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Bess [88]
1 year ago
8

ppositional angle closure in eyes with narrow angles: comparison between the fellow eyes of acute angle-closure glaucoma and nor

motensive case
Mathematics
1 answer:
skad [1K]1 year ago
8 0

Here is the comparision

Purpose:To compare the topologic features of acute primary angle-closure glaucoma eyes before an attack to those of normotensive eyes, assuming that untreated fellow acute primary angle-closure glaucoma eyes are candidates for an acute attack.

Methods:Under dark-room conditions, ultrasound biomicroscopy was used to examine 50 eyes (12 fellow eyes of acute primary angle-closure glaucoma and 38 normotensive cases with a closure-possible narrow angle). Before any surgical or laser intervention, all eyes were examined and found to have normal pupillary response without the use of any topical drugs. Each eye was examined at four predetermined angle locations. The chamber angle configuration parameters were measured and compared between the two groups.

Result:Appositional angle closures were detected in 27 fellow eyes and 48 normotensive eyes with a closure-possible narrow angle. The incidence differed statistically between the two groups (69.2% in fellow eyes and 48% in normotensive eyes). In the fellow eye group, appositional angle closures beginning at the angle's entrance were more frequently detected. The distance between the iris root and the bottom of the angle varied significantly between groups.

Conclusion:Acute primary angle-closure glaucoma fellow eyes have different topologic features than normotensive narrow-angled eyes, as well as a higher incidence of appositional closure, which may predispose these eyes to an impending acute attack.

Learn more about glaucoma here:

brainly.com/question/1318395

#SPJ4

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RoseWind [281]
<h3>Answer:  C) 6</h3>

====================================================

Explanation:

The weird looking E symbol is the greek uppercase letter sigma. It refers to a sum.

It tells us to add up terms in the form (-1)^n*(3n+2) where n is an integer ranging from n = 1 to n = 4.

------------------

If n = 1, then we have

(-1)^n*(3n+2) = (-1)^1*(3*1+2) = -5

Let A = -5 as we'll use it later.

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If n = 2, then

(-1)^n*(3n+2) = (-1)^2*(3*2+2) = 8

Let B = 8 since we'll use this later as well

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If n = 3, then

(-1)^n*(3n+2) = (-1)^3*(3*3+2) = -11

Let C = -11

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If n = 4, then

(-1)^n*(3n+2) = (-1)^4*(3*4+2) = 14

Let D = 14.

--------------------

We'll add up the values of A,B,C,D to get the final answer

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\displaystyle \sum_{n=1}^{4}(-1)^n(3n+2) = 6

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3 years ago
in a game involving a pair of fair dice a player rolls the dice until he gets a sum of 7 let z equal the number of rolls it take
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Answer:

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Step-by-step explanation:

A die has 6 possible outcomes, which sums to 36 for two dice for every value on both dice.

The outcomes for rolling both dice for 36 times gives 6 possible outcomes summing to 7, that is, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1).

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= 6 / 36 = 1/6

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3 years ago
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