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3 years ago

6

A spinner has 5 equal sections numbered 1 to 5. What is the probability of the spinner stopping on a number that is a multiple o

f 2 or is less than 3?
help!!!!!!!!!plz.....

1 answer:

Kruka [31]3 years ago

4 0

Answer: 1/5, .2, and or 20%

Step-by-step explanation:

The only multiple of 2 that is less than 3 and greater than 1 is 2, so there is a 1/5, .2 and or 20% chance of this happening

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What is the answer to Y=-2x-1/2

ruslelena [56]

Answer:

Is this slope-intercept form?

Step-by-step explanation:

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3 years ago

Can you help me i Don’t understand

snow_tiger [21]

Answer:

A. {-1, 2, 7}

Step-by-step explanation:

For a set of ordered pairs, the domain is the set of first-numbers (x). The range is the set of second numbers (y, or g(x)).

The first step here is to find the g(x) values for each of the listed x-values.

The attachment shows a table.

Often, we like to list the domain and/or range values in numerical order, eliminating any duplicates. That is, the set of range values {2, -1, -1, 7} would be listed as ...

{-1, 2, 7} . . . . . . matches choice A

_____

You find the g(x) values by substituting for x and doing the arithmetic.

g(-2) = (-2)² -2 = 4 -2 = 2, for example

7 0

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.

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3 years ago

When I use 14 cat lovers to pet 350 cats, it takes 15 days. How long would it take to pet 480 cats if I use 18 cat lovers?

nika2105 [10]

I think it would take 16 days

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3 years ago

50 PTS!!! When 1250^3/4 is written in simplest radical form, which value remains under the radical?

rewona [7]

4 radical 1250^3.
The 3 stays on the inside, hope this helps.

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3 years ago