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2 years ago

8

Suppose that a population parameter is 0.3, and many samples are taken from the population. As the size of each sample increases

, the mean of the sample proportions would approach which of the following values? O A. 0.4 O B. 0.1 O C. 0.3 O D. 0.2 SUBMIT

1 answer:

11111nata11111 [884]2 years ago

8 0

Answer: C. 0.3

Step-by-step explanation:

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find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

What is the LCM of 56 and 504?

tatuchka [14]

The LCM is 2x2x2x7 factorized

6 0

2 years ago

Read 2 more answers

Can someone please help me factor and write the equations for these two problems ?

prisoha [69]

Answer:

Step-by-step explanation:

Asymptotes 3

g(x) = $\frac{3x}{x^2-3x-10}$

Factors of denominator will be,

x² - 3x - 10 = x² - 5x + 2x - 10

= x(x - 5) + 2(x - 5)

= (x + 2)(x - 5)

Therefore, factored form of g(x) will be,

g(x) = $\frac{3x}{(x + 2)(x - 5)}$

Asymptotes 4

h(x) = $\frac{(x-5)}{x^{2} + 14x + 40}$

= $\frac{x-5}{x^{2}+10x+4x+40}$

= $\frac{x-5}{x(x+10)+4(x+10)}$

= $\frac{x-5}{(x+10)(x+4)}$

5 0

2 years ago

100 is the regular price if you buy one and get the other half price. Explain how this different from getting a 50% discount?

Vinvika [58]

100 = x + x/2

3x/2 = 100

x = 200/3

if there was 50% discount:

buying 1 would be $33.33

buying 2 would be $66.67

4 0

2 years ago

Plz help ASAP !!!!! Plzzz

adell [148]

Answer:

The second one

Step-by-step explanation:

She started with x dollars and then used 8 dollars to buy a football game ticket, so x-8. Then, she is left with 56 dollars, so x-8=56. Therefore, the second story represents the equation.

5 0

2 years ago