A school band found they could arrange themselves in rows of 6, 7, or 8 with no one left over. What is the minimum number of students in the band?

Answer:

x = 2

x =-2

x =1 +√3i

x =1 -√3i

x =-1 +√3i

x =-1 -√3i  

Step-by-step explanation:

[tex]x^{6} -64 = 0\\(x^{3} -8)(x^{3} + 8) = 0\\ \\( x - 2) (x^{2} + 2x + 4)( x + 2) (x^{2} -2x + 4) = 0\\[/tex]

x = 2

x =-2

x =1 +√3i

x =1 -√3i

x =-1 +√3i

x =-1 -√3i  

Answer:

35x^2 - 28x - 7

Step-by-step explanation:

Answer:

[tex]P(3) = ^9C_3 * 0.5^3 *0.5^6[/tex]

Step-by-step explanation:

Given

[tex]n = 9[/tex] --- number of flips

Required

[tex]P(x = 3)[/tex]

The probability of getting a head is:

[tex]p = \frac{1}{2}[/tex]

[tex]p = 0.5[/tex]

The distribution follows binomial probability, and it is calculated using:

[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x}[/tex]

So, we have:

[tex]P(3) = ^9C_3 * 0.5^3 * (1 - 0.5)^{9-3}[/tex]

[tex]P(3) = ^9C_3 * 0.5^3 *0.5^6[/tex]

Answer:

Aron flips a penny 9 times. Which expression represents the probability of getting exactly 3 heads?

Answer: A

Step-by-step explanation:

Answer:

3/20

Step-by-step explanation:

By question it's given that ,

[tex]\implies \dfrac{a}{b}=\dfrac{2}{5}[/tex]

[tex]\implies \dfrac{b}{c}=\dfrac{3}{8}[/tex]

And we need to find out the value of a/c .For that Multiply both of them , we have ;

[tex]\implies \dfrac{a}{b} \times\dfrac{b}{c}=\dfrac{2}{5}\times \dfrac{3}{8} \\\\\implies \dfrac{a}{c}= \dfrac{3}{20}[/tex]

Hence the required answer is 3/20 .

Answer:

[tex]4 {x}^{2} - 4x - 8 - (2 {x}^{2} + 3x - 6) = 4 {x}^{2} - 4x - 8 - 2 {x}^{2} - 3x + 6 = 2 {x}^{2} - 7x - 2[/tex]

Answer:

0.663 ( c )

Step-by-step explanation:

U1 = 0.1 , U2 = 0.8

using the "cosine" version of Box-Muller to generate a single Nor(-1,4) random variable

first step : generate single obsⁿ from N ( -1,4 )

attached below is the detailed solution

Answer:

Step-by-step explanation:

Um where is the diagrahm

Answer:

2.2763 x 10 to the power of 4

for some reason it doesn't let me put in the explanation

Answer:

Tenemos dos propiedades de la potencia en este caso:

Para un numero real A:

[tex]A^0 = 1[/tex]

[tex](A^n)^m = A^{n*m}[/tex]

En este caso nuestra ecuación es:

[tex][ [(\frac{0.1234}{-3.2098})^4]^3]^0[/tex]

usando la segunda propiedad, podemos reescribir como:

[tex][ [(\frac{0.1234}{-3.2098})^4]^3]^0 = (\frac{0.1234}{-3.2098})^{4*3*0} = (\frac{0.1234}{-3.2098})^0[/tex]

Y acá tenemos un numero real a la potencia 0, sabemos que esto es igual a 1, entonces:

[tex](\frac{0.1234}{-3.2098})^0 = 1[/tex]

Answer:

36281629273781646181993836619946527189119292937467482919198$7473828191927364732818919283838292927383883829118661552621718919191019284746617171819001187373765252728

Step-by-step explanation:

173899918377+28910873638282

Answer:

9a²b

Step-by-step explanation:

Hi there!

