To find the standard equation for a circle centered at the origin, we use the distance formula, since the radius measures?
2 answers:
Answer: The answer is (a). the distance from any point in the circle to the origin.
Step-by-step explanation: We know that to find the standard equation of a circle centred at the origin, we always use the distance formula. We are to select the correct measurement for the radius from the options given.
In a circle, as shown in the attached figure, the radius 'r' is the distance from any point 'P' on the circle to the centre 'O' of the circle. The centre of the circle in standard form is usually the origin.
So, we can say that radius measures the distance from any point on the circle to the origin.
Thus, (a) is the correct option.
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Answer:
The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 45 - 1 = 44
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0141
The margin of error is:
M = T*s = 2.0141*170.5 = 343.4
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 390.47 - 343.40 = 47.07 units per month
The upper end of the interval is the sample mean added to M. So it is 390.47 + 343.40 = 733.87 units per month
The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87
Answer:
Shown
Step-by-step explanation:
Given that twelve basketball players, whose uniforms are numbered 1 through 12, stand around the center ring on the court in an arbitrary arrangement.
Let us consider consecutive numbers in this set.
After this we find the totals are more than 20.
When 1 to 12 are arbitrarily arranged, there are chances that numbers from 6 and above are having consecutive numbers.
These totals are greater than 20
Hence shown that some three consecutive players have the sum of their numbers at least 20.
(i.e. starting from if we take)