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Sonja [21]
3 years ago
8

3. write an equation of the circle whose center is (-5,2) and whose radius length is 4 express your answer in standard form. I N

EED MY WORK TO BE SHOWN PLEASE HURRY I AM BEING TIMED!!!!!
Mathematics
1 answer:
marin [14]3 years ago
6 0
Circle = (x-h)^2 + (y-k)^2 = r^2

Center is (h,k) h = -5, k = 2
Radius is 4, r = 4

(x - -5)^2 + (y - 2)^2 = 4^2
(x + 5)^2 + (y - 2)^2 = 16
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Is 12 a solution to the equation × - 8 = 20? Explain how you know.
jeyben [28]

We have the following:

x-8=20

solving for x

\begin{gathered} x=20+8 \\ x=28 \end{gathered}

The solution of x is equal to 28, therefore 12 is not a solution of the equation

5 0
1 year ago
The 8 red markers in a package account for 4% of all markers in the package. How many markers are in the package?
Vlad [161]

Answer:

192 markers

Step-by-step explanation:

if 8 = 4% then 2=1%

so we can assume there are 200markers in the package.

8-200= 192 markers

6 0
2 years ago
In the function y = 1.25x + 5, y represents the cost of a foot of cable and x represents the number of feet. How much does the c
klemol [59]
It increases by B $1.25 every time
7 0
3 years ago
Read 2 more answers
F
oksian1 [2.3K]

Answer:

-9

Step-by-step explanation:

f(2) - f(4) / 2-4

[2²+3(2)] - [4²+3(4)] / -2

(10 - 28) / 2

= -18/2

= -9

6 0
3 years ago
Read 2 more answers
A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part
Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range          

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{8726}{20} = 436.3

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

S.D = \sqrt{\frac{75924.2}{19}} = 63.21

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Median = \frac{431 + 437}{2} = 434

Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482

IQR = 482 - 395 = 87

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

Mean =\displaystyle\frac{8406}{20} = 420.3

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

S.D = \sqrt{\frac{247636.2}{19}} = 114.16

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

Median = \frac{431 + 437}{2} = 434

Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482

IQR = 482 - 395 = 87

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0
3 years ago
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