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3 years ago

How to do tree diagrams

Mathematics

2 answers:

denpristay [2]3 years ago

7 0

In probability?
each arrow shows what % chance it is.
uhhh here’s a pic off google to help

it can be more than 2 arrows and as many times as you like.

Art [367]3 years ago

4 0

Let's take a look at a simple example, flipping a coin and then rolling a die. We might want to know the probability of getting a Head and a 4.
If we wanted, we could list all the possible outcomes:

(H,1) (H,2) (H,3) (H,4) (H,5) (H,6)
(T,1) (T,2) (T,3) (T,4) (T,5) (T,6)

Probability of getting a Head and a 4:
P(H,4) = 112

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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

2 years ago

What value of x makes this proportion true?<br> O A. 20<br> OB. 5<br> O C. 6<br> O 0.9

Archy [21]

B. 25/20 using gcf simplified is 5/4

5 0

3 years ago

Read 2 more answers

Can someone please help me with this question?

Charra [1.4K]

So what I would do is first multiply the mean by 4 and that is the total u need, 146.5, to have a mean of 36 and 5/8. Next I add the lengths given, which is 110.625, and then I would subtract it from 146.5 which is 35.875 which is 35 and 7/8. You can also check the answer by finding the mean of the 4 numbers and determining if it comes out to 36 and 5/8. The answer is C. Hope this helps!

4 0

3 years ago

there are 30 students in a math class of the 30 students 30% have an a in math 40% have a b in math how many students have a c i

stepladder [879]

the answer is 9

30 percent of 30 is 9

40 percent of 30 is 12

12+9 =21

30-21 =9

3 0

3 years ago

Read 2 more answers

bonufazy [111]

A paln1 charges $0.35 per minute with no monthly fee (not sure )

4 0

3 years ago