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4 years ago

Circumcenter of triangle with vertices of (1,1) (1,-3) (7,-3)

Mathematics

1 answer:

Natali [406]4 years ago

7 0

Answer:

The circumcenter is (-17/2, -15/2)

Step-by-step explanation:

To find the circumcenter, solve any two bisector equations and find out the intersection points. The given are A(1,1), B(0,2), and C(3,-2).Midpoint of AB = (1/2, 3/2) - You can get the midpoint by getting the average of x-coordinates and y-coordinates. Slope of AB = -1Slope of perpendicular bisector = 1Equation of AB with slope 1 and the coordinates (1/2, 3/2) isy - (3/2) = (1)(x - 1/2) y = x+1Do the same for ACMidpoint of AC = (2, -1/2)Slope of AC = -3/2Slope of perpendicular bisector = 2/3Equation of AC with slope 2/3 and the coordinates (2, -1/2) isy - (-1/2) = (2/3)(x - 2) y = -11/6 + 2x/3So the perpendicular bisectors of AB and BC meety = x+1y = -11/6 + 2x/3To solve for x,(-11/6 + 2x/3) = (x+1)x= -17/2Now get y by substituting y = (-17/2) + 1y = -15/2

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3 years ago

Joseph has a collection of stickers. When the stickers are aranged in piles of 3, there are 2 stickers left over. When the stick

Elodia [21]

Answer:

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Step-by-step explanation:

To answer the question, all you need to do is figure out at what value of stickers would the remaining stickers be equal to 2 if the piles are made up of either 3 to a set or 4 to a set. To do this, we need to figure out the size of the whole set of stickers.

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Lets test the second condition. If we divide it in piles of 4, we will again have 12 stickers (3 piles of 4 stickers each) with 2 stickers remaining.

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3 years ago

A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of

GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that the t-distribution is used 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 = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.989.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

$M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23$

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

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3 years ago

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Answer:

A-33(PAY) OR 33(P)

B-33 x pay or 33 x p

C-hours(33)or H(33)

D-pay(33) or p(33)

its D is correct hope this helps

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3 years ago