F(x)=5(1+0.25)^x,f(4) Please show step by step of how you got the answer
2 answers:
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Answer:
0.375 feet-lb
Step-by-step explanation:
We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.
We can represent our given information as:
We will use Hooke's Law to solve our given problem.
Substituting this value in our integral, we will get:
Using power rule, we will get:
We know that 6 inches is equal to 0.5 feet.
Work needed to stretch it beyond 6 inches beyond its natural length would be
Using power rule, we will get:
Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.
Using the t-distribution, we have that the 95% confidence interval for the true mean number of pushups that can be done is (9, 21).
For this problem, we have the <u>standard deviation for the sample</u>, thus, the t-distribution is used.
- The sample mean is of 15, thus
.
- The sample standard deviation is of 9, thus
.
- The sample size is of 10, thus
.
First, we find the number of degrees of freedom, which is the one less than the sample size, thus df = 9.
Then, looking at the t-table or using a calculator, we find the critical value for a 95% confidence interval, with 9 df, thus t = 2.2622.
The margin of error is of:
Then:
The confidence interval is:
Then
The confidence interval is (9, 21).
A similar problem is given at brainly.com/question/25157574