Answers:
length of rope = 2 meters
perimeter of meadow = 12.56 square meters
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Explanation:
Assuming there are no structures (buildings, fences, etc) to get in the way of the goat, this means the goat's area that it can roam is a circle with radius r
The area of the circle is pi*r^2 which is equal to 12.56 square meters
pi*r^2 = 12.56
3.14*r^2 = 12.56
r^2 = 4
r = sqrt(4)
r = 2
The radius of the circle is 2 meters. This is the length of the rope. If the goat goes out as far as it can go, and tries to move left/right, then it will trace out a circle of radius 2. This is assuming the rope is kept fully taut the entire time.
The perimeter of the circle is the same as the circumference
circumference = 2*pi*r
circumference = 2*pi*2
circumference = 4*pi
circumference = 4*3.14
circumference = 12.56
The circumference and area are the same because r^2 = 4 and 2*r = 4 when r = 2. For any other positive r value, the area and circumference will be different values.
So in short, if a friend says that the area of a circle is the same as its perimeter, then you can immediately conclude the radius is 2.
Side note: throughout all of this solution, I used the approximation pi = 3.14 though pi has more decimal digits
Answer:
Only B and C are always true.
Step-by-step explanation:
To analyse which statements are always true, we go through the process of finding confidence intervals for sample means, from the start.
Confidence Interval = (Sample mean) ± (Margin of error)
From this expression, it is evident that the margin of error determines how wide the confidence interval would be.
Sample Mean = 110 (given)
Margin of Error = (Critical value) × (Standard deviation of the distribution of sample means)
Since no information about the population standard deviation is provided, the critical value is obtained using t-distribution.
The critical value usually varies at different confidence levels and degree of freedoms.
The higher the confidence level, the higher the critical value and the higher the margin of error leading to a wider range.
Hence, a confidence interval of 95% will have a higher critical value than a confidence interval of 90%. Hence, statement C is proved once that 'for n = 100, the 95% confidence interval will be wider than the 90% confidence interval'.
After obtaining the critical value, we then obtain the standard deviation of the distribution of sample means or simply the standard error of the mean. This is given as
σₓ = σ/√n
where σ = standard deviation; which isn't given. The standard deviation might be high enough to guarantee that the Margin of error is high too for the confidence interval to contain 115 or low enough to ensure that the Margin of error is very small and the confidence interval will not contain 115.
Or the sample size might be high enough to make the standard error of the mean to be eventually small and lead to a small margin of error and the condidence interval will not contain 115.
The point is, it isn't always sure that the resulting interval.would contain 115. So, statement A isn't always true.
Then from σₓ = σ/√n,
n = sample size, a large sample size means a more narrow confidence interval and a small sample size means a wider sample size. This proves statement C.
The 95% confidence interval for n = 100 will be more narrow than the 95% confidence interval for n = 50.
Hence, Only B and C are always true.
Hope this Helps!!!
1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5