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erastova [34]
3 years ago
15

a bag of marbles contains 8 red marbles, 4 blue marbles, and 5 white marbles. tom picks a marble at random. is it more likely th

at he picks a red marble or a marble of another color?
Mathematics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

It is more likely of a diffrent color.

Step-by-step explanation:

8 to 9

Red to other.

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Find the perimeter of a rhombus whose diagonals are
mojhsa [17]

Answer:

Rhombus Side = sqr root [(34 / 2)^2 + (30 / 2)^2]

Rhombus Side = sqr root [17^2 + 15^2]

Rhombus Side = sqr root [289 + 225}

Rhombus Side = sqr root [514]

Rhombus Side = 22.6715680975

Rhombus Perimeter = side * 4 = 90.68627239  cm

Step-by-step explanation:

5 0
3 years ago
A rectangular prism has a length of 4.2 cm, a width of 5.8 cm, and a height of 9.6 cm. A similar prism has a length of 14.7 cm,
solong [7]
They are each multiplied by a factor of 3.5
You can calculate this by taking the length of the similar prism and divide it with the first prism 14.7/4.2 = 3.5
Do this with the width and height to be sure
20.3/5.8 = 3.5
33.6/9.6 = 3.5
5 0
3 years ago
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Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o
ivolga24 [154]
\bf (2x+3)^5\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^5(+3)^0\\
2&+5&(2x)^4(+3)^1\\
3&+10&(2x)^3(+3)^2\\
4&+10&(2x)^2(+3)^3\\
5&+5&(2x)^1(+3)^4\\
6&+1&(2x)^0(+3)^5
\end{array}

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element?  well, 5*4/2
how did we get 10 for the fourth element?  well, 10*2/4


\bf (2x-3y)^4\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^4(-3y)^0\\
2&+4&(2x)^3(-3y)^1\\
3&+6&(2x)^2(-3y)^2\\
4&+4&(2x)^1(-3y)^3\\
5&+1&(2x)^0(-3y)^4
\end{array}


\bf (3a+4b)^8\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(3a)^8(+4b)^0\\
2&+8&(3a)^7(+4b)^1\\
3&+28&(3a)^6(+4b)^2\\
4&+56&(3a)^5(+4b)^3\\
5&+70&(3a)^4(+4b)^4\\
6&+56&(3a)^3(+4b)^5\\
7&+28&(3a)^2(+4b)^6\\
8&+8&(3a)^1(+4b)^7\\
9&+1&(3a)^0(+4b)^8
\end{array}

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶
7 0
3 years ago
Read 2 more answers
Solve pls brainliest
Natalija [7]
Answer:

1/4 - 1/8 = 2/8 - 1/8 = 1/8
5 0
2 years ago
Read 2 more answers
P is inversely proportional to the cube of (q-2) p=6 when q=3 find the value of p when q is 5
sveticcg [70]
\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
\\
&&y=\cfrac{{{  k}}}{x}
\end{array}\\\\
-----------------------------\\\\
\textit{p is inversely proportional to the cube of (q-2)}\implies p=\cfrac{k}{(q-2)^3}
\\\\\\
now \quad 
\begin{cases}
p=6\\
q=3
\end{cases}\implies 6=\cfrac{k}{(3-2)^3}

solve for "k", to find k or the "constant of variation"

then plug k's value back to \bf p=\cfrac{k}{(q-2)^3}

now.... what is "p" when q = 5?  well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
4 0
3 years ago
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