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malfutka [58]
2 years ago
11

Suppose that 35 people are divided in a random mannerinto two teams in such a way that one team contains10 people and the other

team contains 25 people.what isthe probability that two particular people a and b will beon the same team
Mathematics
1 answer:
Alisiya [41]2 years ago
7 0

To be precise, the size of your sample space is <span><span>(<span>2410</span>)</span><span>(<span>2410</span>)</span></span>. This number does go on the bottom of the fraction, and what goes on top is the size of the event. Break up the event into independent events 1. choose the 2 defective bulbs, and 2. choose the remaining 8 bulbs. I don't have much choice in event 1. There is only one way to choose both of the defective balls. In other words, <span><span>(<span>22</span>)</span><span>(<span>22</span>)</span></span> (choosing 2 defective bulbs from a set of 2 defective bulbs). For event 2, there are <span><span>24−2=22</span><span>24−2=22</span></span> nondefective bulbs, and I must choose <span>88</span> of them, so that's <span><span>(<span>228</span>)</span><span>(<span>228</span>)</span></span>. Finally, since events 1 and 2 are independent, we multiply the answers for the combined event: <span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span></span>

<span><span>P=<span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span><span>(<span>2410</span>)</span></span></span><span>P=<span><span><span>(<span>22</span>)</span><span>(<span>228</span>)</span></span><span>(<span>2410</span>)</span></span></span></span>

Or, since <span><span><span>(<span>22</span>)</span>=1</span><span><span>(<span>22</span>)</span>=1</span></span>,

<span><span>P=<span><span>(<span>228</span>)</span><span>(<span>2410</span>)</span></span></span><span>P=<span><span>(<span>228</span>)</span><span>(<span>2410</span>)</span></span></span></span>

Hope this helps!

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Five-sixths of the boats in the marina are white, 2 5 of the remaining boats are blue, and the rest are red. If there are 12 red
Inga [223]

Answer: There are 120 boats in the marina.

Step-by-step explanation:

Let's suppose that there are N boats

We know that 5/6 of them are white, then we have: (5/6)*N white boats.

The remaining is: N - (5/6)*N = (1/6)*N

We know that 2/5 of the remaining are blue, then we have:

(2/5)*(1/6)*N blue boats

And the rest are red, the rest is:

(1/6)*N - (2/5)*(1/6)*N = (3/5)*(1/6)*N

Then we have (3/5)*(1/6)*N red boats.

And we know that there are 12 red boats, then:

(3/5)*(1/6)*N = 12

Now we can solve this for N, which is the total number of boats in the marina.

N = 12*(6)*(5/3) = 120

There are 120 boats in the marina.

4 0
2 years ago
Find the value of y when x equals 18.<br><br> x - 7y= -17
galina1969 [7]

Answer:

y = 5

Step-by-step explanation:

<u>Step 1:</u>

(18) -7y = -17

-18           -18

<u>Step 2:</u>

-7y = -35

÷-7     ÷-7

y = 5

5 0
2 years ago
Solve the system of equations algebraically
EleoNora [17]

Answer:

c

Step-by-step explanation:

4 0
3 years ago
A race car driver won a 300 mile race with a speed of 194 miles per hour. Find the drivers time.
agasfer [191]
Time = distance / speed

time = 300/194
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8 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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