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3 years ago

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A radio disc jockey has 7 songs on this upcoming hour's playlist: 2 are rock songs, 2 are reggae songs, and 3 are country songs.

The disc jockey randomly chooses the first song to play, and then she randomly chooses the second song from the remaining ones. What is the probability that the first song is a country song and the second is a reggae song? Write your answer as a fraction in simplest form.

1 answer:

Eddi Din [679]3 years ago

7 0

Probability first song is country is 3/7
probability second song is reggae is 2/6 which reduces to 1/3
probability of both is : 3/7 * 1/3 = 3/21 which reduces to 1/7 <==

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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:

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

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beks73 [17]

Answer:

The Value of $a=\frac{15}{4}$ .

Step-by-step explanation:

We have Named the figure please find the attachment for your reference.

Given:

PR = y

QR = a

RS = b

PS = z

PQ = x

QS = 15

∠P = 90°

∠R = 90°

∠Q = 60°

∠S = 30°

We need to find the Value of 'a'.

Solution:

Now we know that:

In Δ PQS

∠P = 90°

∠S = 30°

Now we know that;

$sin\ \theta = \frac{opposite\ side}{Hypotenuse}$

$sin \ S= \frac{PQ}{QS}$

Substituting the given values we get;

$sin\ 30\°=\frac{x}{15}$

Now we know that;

$sin\ 30\° = \frac12$

So we can say that;

$\frac{1}{2}=\frac{x}{15}\\\\x=\frac{15}{2}$

Now In Triangle PQR.

∠R = 90°

∠Q = 60°

So we can say that;

$Cos \theta = \frac{adjacent \ Side}{Hypotenuse}\\$

$Cos\ Q = \frac{QR}{PQ}$

Substituting the given values we get;

$cos 60\°= \frac{a}{x}$

Now we know that;

$cos 60\°= \frac12$

$x=\frac{15}{2}$

So substituting the values we get;

$\frac{1}{2}=\frac{a}{\frac{15}{2}}$

By Using Cross Multiplication we get;

$a= \frac{1}{2}\times\frac{15}{2}\\\\a=\frac{15}{4}$

Hence The Value of $a=\frac{15}{4}$ .

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