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iren [92.7K]
3 years ago
13

In a group of 101 students 40 are juniors, 50 are female, and 22 are female juniors. Find the probability that a student picked

from this group at random is either a junior or female.
a) 90/101
b) 22/101
c) 68/101
d) 40/101 ​
Mathematics
2 answers:
GREYUIT [131]3 years ago
5 0
I think it might be d
Goshia [24]3 years ago
4 0

Answer:

I would say A. as a best guess but the question is odd it does not add up to 101 students it adds up to 101 plus all of them are female juniors so they would be picked every time it would be 101/101

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Please help I'm begging!!
Orlov [11]

Answer:

All you have to do is subtract 999 from both sides of the equation.

999 + k = 1,721

999 - 999 + k = 1,721 - 999

k = 722

The answer is C.) k = 722

Hope this helps you!


6 0
3 years ago
Read 2 more answers
Suppose you invest &10410 in equipment to manufacture a new board game. Each game costs $2.65 to manufacture and sells for $
lidiya [134]

Answer:

You must make 3928 and sell 600 games before the business breaks even

Step-by-step explanation:

Since $10410 was invested and cost of making one game is $2.65 then you make =10410/2.65= 3928 games

The profit gained in making one game= $20- $2.65

=$17.35

The number of games that needs to be sold to break even = 10410/17.35= 600 games

5 0
3 years ago
I need help- please help meh :')
Slav-nsk [51]

Answer:

1 equals 2.6

2 equals  1.8

Step-by-step explanation:

just divide

3 0
2 years ago
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Find the product 0.04 x 0.58
Charra [1.4K]
0.0232 is the answer lol
7 0
3 years ago
Is my answer correct, if now can you explain​
ludmilkaskok [199]

Yes. The first step is to set up the equations so that one unknown cancels out: here, the z's will cancel out:

5z+4y=-14

3z+6y=-6

-15z-12y=42

15z+30y=-30

8y=12

y=1.5

3z+6y=-6

3z+6(1.5)=-5-9

z=-14/3

6 0
3 years ago
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