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EastWind [94]
3 years ago
14

The table shows the outcome of rolling a number cube (labeled 1 through 6) 200 times.

Mathematics
2 answers:
dybincka [34]3 years ago
8 0

From the table, the  experimental Probability of rolling a 1 = 28/200 = 7/50

So for 25 rolls we would expect 7 * 25 / 50  =  3.5  

answer:   4 times

denis23 [38]3 years ago
5 0

I'm pretty sure it would be 4

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Answer:

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Step-by-step explanation:

10=7-m

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Please do #4 It’s it the one on the bottom
spin [16.1K]

We need to graph this equation:

16x+2y=300

Its solutions are the points through which it graph passes. Since it's a linear equation its graph is a straight line and we only need two of its points to draw it. But before graphing let's re-write the equation. We can substract 16x from both sides:

\begin{gathered} 16x+2y=300 \\ 16x+2y-16x=300-16x \\ 2y=300-16x \end{gathered}

And we divide both sides by 2:

\begin{gathered} \frac{2y}{2}=\frac{300-16x}{2}=\frac{300}{2}-\frac{16x}{2} \\ y=150-8x \end{gathered}

So now with this equation if we pick two random x values we'll get their corresponding y values. This way we'll find two points that are part of the graph which is the line that passes through both. We can begin with x=0:

y=150-8\cdot0=150

So the first point is (0,150). Then we can take x=10 and we get:

y=150-8\cdot10=150-80=70

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X=2y+7 3x−2y=3 solve by substitution
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Step-by-step explanation:

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Answer:

The two roots of the quadratic equation are

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Step-by-step explanation:

Original quadratic equation is 9x^{2}-3x-2=0

Divide both sides by 9:

x^{2} - \frac{x}{3} - \frac{2}{9}=0

Add \frac{2}{9} to both sides to get rid of the constant on the LHS

x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}  ==> x^{2} - \frac{x}{3}=\frac{2}{9}

Add \frac{1}{36}  to both sides

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}

This simplifies to

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = \frac{1}{6}\right) we can see that

\left(x - \frac{1}{6}\right)^2 = x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}

So

\left(x - \frac{1}{6}\right)^2=\frac{1}{4}

Taking square roots on both sides

\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}

So the two roots or solutions of the equation are

x - \frac{1}{6}=-\sqrt{\frac{1}{4}}  and x - \frac{1}{6}=\sqrt{\frac{1}{4}}

\sqrt{\frac{1}{4}} = \frac{1}{2}

So the two roots are

x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}

and

x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}

7 0
1 year ago
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