A) Explain...
1 answer:
<span>a) The cell corresponding to a positive diagnosis when the
person has antibodies present as well as the cell for a negative diagnosis when
the person has no antibodies present are the cells representing a CORRECT
diagnosis.
</span>
The cell corresponding to a negative diagnosis when you
actually have antibodies present is called a FALSE NEGATIVE and is considered
as a TYPE II error.
The cell corresponding to a positive diagnosis when you
actually don't have antibodies present is called a FALSE POSITIVE and is considered
as a TYPE I error.
For the tree diagram for the questions below, refer to the
picture attached.
b) To know the probability that a person without HIV will be diagnosed positive (false positive), we just trace the tree diagram from population to "antibodies not present" to "positive". The tree diagram will give us a value of 0.00588.
ANSWER: The probability of a false positive is 0.00588 or 0.588%.
c) We do the same thing as the previous subproblem to determine the probability that a person with HIV will be diagnosed as negative. We trace the tree diagram from population to "antibodies present" to "negative". The tree diagram will give us a value of 0.003.
ANSWER: The probability of a false negative is 0.003 or 0.3%.
d) Same thing for this subproblem. We trace the tree diagram from population to "antibodies present" to "positive" to know that the value is 0.01997.
ANSWER: The probability that a person with HIV will be diagnosed as positive is 0.01997 or 1.997%.
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Answer:
19.2
Step-by-step explanation:
Add 12 squared to 15 squared(144+225)
Square root the answer()
Round to the nearest tenth(19.209)
Answer:
The gain is of $17.50
Step-by-step explanation:
When you buy the bond the exchange rate is $1.50 = £1.
You pay £45 = =$67.50 for the British bond.
No, you sell the bond for £50 and the exchange rate is $1.70 = £1.
So, you earned dollars
Your gain is = $17.50
<h3>Answer: Venn Diagram is below</h3>
The rectangle represents the universal set (marked as "set U"). Everything is contained in this rectangle.
Circle A represents set A = {a, b, e, f, h, j}
Circle B represents set B = {b, c, d, f, j}
The common elements between the two sets are {b, f, j}, so we'll write these three values in the overlapping portion of the two circles.
The items {a,e,h} will go in circle A, but not in the overlapping portion since they are found only in set A.
Similarly we'll have {c,d} go in circle B but not in the overlapping portion since these items are only found in set B.
The remaining terms {g, i, k} go outside both circles, but inside the rectangle, because neither of these elements are in set A or set B.