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Black_prince [1.1K]
3 years ago
13

A box contains ten cards labeled Q, R, S, T, U, V, W, X, Y, and Z. One card will be randomly chosen.

Mathematics
1 answer:
Triss [41]3 years ago
8 0

Answer: 2/5

Step-by-step explanation: The total number of letters available to choose from is 10. You want to know the probability of 4 of those 10. So you divide 4/10. When you simplify that fraction it becomes 2/5.

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Factorise<br><br> x^3 - 2x^2 -x+2
tankabanditka [31]

Answer:

Solution given:

x^3 - 2x^2 -x+2

take common from two each term

x²(x-2)-1(x-2)

take common again and keep left one on other bracket

<u>(x-2)(x²-1) or (x-1)(x+1)(x-2)</u> is a required answer.

note:using formula a²-b²=(a+b)(a-b) for x²-1.

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\bf n^{th}\textit{ term of an arithmetic sequence}\\\\&#10;a_n=a_1+(n-1)d\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;d=\textit{common difference}\\&#10;----------\\&#10;n=12\\&#10;d=-10\\&#10;a_{12}=13&#10;\end{cases}&#10;\\\\\\&#10;a_{12}=a_1+(12-1)d\implies 13=a_1+(12-1)(-10)&#10;\\\\\\&#10;13=a_1-110\implies \boxed{123=a_1}

\bf \\\\&#10;-------------------------------\\\\&#10;\textit{sum of a finite arithmetic sequence}\\\\&#10;S_n=\cfrac{n}{2}(a_1+a_n)\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;a_n=\textit{value of the }n^{th}\ term\\&#10;------------\\&#10;n=12\\&#10;a_1=123\\&#10;a_{12}=13&#10;\end{cases}&#10;\\\\\\&#10;S_{12}=\cfrac{12}{2}(a_1+a_{12})\implies S_{12}=\cfrac{12}{2}(123+13)

and surely you know how much that is.
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3 years ago
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