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lianna [129]
3 years ago
15

A box contains10 pens, 5 of the pens are red, 4 are balck and 1 are bluem What is the probability of pulling out a black pen fir

st, putting it back into the box, and then pulling a red pen ?
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
6 0
Pulling out a black pen first will grant you a 40% chance of that occurring, then when you put it back the probability of you pulling out a red pen would be 50% so in decimal form these two values would be .40 and .50 multiply these two together you should get .20 or a 20% chance. Also a written fraction would be 1/5, all of which are valid answers.
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More than 450 students traveled to a state park for a field trip. The school allowed 6students to travel by car, and the rest tr
ki77a [65]

Answer:


Step-by-step explanation:

Let b= number of people per bus

 

10b + 5 > 450

 

10b > 445

b > 44.5

Minimum per bus is 45 students

6 0
3 years ago
Ramon plays a game which his older sister he first gains 5 points then loses 8 points and finally gains 10 points what is his fi
vladimir1956 [14]

Answer:

7 points

Step-by-step explanation:

started out w/ 5 points, then loses 8

5 - 8 = -3

gains 10 points

-3 + 10 = 7

3 0
3 years ago
The number of donuts produced by a machine is directly proportional to the time the machine runs . The machine produces 380 donu
mrs_skeptik [129]

Answer:

855 donuts

Step-by-step explanation:

380/20 = 19 donuts per minute

19 × 45 = 855

5 0
3 years ago
Eight rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other ro
Brums [2.3K]

Answer:

9.11*10^-4 %

Step-by-step explanation:

To find the probability, you simply need to find the possible outcomes that allows no rooks to be in danger, and the possible amount of ways to place the rooks.

For the first outcome, you start by putting 1 rook in the first columns, you have 8 possible rows to do this. The next rook in the next column will only have 7 possible rows, as you have to exclude the one where the previous rook is located. The next rook, 6 possibilities, the next 5, and so on. So we conclude that the total amount of ways so that none of the rooks can capture any of the other rooks is 8*7*6*5*4*3*2*1 = 8! = 40320

In order to find the total amount of ways to place the rooks, you can just use a combinatoric:

\left[\begin{array}{ccc}64\\8\end{array}\right]= \frac{64!}{8!(64-8)!} = 4.43*10^9

Then:

P = \frac{40320}{4.43*10^9}*100\%=9.11*10^{-4}\%

5 0
3 years ago
What is the sum of the first five terms of a geometric series with a1 =20 and r =1/4?
drek231 [11]
The sum of the first n terms in a geometric sequence given the first term (a1) and the common ratio (r) is calculated through the equation,

                            <span>Sn </span>= (<span><span><span>a1</span>(1−<span>r^n</span>) / (</span><span>1−r)

Substituting the known terms,
                           Sn = (20)(1 - (1/4)^4)) / (1 - 1/4)
                           Sn = 26.5625

Thus, the sum of the first four terms is 26.5625. </span></span>
5 0
3 years ago
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