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Usimov [2.4K]
3 years ago
15

Five balls are numbered 1 through 5 and placed in a bowl. Josh will randomly choose a ball from the bowl, look at its number and

then put it back into the bowl. Then Josh will again randomly choose a ball from the bowl and look at its number. What is the probability that the product of the two numbers will be even and greater than 10
Mathematics
1 answer:
andrezito [222]3 years ago
4 0

Answer:

1/5

Each of the 5 balls has 5 possible combinations, as each one (1 - 5) can be paired with any of the 5. Therefore, the total number of combinations is 25.

For the product to be even, one of the balls has to be a 2 or 4. However, 2 multiplied by any number 1 - 5 is not greater than 10, so 2 is not an option. This means one of the balls has to be a 4.

There are only 3 numbers of 1 - 5 that, when multiplied by 4, are greater than 10. These are 3, 4, and 5.

This means that there are 5 combinations that work:

3 × 4 = 12

4 × 3 = 12

4 × 4 = 16

4 × 5 = 20

5 × 4 = 20

This means that 5 out of 25 combinations produce a result that is both even and greater than 10.

5/25 is then simplified to 1/5

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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.

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Robert takes a roast out of the oven when the internal temperature of the roast is 165°F. After 15 minutes, the temperature of t
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Step-by-step explanation:

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Step-by-step explanation:

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