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kozerog [31]
3 years ago
5

Please help please please

Mathematics
1 answer:
alexgriva [62]3 years ago
8 0
Two sets (or three technically)
sets {2, 4, 6, 8, 10} & {8,9,10}

The probability of one of the above numbers because it is a union of those two vars/sets so numbers from either set go

{2, 4, 6, 8, 9, 10}
Thats 6 of the 10 numbers 

6/10
.6

If i'm wrong, sorry, haven't done this kind of stuff in a while
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Emily has 300 books. if frank were to double the number of books that he now owns, he would still have fewer than Emily has
Alex_Xolod [135]
This statement is false. 
If he were to double the amount that Emily has to get franks, franks would have more. 
Look :)
Emily's = 300
Frank's = 300 x 2 = 600

now...
Emily's = 300
Frank's = 600

So frank has more! NOT fewer. 

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Lisa used the rule "Red, Yellow, Yellow, Yellow" to make a bracelet with a repeating pattern. She
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Answer:

28 beads were used to make the bracelet. 7 beads were not yellow.

Step-by-step explanation:

3/4 beads used to make the bracelet are yellow

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Which division expression is equivalent to 4 1/3 ÷ -5/6
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4\dfrac{1}{3}\div\left(-\dfrac{5}{6}\right)\qquad(*)\\\\\text{Convert the mixed number to the improper fraction:}\\\\4\dfrac{1}{3}=\dfrac{4\cdot3+1}{3}=\dfrac{12+1}{3}=\dfrac{13}{3}\\\\(*)=\dfrac{13}{3}\div\left(-\dfrac{5}{6}\right)\\\\\text{By dividing by a fraction, we multiply by its reciprocal.}\\\\=\dfrac{13}{3}\cdot\left(-\dfrac{6}{5}\right)\qquad\text{simplify}\\\\=\dfrac{13}{3\!\!\!\!\diagup_1}\cdot\left(-\dfrac{6\!\!\!\!\diagup^2}{5}\right)=-\dfrac{13\cdot2}{1\cdot5}=-\dfrac{26}{5}=-\dfrac{25+1}{5}=-5\dfrac{1}{5}

5 0
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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

8 0
3 years ago
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