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oksano4ka [1.4K]
4 years ago
14

What is the probability of rolling an even number on a six sided die and then rolling 4?

Mathematics
1 answer:
Ivahew [28]4 years ago
3 0

Answer:

B. 1/12.

Step-by-step explanation:

There are 3 even numbers on the die so Probability(an even) = 3/6 = 1/2.

Probability (rolling a 4) = 1/6.

Probability ( rolling an even then a 4) = 1/2 * 1/6 = 1/12.

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grin007 [14]
Let's set up this problem.
k+7=34
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k=27
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3 years ago
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A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$$

$$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$$

$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}

$$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$$

$$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$$

$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square

8 0
2 years ago
Help me out here real quick please
kirill115 [55]

Answer:

25

Step-by-step explanation:

10÷1/4=25

3 0
3 years ago
Need help ASAP <br> Thanks+ BRAINLIST only for correct answers (the answer negative is wrong)
saw5 [17]
None those two don’t correlate so I’m pretty sure it’s none
3 0
3 years ago
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Two years of local Internet service costs $685,including the installation fee of $85.What is the monthly fee?
hjlf

Answer:

$25 per month

Step-by-step explanation:

$685 - $85 (installation fee) = $600

$600 / 24 (12 months in a year x2 for 2 years)

$25 per month

7 0
3 years ago
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