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yanalaym [24]
2 years ago
12

Juanita is 13 years older than her cousin. The sum of their ages is no less than 103 years.

Mathematics
1 answer:
Ann [662]2 years ago
4 0
Juanita = x+13
Cousin = x

No less than is the same as greater than or equal to .  If I am wrong on this the answer is 46 instead of 45.

(x+13) + x >= 103
2x + 13 >= 103
2x >= 90
x >= 45
Juanita's cousin must be at least 45 years old.
Juanita is at least 58 years old.

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7 0
2 years ago
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Compute $14A6_{12} - 5B9_{12}$. Give your answer as a base $12$ integer.
avanturin [10]

Answer:

you need to 12 minus 5b9 as base and don’t forget to mention the integer

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2 years ago
What is the value of x in the equation 4 x plus 8 y equals 40, when y equals 0.8? 4.6 8.4 10 12What is the value of a in the equ
Mama L [17]

Answer:

1. 8.4 2. 15

Step-by-step explanation:

1. 4x+8y= 40

wen y= 0.8

4x+8×0.8=40

4x=40-8×0.8

4x= 40- 6.4= 33.6

....x= 33.6/4 = 8.4

2. 3a+b= 54

wen b= 9

3a+ 9= 54

3a= 54- 9

3a=45

.....a = 45/3 = 15

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8 0
2 years ago
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Which equations are correct? select each correct answer. 8x3(3x2)=24x6 4x3(6x2)=24x6 6b3(−2b3)=−12b6 3a3(−3a4)=−9a7?
nlexa [21]
<span>8x3(3x2)=24x6 = True
</span><span>4x3(6x2)=24x6 = True
</span><span>6b^3(−2b3)=−12b^6 = Ture
</span><span>3a^3(−3a^4)=−9a^7 = True 

They are all correct</span>
7 0
2 years ago
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7 girls audition for 12 roles in a school play. What is the probability that at least 2 of the girls audition for the same part?
Sophie [7]

Answer:

Step-by-step explanation:

This is one minus the probability that all the girls audition for different roles. The total number of ways of assigning roles to the girls is 12^7, because to each of the 7 girls, you have a choice of 12 roles.

Then if each girl is to receive a different role, then there are 12!/5! possibilities for that. If you start assigning roles to the girls, then for the first girl, there are 12 choices, but for the next you have to choose one of the 11 different ones, so 11 for the next, and then one of the 10 remaining for the next etc. etc., and this is 12*11*10*...*6 = 12!/(12-7)! =12!/5!

The probability that a random assignment of one of the 12^7 roles would happen to be one of the 12!/5! roles where each girl has a different role, is

(12!/5!)/12^7 = 12!/(12^7 5!)

Then the probability that two or more girls addition for the same part is the probability that not all the girls are assigned different roles, this is thus:

1 - 12!/(12^7 5!)

6 0
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