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CaHeK987 [17]
2 years ago
13

What is the standard equation of a circle with center (-2,0) and radius 4?

Mathematics
1 answer:
Fed [463]2 years ago
5 0

Answer:

B(x+2)^2+y^2=16

Step-by-step explanation:

since you have an x-coordinate only and it is negative and a y coordinate which is 0..basically x will appear positive with its conjecture x^2 and the y value will be independent...y^2...bare in Mind that radius is always squared hence it is 16 instead of 4

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ANSWER: the amount an individual earns after subtracting all expenses, taxes and costs.

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What kind of change occurs when a figure is translated?
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It depends on what way the figure is translated, but it's possible it could be a dilations, reflection, rotation, etc.
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The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use t
astraxan [27]

Answer:

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Step-by-step explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)=\frac{ME*\sqrt{N}}{s } where

  • Me is the margin of error from the mean
  • s is the standard deviation of the population
  • N is the sample size

a.

z(p)=\frac{8*\sqrt{30}}{50 } ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)=\frac{8*\sqrt{50}}{50 } ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)=\frac{8*\sqrt{100}}{50 } ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥(\frac{z*s}{ME} )^2 where

  • N is the sample size
  • z is the corresponding z-score in 95% probability (1.96)
  • s is the standard deviation (50)
  • ME is the margin of error (8)

then N≥(\frac{1.96*50}{8} )^2 ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

7 0
3 years ago
Round your answers to one decimal place, if necessary.
Marina CMI [18]

Answer: 11

Step-by-step explanation:

Add the numbers together to get 68, then divide it by 6 to get your answer! Sorry if I'm wrong!

5 0
3 years ago
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In a group of 10 ​kittens, 5 are female. Two kittens are chosen at random. ​a) Create a probability model for the number of male
olya-2409 [2.1K]

Answer:

(b) Expected number of male kitten E(x) = 1

(c) Standard deviation is ±\frac{2}{3}.

Step-by-step explanation:

Given that,

Number of kittens in a group are 10. Out of these 5 are female.

To find:- (a) create a probability model of male kitten chosen.

    So,  total number of kitten = 10

           number of female kitten = 5    

then, Number of male kitten = 10-5= 5

Now total number of ways to choosing two kittens from group of 10 = ^{10} C_{2}

                                                                                                                 = \frac{10!}{2!\times8!}

                                                                                                                 = 45

for choosing (i) No male P(x=0) = P(Two female) =  \frac{^{5} C_{2}}{45}

                                                                           = \frac{2}{9}

                     (ii) 1 male P(x=1) = P(1 male and 1 female) = \frac{^{5} C_{1}\times^{5} C_{1}}{45} =  \frac{25}{45} =\frac{5}{9}

                     (iii) 2 male P(x=2)= P(2 male) =\frac{^{5} C_{2}}{45} =  \frac{2}{9}

      x_{i}                             0                     1                      2

P(X=x_{i})                           \frac{2}{9}                      \frac{5}{9}                      \frac{2}{9}

(b) what is expected number of male kittens chosen ?

    Expected number of male kitten E(x) =  o\times \frac{2}{9} + 1\times \frac{5}{9} + 2\times\frac{2}{9}

                                                                  = 1

(c) what is standard deviation ?

                              \sigma(x) = \sqrt{ (E(x^{2} )-(Ex)^{2})

No,                          

                               Ex^{2} =o\times \frac{2}{9} + 1\times \frac{5}{9} + 4\times\frac{2}{9}

                                       = \frac{13}{9}

                               \sigma(x)=\sqrt{\frac{13}{9} - 1 }=\sqrt{\frac{4}{9} }  

                                       = ± \frac{2}{3}

                               

6 0
2 years ago
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