Does nausea appear to be a problematic adverse reaction? Since the rate of nausea appears to be relatively it a problematic adverse reaction.
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A: Exposure time (Minutes) Ripening time (days) 10 4.3 15 3.2 20 2.7 25 2.1 30 1.3
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Q: 1. How is the average rate of change found?
A:
Q: How can the standard error of estimate be interpreted?
A: here use basic of standard error of estimate
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A: The average MPG of cars that weigh greater than 2,500 lb. is, Thus, the average MPG of cars that…
Q: how to find the average rate of change?
A: To find the average rate of change of a function, we use the formula
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A: Yes, the rate of change of cost is constant with the rate of change of pencil.
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A: To write about the graph
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A: From the given information,The correlation between weight and height for adults is 0.80
Q: What is the instantaneous rate of change?
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Q: normal line he
A: we have to find the equation of normal line here in the given equation.x3+y2+2x=6 at P=(-1,3)
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- A researcher was interested in how the effects of stress were different in people with our without military training. A equal number of participants were recruited that either had no military experience or had completed military experience. All participants were asked to give a speech without preparation while being judged by a panel of experts. Their heart rates were monitored during the speech and an average heart rate was calcuated for each participant. Which statistical test would be best to determine if there is a significant difference between groups? one-sample t-test dependent samples t-test independent samples t-test ANOVAA health officer thinks that there is an association between diabetes (DM) and hypertension (HPT). He believes that those with DM are more likely to have HPT as well. For his research, from the hospital records, he selected a random sample of 400 patients, 300 with DM and 100 without DM. Out of the 300 DM patients, 60 had HPT as well and out of the 100 non DM patients 10 had HPT. He also recorded the patients’ gender. With the data, he performed two analyses, one for all policyholders and the other by gender. The following are some of the output from SPSS. Must the manager consider gender as well in this case? If yes, stating the evidence, explain why. If you are the health officer, how would the results above help you in making decision? What other variables, if any, should be considered as well?A health officer thinks that there is an association between diabetes (DM) and hypertension (HPT). He believes that those with DM are more likely to have HPT as well. For his research, from the hospital records, he selected a random sample of 400 patients, 300 with DM and 100 without DM. Out of the 300 DM patients, 60 had HPT as well and out of the 100 non DM patients 10 had HPT. He also recorded the patients’ gender. With the data, he performed two analyses, one for all policyholders and the other by gender. The following are some of the output from SPSS. State the percentage of DM and non DM patients who had State the test that was used in these Test if there is an association between DM and HPT
- An antidepressant has historically been known to relieve depression for 65% of patients who use the drug. The drug's formula was changed recently to include a smaller dose of its main, active ingredient. Due to this change, a medical researcher suspects that among all people with depression who use the new version of the drug, the proportion, p, of people who find relief is lower than 65%. A random sample of 215 patients who use the new version is selected by the researcher, and 125 of them find relief from depression. Based on these data, can we support the researcher's claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)A survey claims that 9 out of 10 doctors (i.e., 90%) recommend brand Z for their patients who have children. To test this claim against the alternative that the actual proportion of doctors who recommend brand Z is less than 90%, a random sample of 100 doctors results in 83 who indicate that they recommend brand Z. Find the test statistic.An antidepressant has historically been known to relieve depression for 65% of patients who use the drug. The drug's formula was changed recently to include a smaller dose of its main, active ingredient. Due to this change, a medical researcher suspects that among all people with depression who use the new version of the drug, the proportion, p, of people who find relief is lower than 65%. A random sample of 245 patients who use the new version is selected by the researcher, and 139 of them find relief from depression. Based on these data, can we support the researcher's claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) (a) State the null hypothesis Ho and the alternative hypothesis H₁. Ho :O H₁ :0 (b) Determine the type of test statistic to use. (Choose one) ▼ (c) Find the…
- In a study conducted by the U.S. Department of Health and Human Services, a sample of 546 boys aged 6–11 was weighed, and it was determined that 87 of them were overweight. A sample of 508 girls aged 6–11 was also weighed, and 74 of them were overweight. Can you conclude that the proportion of boys who are overweight differs from the proportion of girls who are overweight? Please show calculInternet addiction has been described as excessive and uncontrolled Internet use. The authors of a research article used a score designed to measure the extent and severity of Internet addiction in a study of 837 male and 874 female sixth-grade students in China. Internet Addiction was measured using Young's Internet Addiction Diagnostic Test. The lowest possible score on this test is zero, and higher scores indicate higher levels of Internet addiction. For the sample of males, the mean Internet Addiction score was 1.51 and the standard deviation was 2.02. For the sample of females, the mean was 1.01 and the standard deviation was 1.67. For purposes of this exercise, you can assume that it is reasonable to regard these two samples as representative of the population of male Chinese sixth-grade students and the population of female Chinese sixth grade students, respectively. The standard deviation is greater than the mean for each of these samples. Explain why it is not reasonable to…An antiaepressant has historically been known to relieve depression for 80% of patients who use the drug. The drug's formula was changed recently to include a smaller dose of its main, active ingredient. Due to this change, a medical researcher suspects that among all people with depression who use the new version of the drug, the proportion, p, of people who find relief is lower than 80%. A random sample of 175 patients who use the new version is selected by the researcher, and 135 of them find relief from depression. Based on these data, can we support the researcher's claim at the 0.10 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H.. H, : (b) Determnine the type of test statistic to use. C=D ロS口 (Choose one) (c) Find the…
- Internet addiction has been described as excessive and uncontrolled Internet use. The authors of a research article used a score designed to measure the extent and severity of Internet addiction in a study of 837 male and 874 female sixth-grade students in China. Internet Addiction was measured using Young's Internet Addiction Diagnostic Test. The lowest possible score on this test is zero, and higher scores indicate higher levels of Internet addiction. For the sample of males, the mean Internet Addiction score was 1.51 and the standard deviation was 2.02. For the sample of females, the mean was 1.01 and the standard deviation was 1.67. For purposes of this exercise, you can assume that it is reasonable to regard these two samples as representative of the population of male Chinese sixth-grade students and the population of female Chinese sixth grade students, respectively.Internet addiction has been described as excessive and uncontrolled Internet use. The authors of a research article used a score designed to measure the extent and severity of Internet addiction in a study of 835 male and 873 female sixth-grade students in China. Internet Addiction was measured using Young's Internet Addiction Diagnostic Test. The lowest possible score on this test is zero, and higher scores indicate higher levels of Internet addiction. For the sample of males, the mean Internet Addiction score was 1.53 and the standard deviation was 2.02. For the sample of females, the mean was 1.07 and the standard deviation was 1.68. For purposes of this exercise, you can assume that it is reasonable to regard these two samples as representative of the population of male Chinese sixth-grade students and the population of female Chinese sixth grade students, respectively. (a) The standard deviation is greater than the mean for each of these samples. Explain why it is not reasonable…An antidepressant has historically been known to relieve depression for 80% of patients who use the drug. The drug's formula was changed recently to include a smaller dose of its main, active ingredient. Due to this change, a medical researcher suspects that among all people with depression who use the new version of the drug, the proportion, p, of people who find relief is lower than 80%. A random sample of 215 patients who use the new version is selected by the researcher, and 159 of them find relief from depression. Based on these data, can we support the researcher's claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) (a) State the null hypothesis H and the alternative hypothesis H₁. H:0 H₁ :0 (b) Determine the type of test statistic to use. (Choose one) ▼ (c) Find the value…