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algol [13]
3 years ago
8

If you have a bag that contains 24 marbles: 4 red, 12 blue, 6 black, 2 green. what is the probability of randomly picking one ma

rble that is black?
Mathematics
1 answer:
seropon [69]3 years ago
4 0

24 total marbles

6 are black

 so you have a 6/24, reduced to 1/4 probability of picking a black one.

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-3x + 22 greater than equal to -5
melamori03 [73]

Answer:

x is less than or equal to 9  

which is x ≤ 9

Step-by-step explanation:

8 0
2 years ago
Elyse uses 42 gallons of water in a 10-minute shower. Write and solve a prediction problem that uses this information
kvv77 [185]

Let

x-----> the time in minutes

y------> the number of gallons of water

we know that

the unit rates is equal to

\frac{y}{x} = \frac{42}{10}=4.2\frac{gal}{min}

the unit rate is equal to the slope of the linear equation

then

y=4.2x

<u>Problem</u>

1) How many gallons of water will Elyse use in three minutes in the shower?

For x=3 minutes

substitute in the equation above

y=4.2x

y=4.2*3

y=12.6 gal

<u>the answer is</u>

12.6 gal

7 0
3 years ago
Consider the graph shown. Which of the following does not represent the rate of change found when using similar triangles?
ycow [4]

Answer:

im doing this test too

Step-by-step explanation:

please tell me the answer

3 0
3 years ago
Read 2 more answers
Can need help again please anyone
Tom [10]

Answer:

Put one point on (0, -4)

Put the other on (3, -2)

Step-by-step explanation:

That's how graphing works

7 0
3 years ago
The position function of a particle moving along a coordinate line is given, wheresis in feet andtis in seconds.
Sati [7]

Answer:

a) s = (4-t)/(t^2+4)^2,   a(t) = (2t^3-24t)/(t^2+4)^3

b) s = 0.2ft, v = 0.12 ft/s,  a = -0.176 ft/s^2

c) t = 2s

d) slowing down for t < 2, speeding up for t > 2

e) 0.327 ft

Step-by-step explanation:

The position function of a particle is given by:

s(t)=\frac{t}{t^2+4},\ \ \ t\geq  0   (1)

a) The velocity function is the derivative, in time, of the position function:

v(t)=\frac{ds}{dt}=\frac{(1)(t^2+4)-t(2t)}{(t^2+4)^2}=\frac{4-t^2}{(t^2+4)^2}   (2)

The acceleration is the derivative of the velocity:

a(t)=\frac{dv}{dt}=\frac{(-2t)(t^2+4t)^2-(4-t^2)2(t^2+4)(2t)}{(t^2+4)^4}\\\\a(t)=\frac{(-2t)(t^2+4)-4t(4-t^2)}{(t^2+4)^3}=\frac{2t^3-24t}{(t^2+4)^3} (3)

b) For t = 1 you have:

s(1)=\frac{1}{1+4}=0.2\ ft\\\\v(1)=\frac{4-1}{(1+4)^2}=0.12\frac{ft}{s}\\\\a(1)=\frac{2-24}{(1+4)^3}=-0.176\frac{ft}{s^2}

c) The particle stops for v(t)=0. Then you equal equation (2) to zero ans solve the equation for t:

v(t)=\frac{4-t^2}{(t^2+4)^2}=0\\\\4-t^2=0\\\\t=2

For t = 2s the particle stops.

d) The second derivative evaluated in t=2 give us the concavity of the position function.

\frac{d^2s}{dt^2}=a(2)=\frac{2(2)^3-24(2)}{(2^2+4)^3}=-0.062

Then, the concavity of the position function is negative. For t=2 there is a maximum. Before t=2 the particle is slowing down and after t=2 the particle is speeding up.

e) Due to particle goes and come back. You first calculate s for t=2, then calculate for t=5.

s(2)=\frac{2}{2^2+4}=0.25\ ft

s(5)=\frac{5}{5^2+4}=0.172\ ft

The particle travels 0.25 in the first 2 seconds. In the following three second the particle comes back to the 0.172\ ft. Then, in the second trajectory the particle travels:

0.25 - 0.127 = 0.077 ft

The total distance is the sum of the distance of the two trajectories:

s_total = 0.25 ft + 0.077 ft = 0.327 ft

6 0
3 years ago
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