A diver begins at 150 feet below sea level, descends at a steady rate of 5 feet per minute for 3.5 minutes, and then ascends 122
2 answers:
Answer:
The answer is -45.3 feet depth. while minus sign shows the descend under the water with water surface as starting point.
Step-by-step explanation:
Initial depth of diver under the water
Diver descends 5 feet per minute for 3.5 minutes
Note: minus sign shows that the diver is under the water.
Diver ascends
Now adding all the terms a below as a numerical expression.
Current Depth of Diver after he descends 5 feet for 3.5 minutes and ascend 122.2 feet after
By taking the initial depth of diver from which we add or subtract the changes made when he ascends or descends in the water to get the current depth.
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In ΔWXY, x = 680 inches, w = 900 inches and ∠W=157°.
To find:
The all possible values of ∠X, to the nearest degree.
Solution:
Law of Sines:
For ΔWXY,
Now,
Therefore, the value of ∠X is 17 degrees.
The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.
<h3>When do we use two-sample t-test?</h3>The two-sample t-test is used to determine if two population means are equal.
A nutritionist has developed a diet that she claims will help people lose weight. In this,
- Twelve people were randomly selected to try the diet.
- Their weights were recorded prior to beginning the diet and again after 6 months.
Here are the original weights, in pounds, with the weight after 6 months in parentheses.
- Before 192 212 171 215 180 207 165 168 190 184 200 196
- After 183 196 174 211 160 191 162 175 190 179 189 195
The mean of the weights before 6 moths is,
The mean of the weights after 6 months is,
Standard deviation of both the data is 16.9 and 14.7.
1. Null and Alternative Hypotheses.
The following null and alternative hypotheses need to be tested:
This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.
- (2) Rejection Region
Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:
Hence, it is found that the critical value for this right-tailed test is
, for α=0.05 and df=22
The rejection region for this right-tailed test is,
(3) Test Statistics
Since it is assumed that the population variances are equal, the t-statistic is computed as follows:
- (4) Decision about the null hypothesis
Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721 it is concluded that the null hypothesis is not rejected.
(5) Conclusion
It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is greater than μ2, at the α=0.05 significance level.
Confidence Interval
The 95% confidence interval is −7.16<μ<19.66
Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.
Learn more about the two-sample t-test here;
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