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Archy [21]
3 years ago
5

Which expression is equal to 3/4

Mathematics
1 answer:
Vitek1552 [10]3 years ago
6 0
3 x 1/4= 3/4
3/3 x 4/4=1
1/4 x 3/4=3/16
4 x 1/3=4/3
The answer is A 3 x 1/4
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It is: 5.61 times 0.15 = 0.8415
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4x+12=-7y<br> -y+12=4x<br><br> PLS HELP!!!!<br> Addition/ Elimination method
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4×+7y=-12 __ -4x-y=-12 ________ 0x+6y=-24 , so you eleminated by it self 6y/6=-24/6 , you get : y=-4 and then you plug in 4x+7(-4)=-12 , that's your answer 4x-28 =-12 , you divided in both side 4x=-12+28 , 4x=16 , and then x=16/4 =4 , so x=4
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A regular six-sided die is rolled 1000 times. Use the binomial distribution to determine the standard deviation for the number o
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The standard deviation for the number of times an odd number is rolled is 15.8

<h3>How to determine the standard deviation?</h3>

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Die = regular six-sided die

n = 1000

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8 0
2 years ago
Which is the graph of f(x) = x2 - 2X + 3?
Svet_ta [14]
The graph is the one with coordinates (1,2) as the center.

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3 years ago
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Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis
Tasya [4]
Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5


A.
\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx

B.
sepearte the integrals
\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}

next one
\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}

the last one you can do yourself, it is \frac{50}{3}
the sum is \frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}


so the area under the curve is \frac{233}{3}
6 0
3 years ago
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