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3 years ago

8

a glass of jar contains 1 red, 3 green and 2 blue marbles. asingle marble is chosen at random from the jar and its color is reco

rded.give your answe using set notation.

1 answer:

lubasha [3.4K]3 years ago

4 0

hi,

first let's count the marbles : 1+3+2 = 6

so picking a red is 1/6

a green is : 3/6

a bleu is : 2/6

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Here is a parallelogram with its diagonals shown.

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A. False

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3 years ago

The Hudson high school wrestling team just won the state tournament and have been awarded a triangular pennant to hang on the wa

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Answer:

3 feet.

Step-by-step explanation:

Base of the triangular pennant = 1.5 feet long.
Area of the triangular pennant = 2.25 Square feet.

Now, we know that:

Area of a Triangle $=\frac{1}{2}$ X Base X Height$

Substituting the given values, we have:

$2.25=\dfrac{1}{2}$ X 1.5 X Height\\\\2.25=$\dfrac{1.5}{2}$ X Height\\\\Cross multiply\\\\2.25*2=1.5 X Height\\\\Divide both sides by 1.5 to solve for the height\\\\\dfrac{2.25*2}{1.5}=\dfrac{1.5 X Height}{1.5}\\\\$Height of the Triangle =3 feet$

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3 years ago

What is the expansion of (3+x)^4

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Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

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3 years ago

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Answer:

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Step-by-step explanation:

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