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miv72 [106K]
3 years ago
15

Write the standard equation of a circle that passes through (-5 5) with center (-10 -5) brainly

Mathematics
1 answer:
timama [110]3 years ago
8 0

Answer:

The equation of the circle is (x + 10)² + (y + 5)² = 125 in standard form

Step-by-step explanation:

* lets study the standard form of the equation of a circle

- If the coordinates of the center of the circle are(h , k) and its radius

 is r, then the standard equation of the circle is:

 (x - h)² + (y - k)² = r²

* Now lets solve the problem

∵ The coordinates of the center of the circle are (-10 , -5)

∵ The standard form of the equation is (x - h)² + (y - k)² = r²

∵ h , k are the coordinates of the center

∴ h = -10 , k = -5

∴ The equation of the circle = (x - -10)² + (y - -5)² = r²

∴ The equation of the circle = (x + 10)² + (y + 5)² = r²

- To find the value of the radius lets use the point (-5 , 5) to

  substitute their coordinate instead of x and y in the equation

∵ The circle passes through point (-5 , 5)

∵ (x + 10)² + (y + 5)² = r²

- Use x = -5 and y = 5

∴ (-5 + 10)² + (5 + 5)² = r² ⇒ simplify

∴ (5)² + (10)² = r²

∴ 25 + 100 = r²

∴ r² = 125

* Now lets write the equation in standard form

∴ (x + 10)² + (y + 5)² = 125

* The equation of the circle is (x + 10)² + (y + 5)² = 125 in standard form

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1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie
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Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

-----------------------

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21
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3 years ago
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Answer:

t = 4

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