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Virty [35]
3 years ago
13

PLS HELP I NEED ANSWER ASAP! PPLS BE RIGHT!

Mathematics
2 answers:
seraphim [82]3 years ago
6 0
There Were 6 Vans And 3 Buses.
6 x 13 = 78
3 x 40 = 120
120 + 78 = 198
Luba_88 [7]3 years ago
4 0

Answer:

They used 6 vans and 3 buses

Step-by-step explanation:

x + y= 9

13x + 40y= 198

x= 9 - y

13(9 - y) + 40y= 198

117 - 13y + 40y= 198

27y= 81

y= 3

9-3= 6

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a baseball pitcher won 80% of the games he played. if he pitched 35 games, how many games did he win?
andrey2020 [161]

Answer:

28 games

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Translate verbal expression to an algebraic expression<br><br> 8 times a number x is subtracted by 4
slamgirl [31]

Answer:

8x - 4

Step-by-step explanation:

8x - 4

3 0
4 years ago
The ratio of red marbles to blue marbles is 3:5. If there are 360 marbles total in the collection, how many of each color are th
kari74 [83]

Answer:

135 red and 225 blue

Step-by-step explanation:

8 0
3 years ago
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A person in a casino decides to play 3 games of blackjack. let w denote a win and l denote a loss. define the event a as "the pe
Ann [662]

Answer: The answer is {(w, l, l), (w, w, l), (w, l, w), (w, w, w), (l, l, w), (l, w, l), (l, w, w)}.



Step-by-step explanation: Given that a person in the casino is going to play 3 games of blackjack. Here, 'w' denotes a win and 'l' denotes a loss.  

Also, let 'A' be the event that the person wins at least one game of blackjack. Then, the outcomes of Event 'A' are

(i) win in 1st game     loss in 2nd game       loss in 3rd game

(ii) win in 1st game     win in 2nd game        loss in 3rd game

(iii) win in 1st game     loss in 2nd game       win in 3rd game

(iv) win in 1st game     win in 2nd game        win in 3rd game

(v) loss in 1st game     loss in 2nd game       win in 3rd game

(vi) loss in 1st game     win in 2nd game        loss in 3rd game

(vii) loss in 1st game     win in 2nd game       win in 3rd game

Thus, the outcomes of the event A are (w, l, l), (w, w, l), (w, l, w), (w, w, w), (l, l, w), (l, w, l) and (l, w, w).

Thus, A = {(w, l, l), (w, w, l), (w, l, w), (w, w, w), (l, l, w), (l, w, l), (l, w, w)}.


5 0
3 years ago
Dawn comma Sergio comma Tyrone comma and Jim have all been invited to a dinner party. They arrive randomly and each person arriv
professor190 [17]

Answer:

a. 24

b. 2

c. 0.0833 = 8.33%

Step-by-step explanation:

a.

The first "slot" of person to arrive has 4 possibilities, then the second "slot" will have 3 possibilities, as one has already arrived, then the third "slot" has 2 possibilities, and the fourth "slot" has just 1 possibility.

So, multiplying all these combinations, we have 4*3*2*1 = 24 possible ways they can arrive

b.

If the first and the last person are already "locked", we just have possibilities for the second and third person. The second will have 2 possibilities (Sergio or Tyrone), and the third will have only 1 (the person that wasn't the second between Sergio and Tyrone). So, the number of possibilities is 2*1 = 2

c.

If we have 2 cases where Dawn is first and Jim is last, from a total of 24 possible cases, the probability is 2/24 = 1/12 = 0.0833 = 8.33%

3 0
3 years ago
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