{5}{7}  \div  \frac{2}{5}  \\  \frac{5}{7}  \times  \frac{5}{2}  \\  \frac{25}{14}

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