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4 years ago

Find the quotient 5/18 divided by 2/6

Mathematics

1 answer:

Gnesinka [82]4 years ago

7 0

5/18 diveded by 2/6 is 5/18*6/2 then its 5/18 * 3 which gives 15/18 or 5/6

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What's the slope of x + 2y = 12

topjm [15]

Answer:-1 over 2.

Step-by-step explanation:

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3 years ago

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Solve for y... put 2y-4x=8 into slope intercept form : y=mx+b

Arisa [49]

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4 years ago

"Reduce Fractions to lowest terms"<br><br> 6 divided by 3.66

frozen [14]

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Step-by-step explanation:

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3 years ago

Need help pleaseI was bad at math in school so lwant to learn

aleksklad [387]

The probability of an event is expressed as

$Pr(\text{event) =}\frac{Total\text{ number of favourable/desired outcome}}{Tota\text{l number of possible outcome}}$

Given:

$\begin{gathered} \text{Red}\Rightarrow2 \\ \text{Green}\Rightarrow3 \\ \text{Blue}\Rightarrow2 \\ \Rightarrow Total\text{ number of balls = 2+3+2=7 balls} \end{gathered}$

The probability of drwing two blue balls one after the other is expressed as

$Pr(\text{blue)}\times Pr(blue)$

For the first draw:

$\begin{gathered} Pr(\text{blue) = }\frac{number\text{ of blue balls}}{total\text{ number of balls}} \\ =\frac{2}{7} \end{gathered}$

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).

Thus,

$\begin{gathered} Pr(\text{blue)}=\frac{number\text{ of blue balls left}}{total\text{ number of balls left}} \\ =\frac{1}{6} \end{gathered}$

The probability of drawing two blue balls one after the other is evaluted as

$\begin{gathered} \frac{1}{6}\times\frac{2}{7} \\ =\frac{1}{21} \end{gathered}$

The probablity that none of the balls drawn is blue is evaluted as

$\begin{gathered} 1-\frac{1}{21} \\ =\frac{20}{21} \end{gathered}$

Hence, the probablity that none of the balls drawn is blue is evaluted as

$\frac{20}{21}$

8 0

1 year ago

Find r(t) if r'(t) = 5t4i + 6t5j + t k and r(1) = i + j.

Vladimir [108]

$\mathbf r(t)=\displaystyle\int(5t^4\,\mathbf i+6t^5\,\mathbf j+t\,\mathbf k)\,\mathrm dt$
$\mathbf r(t)=(t^5+C_1)\,\mathbf i+(t^6+C_2)\,\mathbf j+\left(\dfrac12t^2+C_3\right)\,\mathbf k$

Given $\mathbf r(1)=\mathbf i+\mathbf j$ , we have

$\begin{cases}1^5+C_1=1\\\\1^6+C_2=1\\\\\dfrac121^2+C_3=0\end{cases}\implies C_1=0,C_2=0,C_3=-\dfrac12$

so that the particular solution is

$\mathbf r(t)=t^5\,\mathbf i+t^6\,\mathbf j+\dfrac12(t^2-1)\,\mathbf k$

8 0

3 years ago