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yKpoI14uk [10]
2 years ago
6

Suppose two drugs are routinely used for the treatment of thyroid dysfunction. Drug X is known to cure the disorder 75% of the t

ime and costs $88.
Drug Y is known to cure the disorder 60% of the time and costs $75. The two drugs work independent of each other. The two treatment plans are as
follows:
Plan A: Treatment with Drug X-if not effective treatment with Drug Y
Plan B: Treatment with Drug Yếif not effective, treatment with Drug X
28
Select the correct answer.
From the perspective of an insurance company paying for the treatment, which is the best decision?
ОА.
Based on the overall probability of a cure, plan A should be selected over plan B.
ОВ.
Based on the overall probability of a cure, plan B should be selected over plan A.
OC. Based on the overall cost of treatment plan A should be selected over plan B.
OD
Based on the overall cost of treatment plan B should be selected over plan A.
SAT
1 answer:
Shtirlitz [24]2 years ago
3 0

Answer:

The correct answer is A)

Explanation:

This is because the chances that the company will prevent additional expenses on the treatment of the ailment is 75% which is higher than that of the other drug whose effectiveness is at best 40% less than the expected results and 15% reduced than drug A.

In other words, if the drug b is used first, there is an increased chance that the company will spend more money to treat the ailment.

Cheers

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At 900 ∘c,kc=0. 0108 for the reaction caco3(s)←→cao(s)+co2(g) a mixture of caco3, cao, and co2 is placed in a 10. 0-l vessel at
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The concentration of CaCO3 increases in all cases because Q> K.

<h3>What is equilibrium?</h3>

A chemical reaction is said to have attained equilibrium when the rate of forward reaction is equal to the rate of reverse reaction. We are told that the Kc of the reaction is 0. 0108.

In the first case:

CaCO3 - 15.0 g/100g/mol/10 L = 0.015 M

CaO - 15.0 g/56 g/mol/10 L = 0.027 M

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In the second case;

CaCO3 -  2.5 g/100g/mol/10 L =0.0025 M

CaO - 25.0 g/56 g/mol/10 L = 0.045 M

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Q = [0.045 ] [ 0.013]/[0.0025 ]

Q = 0.23

Q > K the concentration of CaCO3 increases

In the third case;

CaCO3 - 30.5 g/100g/mol/10 L =0.031 M

CaO - 25.5 g/56 g/mol/10 L =0.046 M

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Q = [0.046] [ 0.015]/[0.031]

Q = 0.022

Q> K hence the concentration of CaCO3 increases

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