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3 years ago

7

At a fair, each person can spin two wheels of chance. The fist wheel has numbers 1,2, and 3. The second wheel has the letters A

and B. (A) list all the possible outcomes of the computed event. (B) if you spin both wheels, what is the probability that you get either a 1 or an A? Explain.

2 answers:

AlexFokin [52]3 years ago

7 0

The answer to Part (B) is 4/6

nydimaria [60]3 years ago

5 0

(a) There are 3 numbers on the fist well and 2 letters on the second, so number of possible outcomes = 3*2 = 6 possible outcomes

The possible outcomes are 1A, 2A, 3A, 1B, 2B, 3B

So in total there are 6 possible outcomes as shown above if a person can spin the two wheels

(b) From possible outcomes A or 1 can appear four times namely: 1A,2A,3A and 1B

So probability is number of successes/ total number = 4/6

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Answer:

El agente debe disparar a un ángulo aproximado de 87.7° con respecto a la horizontal.

Step-by-step explanation:

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El agente debe disparar a un ángulo aproximado de 87.7° con respecto a la horizontal.

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2 years ago

What’s the axis of symmetry?

Sloan [31]

Answer:

x = -1

Step-by-step explanation:

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The vertex is the part of the line where it "turns" around and starts going the other way. So vertex of this line is at (-1, 3).

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3 years ago

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

$V = \frac{1}{3}\pi r^{2}h$

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

3 0

2 years ago

Help on this one as well ii manged to get an answer but If you understand it or im wrong an you please explain ! thank you !!

AnnZ [28]

Almost got it!

x + 3 = 3(y + 2)/2 [Multiplied both sides by 3]
so x + 3 = (3y + 6)/2
2(x + 3) = 3y + 6 [Multiplied both sides by 2]
2x + 6 = 3y + 6
2x = 3y [Subtracted 6 from both sides]
x = 3y/2 [Divided both sides by 2]
x/3 = y/2 [Divided both sides by 3]

So you wrote y/3 instead of y/2
Hope this helped!

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2 years ago

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120 sixth graders are going on a field trip to the Rock and Roll Hall of Fame and Museum. The museum requests a ratio of student

NARA [144]

Answer: 15

Step-by-step explanation:

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