Answer:
8 miles or 25.133 miles
Step-by-step explanation:
your radius is 4 miles.
The area of a circle is A=
Therefore, the area of this circle is , or 16
Half of this area, because what you're looking for is a semi-circle, is:
8 (exact) or 25.133 (rounded) miles.
Answer:
Length of the rectangle = 9.58 feet and width = 27.14 foot
Step-by-step explanation:
Let the length of the rectangular area is = x feet
and the width of the area = y feet
Area of the rectangle = xy square feet
Or xy = 260
y = -------(1)
Cost to fence the three sides = $3 per foot
Therefore cost to fence one length and two width of the rectangular area
= 3(x + 2y)
Similarly cost to fence the fourth side = $14 per foot
So, the cost of the remaining length = 14x
Total cost to fence = 3(x + 2y) + 14x
Cost (C) = 3(x + 2y) + 14x
C = 3x + 6y + 14x
= 17x + 6y
From equation (1)
C =
Now we take the derivative,
C' = 17 -
To minimize the cost of fencing,
C' = 0
17 - = 0
= 17
x = 9.58 foot
and y =
y = 27.14 foot
Answer:
1. M=108
2. μ=110
3. In the explanation.
4. Test statistic t = -1.05
5. P-value = 0.1597
Step-by-step explanation:
The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.
We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.
1. The mean points is the sample points and has a value of 108 points.
2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.
3. This is a hypothesis test for the population mean.
The claim is that the mean score is significantly less than 110.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=10.
The sample mean is M=108.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.
The estimated standard error of the mean is computed using the formula:
4. Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.1597) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
There is not enough evidence to support the claim that the mean score is significantly less than 110.