We need to find the greatest common factor out of 45a²b³ and 18[tex]a^{4}[/tex]b

We can split apart the monomials to make it easier

45a²b³ is 45*a²b³

18[tex]a^{4}[/tex]b is 18*[tex]a^{4}[/tex]b

First, let's find the GCF out of 45 and 18 (the number coefficients)

we can find all of the multiples of the 2 numbers:

45 is made up of 9 and 5

9 is made up of 3 and 3

so 3*3*5 is 45

18 is made up of 2 and 9

9 is made up of 3 and 3

so 2*3*3 is 18

3*3 is in both 45 and 18, so 9 is the GCF out of 45 and 18

Now let's find the GCF out of a²b³ and [tex]a^{4}[/tex]b

a²b³ made up of a² and b³

so a²b³ is a*a*b*b*b

[tex]a^{4}[/tex]b is made up of [tex]a^{4}[/tex] and b

so [tex]a^{4}[/tex]b is a*a*a*a*b

a*a*b is in both a²b³ and [tex]a^{4}[/tex]b, so the GCF out of a²b³ and [tex]a^{4}[/tex]b is a²b

Now multiply 9 and a²b together, as they are only the GCF of the parts of the monomials

9*a²b=9a²b

there's the greatest common factor of the 2 monomials

Hope this helps!

Answer:

2 Answers. #1. +11. [3+(3/7)] times 7 is 24. DarkBlaze347 May 1, 2015. +5. Good job, DB! civonamzuk May 1, 2015.

35 Online Users.

Step-by-step explanation:

brainliest please and follow:D

Answer: The answer is 2,4. May I have the brainiest? pls I only need one more.

Answer:

24

Step-by-step explanation:

You first find out how many cubes can fit into each measurement, then multiply them. (2*4*3=24)

Answer:

It will take 24 cubes to fill the rectangular prism.

Step-by-step explanation:

Find the volume of a cube with side lengths of 1/2 cm:

1/2^3 = 1/8

1/8 cm^3

Find the volume of the whole rectangular prism (lwh):

1 x 3/2 x 2

= 3/2 x 2

= 3

3 cm^3

Divide the volume of the prism by the bolume of one cube:

3 ÷ 1/8 = 24

Therefore it will take 24 cubes to fill the prism. Hope this helps!

[tex] \sqrt[5]{32} \times 2 \sqrt[3]{81} \\ = {32}^{ \frac{1}{5} } \times 2 {(81)}^{ \frac{1}{3} } \\ = ({{2}^{5}})^{ \frac{1}{5} } \times {2({3}^{3})}^{ \frac{1}{3} } \\ = {2}^{1} \times 2({3}^{1}) \\ = 2 \times 2 \times 3 \\ = 4 \times 3 \\ = 12[/tex]

This is the answer.

Hope it helps!!

16:52

Hope this helps! :)

Answer:

Step-by-step explanation:

Given expressions are,

[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex] and [tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]

Remove the radicals from the denominator from both the expressions.

[tex]p=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}+\sqrt{5}} \times \frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]

  [tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]

  [tex]=\frac{(\sqrt{10}-\sqrt{5})^2}{5}[/tex]

[tex]\sqrt{p}=\sqrt{\frac{(\sqrt{10}-\sqrt{5})^2}{5}}[/tex]

      [tex]=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{5}}[/tex]

[tex]q=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}[/tex]

  [tex]=\frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}-\sqrt{5}}\times \frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}[/tex]

  [tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{(\sqrt{10})^2-(\sqrt{5})^2}[/tex]

  [tex]=\frac{(\sqrt{10}+\sqrt{5})^2}{5}[/tex]

[tex]\sqrt{q}=\sqrt{\frac{(\sqrt{10}+\sqrt{5})^2}{5}}[/tex]

     [tex]=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}[/tex]

[tex]\sqrt{q}-\sqrt{p}-2\sqrt{pq}=\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}}-\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}}-2(\frac{(\sqrt{10}+\sqrt{5})}{\sqrt{5}})(\frac{(\sqrt{10}-\sqrt{5})}{\sqrt{5}})[/tex]

                           [tex]=\frac{1}{\sqrt{5}}(\sqrt{10}+\sqrt{5}-\sqrt{10}+\sqrt{5})-\frac{2}{5}[(\sqrt{10})^2-(\sqrt{5})^2)][/tex]

                           [tex]=\frac{1}{\sqrt{5}}(2\sqrt{5})-\frac{2}{5}(10-5)[/tex]

                           [tex]=2-2[/tex]

                           [tex]=0[/tex]

Step-by-step explanation:

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