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vazorg [7]
3 years ago
8

Chase collects $27.34 for a fundraiser Sydney collects$9.83 more than Chase ally collects 4times as much as Sydney. how much doe

s ally collects. please explain by step by step guide me to
Mathematics
2 answers:
adelina 88 [10]3 years ago
8 0
So Chase has $27.34.
You add $27.34 to $9.83. (Which is $37.17).
That will give you the amount of how much Sydney has.
To find how much Ally has you times $37.17 by 4.
Basically, you are doing $37.17 + $37.17 + $37.17 + $37.17.
When you have finished adding them you will get the answer.
So the answer is $148.68
grigory [225]3 years ago
8 0
Chase collects $27.34 and Sydney got $9.83 more. So you add $27.34 and 9.83 given Sydney a total of $37.17. Therefore if Ally collected 4 times more than Sydney you will multiply $37.17 by 4 given Ally a total of $148.68
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In a small town 68% of the people owned television 72% on radio and 12% owned neither television nor radio (1)represent the info
Naddika [18.5K]

Answer:

1.)

The Venn diagram is in the attachment

2).

percentage of the population that owns only television== 16%

Step-by-step explanation:

The Venn diagram Is represented in the attached photo.

Let's call The total people x

X = people that own only television+ people that own only radio+ people that own both television and radio+ people that own neither the two.

X= (0.68x -y) +(y) +(0.72x-y)+(0.12x)

X= 0.68x-y+y+0.72x-y+0.12x

X= 1.52x-y

Y= 1.52x-x

Y= 0.52x

People that own both= 52%

people that own only television

= (0.68x -y)

= 0.68x-0.52x

= 0.16x

percentage of the population that owns only television= 0.16*100

percentage of the population that owns only television= 16%

4 0
3 years ago
What is the probability of rolling a number greater than or equal to 9 with the sum of two dice, given that at least one of the
olga55 [171]
Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
6 0
3 years ago
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