paul earns 25.50 for every lawn he mows. this week he moved L lawns. he spends 3.50 per lawn on gasoline. write an expression to
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The percentage of the semicircle shaded section is approximately 23,606 %.
The percentage of the area of the semicircle is equal to the ratio of the semicircle area minus the half-cross area to the semicircle area. In other words, we have the following expression:
(1)
Where:
- Area of the half cross, in square centimeters.
- Area of the semicircle, in square centimeters.
- Percentage of the shaded section of the semicircle.
And the percentage of the shaded section is:
The percentage of the semicircle shaded section is approximately 23,606 %.
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Answer:
Part A
The p-value indicates that the largest probability of having the two remedies having a difference as observed or larger than as observed being equal to 0.3822
Part B:
Given that the p-value of 0.3822 is larger than the significance level α of 0.05, we fail to reject the null hypothesis and we conclude that there is enough statistical evidence to suggest that the salvage rates are the same for Remedy A and B
Step-by-step explanation:
The p-value
The test statistic is given as follows;
= 35/220 = 7/44
= 39/230
= 74/220 = 37/110
We get our test statistic z = -0.2996381
Part A
The p-value indicates that the largest probability of having the two remedies having a difference as observed or larger than as observed being equal to 0.3822
Part B:
Given that the p-value of 0.3822 is larger than the significance level α of 0.05, we fail to reject the null hypothesis and we conclude that there is enough statistical evidence to suggest that the salvage rates are the same for Remedy A and B.