Answer:
A.) Think about good feelings he has had
Explanation:
A useful strategy Tomas can use to feel better is to think about good feelings he has had. The other answer choices that were not listed are the following below:
Hanging out with a few friends at a time won't help his case, because he is unwanted among his friends, so we can conclude that pretty much all of his friends won't accept him. Eating nutrient dense high-energy foods will not make a difference because this has to do with his emotions, not his diet. Spending more time alone whenever possible will only make him feel worse and make him prone to doing bad things to relieve his unhappiness. Thinking about the good feelings he has had, on the other hand, will make him feel better. Sometimes when people feel down, they reminisce on the ole days, and that sometimes helps to raise their spirits. So this would be the best strategy Tomas can use to feel better.
Hope this helps y'all :D
Answer:
Think about good feelings he has had
head
ankle
shoulder User: After sperm are produced, they move into a sperm storage area called the _____.
scrotum
seminiferous tubules
epididymis
vas deferens
altered mood and sensory perceptions
reduced heart rate and loss of inhibitions
Answer:
The answer is altered mood an sensory perceptions
Explanation:
Plato
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%.
Answer: maybe boats those fancy boats
Explanation: