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DENIUS [597]
3 years ago
6

A single ball is taken at random from an urn containing 13 balls numbered 1 through 13. What is the probability of obtaining the

following?
a ball different from 1
Mathematics
1 answer:
Maslowich3 years ago
7 0
The probability of obtaining a 1 is 1/13, so the probability of not getting a 1, that is, the probability of getting a ball different from 1, is 1- (1/13)=12/13, which is 0.923
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PLEASE HELP WILL MARK BRAINLIEST!!(plz show work too)
trapecia [35]

\huge \bf༆ Answer ༄

Let the capacity of bus be x students

And van be y students, now ;

From the given statements we get two equations ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \:   \:  \:  \: \:  (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:13x + 5y = 670 \:  \:  \:  \:  \: (2)

multiply the equation (2) with 2 [ it won't change the values ]

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260 \: \:  \:  \:   \:  \:  \:  \:  \: (1)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y = 1340 \:  \:  \:  \:  \: (3)

Now, deduct equation (1) from equation (3)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:26x + 10y - 2x - 10y = 1340 - 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:24x = 1080

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 1080 \div 24

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:x = 45

Therefore each bus can carry (x) = 45 students

Now, plug the value of x in equation (1) to find y ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2x + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(2 \times 45) + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:90 + 10y = 260

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 260 - 90

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:10y = 170

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 170 \div 10

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = 17

Hence, each van can carry (y) = 17 students in total.

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Answer:

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Answer:

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multiplication? then...

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