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VLD [36.1K]
3 years ago
8

A test to detect an infection gives a positive result 98% of the time when an infection is present. It is 97% accurate when an i

nfection is not present.
Mathematics
1 answer:
NNADVOKAT [17]3 years ago
4 0

Answer:

false negatives - 2%

false positives - 3%

Step-by-step explanation:

Here is the complete question :

A test to detect an infection gives a positive result 98% of the time when an infection is present. It is 97% accurate when an infection is not present.

% of results will be false negatives, and % of results will be false positives.

False negative means a negative result which in actuality it is positive

False negative = 100% - percent of positive result

100% - 98% = 2%

False positive result means a positive result which in actuality is negative

False positive = 100% - percent of false positive

100 - 97 = 3%

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Using a linear approximation, estimate f(2.1), given that f(2) = 5 and f'(x) = √3x-1.
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Answer:

f\left( {2.1} \right) \approx 5.22360.

Step-by-step explanation:

The linear approximation is given by the equation

                            {f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).}

Linear approximation is a good way to approximate values of f(x) as long as you stay close to the point x= a, but the farther you get from x=a, the worse your approximation.

We know that,

a=2\\f(2) = 5\\f'(x) = \sqrt{3x-1}

Next, we need to plug in the known values and calculate the value of f(2.1):

{L\left( x \right) = f\left( 2 \right) + f^\prime\left( 2 \right)\left( {x - 2} \right) }=5+\sqrt{3(2)-1}(x-2) =5+\sqrt{5}(x-2)

Then

f\left( {2.1} \right) \approx 5+\sqrt{5}(2.1-2)\approx5.22360.

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