Answer:
the answer is b
Explanation:
got it right on edg.
nformation for how many patients took each drug or combination of drugs is summarized below in the two tables. Use these to answer questions a) -d)
Table 1. Summary of performance of drug A: UTI rates among those taking and not taking drug ADid not take Drug ADid take Drug A
total UTI
759
887
164 No UTI
441 312 753Total 1200 1200
2400Table 2. Summary of performance of drug B and C: recovery status after 1 week of taking medications.
Did not take Drug A
Did take Drug A
Drug B
Drug C
Drug B
Drug C
Recovered
191
209
221
244
Not Recovered
189
170
223
199
Total
380
379
444
443
a. Use the above Table 1 to determine if Drug A was useful in preventing UTIs. In other words, is the proportion of those having taking Drug A but still getting a UTI equal to average rate of UTI for this population (living in an assisted living home) of 74%. Use hypothesis testing to test our hypothesis and use the confidence interval approach with a significance level of α=0.01.
b. Using Table 2, let’s examine the rate of UTI recovery among Drug C (conventional antibiotics). The manufacturer of Drug C claims it has a success rate (recovery within a week) of 55%. Use our data to see if this success rate is true: test if our recovery rate of those taking Drug C, regardless of whether the person took Drug A or not, is the same or different than 55%. Use hypothesis testing and the p-value approach with an α=0.05.
c. Similarly, let’s examine Drug B’s performance. Repeat our hypothesis among Drug B: test if our recovery rate of those taking drug B is different than 55% (regardless of whether the patient took Drug A or not). Use hypothesis testing and p-value approach with an α=0.1.
Answer:
(View Below)
Explanation:
Let's tackle each part of the question step by step:
a. **Testing the Effectiveness of Drug A:**
We want to test if the proportion of patients who took Drug A and still got a UTI is equal to the average rate of UTIs for this population (74%). We can use a hypothesis test for proportions. Here are the hypotheses:
- **Null Hypothesis (H0):** The proportion of patients who took Drug A and got a UTI is equal to 74%.
- **Alternative Hypothesis (Ha):** The proportion of patients who took Drug A and got a UTI is not equal to 74%.
We'll perform a two-tailed test at a significance level of α = 0.01.
Using the provided data:
- Proportion of UTIs among those who took Drug A = 887 / 1200 ≈ 0.7392
- Proportion of UTIs among those who did not take Drug A = 759 / 1200 ≈ 0.6325
We can calculate the standard error for the difference in proportions and perform the hypothesis test. I'll calculate the z-score and p-value for you:
Z = (0.7392 - 0.6325) / √[0.6325 * (1 - 0.6325) / 1200] ≈ 2.8413
Now, looking up the z-score in a standard normal distribution table, we find the critical values for a two-tailed test at α = 0.01 to be approximately ±2.576.
Since our calculated z-score (2.8413) is greater than the critical value (2.576), we can reject the null hypothesis.
Therefore, there is evidence to suggest that Drug A is useful in preventing UTIs because the proportion of patients who took Drug A and still got a UTI is significantly different from the average rate of UTIs for this population.
b. **Testing the Recovery Rate of Drug C:**
We want to test if the recovery rate for Drug C is different from the claimed success rate of 55%. We can use a hypothesis test for proportions. Here are the hypotheses:
- **Null Hypothesis (H0):** The recovery rate of those taking Drug C is equal to 55%.
- **Alternative Hypothesis (Ha):** The recovery rate of those taking Drug C is different from 55%.
We'll perform a two-tailed test at a significance level of α = 0.05.
Using the provided data:
- Proportion of recovery among those taking Drug C = (221 + 244) / 443 ≈ 0.9955
We can calculate the standard error for the proportion and perform the hypothesis test. I'll calculate the z-score and p-value for you:
Z = (0.9955 - 0.55) / √[0.55 * (1 - 0.55) / 443] ≈ 18.3841
The critical values for a two-tailed test at α = 0.05 are approximately ±1.96.
Since our calculated z-score (18.3841) is much greater than the critical value (1.96), we can reject the null hypothesis.
Therefore, there is strong evidence to suggest that the recovery rate for Drug C is different from the claimed success rate of 55%.
c. **Testing the Recovery Rate of Drug B:**
We want to test if the recovery rate for Drug B is different from the claimed success rate of 55%. We'll perform a two-tailed test at a significance level of α = 0.1.
Using the provided data:
- Proportion of recovery among those taking Drug B = (221 + 244) / 444 ≈ 0.9919
We can calculate the standard error for the proportion and perform the hypothesis test. I'll calculate the z-score and p-value for you:
Z = (0.9919 - 0.55) / √[0.55 * (1 - 0.55) / 444] ≈ 17.7503
The critical values for a two-tailed test at α = 0.1 are approximately ±1.645.
Since our calculated z-score (17.7503) is much greater than the critical value (1.645), we can reject the null hypothesis.
Therefore, there is strong evidence to suggest that the recovery rate for Drug B is different from the claimed success rate of 55%.
B. hyperactivity
C. loss of inhibition
D. increased coordination
Answer: D: increased coordination
Explanation: inhalants are vapors or solvents which are inhaled to produce anaesthetic effect.
Effects of inhalants include:
1. Drowsiness
2. Disinhibition
3. Lightheadedness
4. Agitation.
Liquid anesthetic may be administered by mixing the vapors with oxygen or nitrous oxide, then a patient inhale the mixture. This can be administered through a tube or a mask
Static stretches are part of an effective cool down because they can help
increase range of motion
Answer:
Vitamins work with enzymes to initiate chemical reaction within cells
Explanation:
Sometimes, an enzyme requires for its function the presence of non-protein substances that collaborate in catalysis: cofactors. The cofactors can be inorganic ions such as Fe ++, Mg ++, Mn ++, Zn ++. Almost a third of known enzymes require cofactors. When the cofactor is an organic molecule it is called coenzyme. Many of these coenzymes are synthesized from vitamins. In the lower figure we can see a hemoglobin molecule (protein that carries oxygen) and its coenzyme (the heme group). When cofactors and coenzymes are covalently bound to the enzyme they are called prosthetic groups. The catalytically active form of the enzyme, that is, the enzyme linked to its prosthetic group, is called holoenzyme. The protein part of a holoenzyme (inactive) is called apoenzyme .
B) Capillaries
C) Varicose veins
D) Atrioventricular node
Arteries are the blood vessels which use elasticity to push blood further along the bloodstream. Correct answer: A They carry blood away from the heart to the rest of the body.
The middle layer of the arteries has smooth muscle and elastic fibers. The elastic fibers in the arteries expand to accommodate the blood and this way they help in maintaining a high-pressure gradient.
Answer:
They feel good about themselves when they can use their new skills. Their self-esteem grows when parents pay attention, let a child try, give smiles, and show they're proud. As kids grow, self-esteem can grow too. Any time kids try things, do things, and learn things can be a chance for self-esteem to grow.