2(x-3)=7x-31
2x-6=7x-31
-7x -7x
----------------
-5x-6=-31
+6 +6
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-5x=-25
Divide by -5
x=5
3x-2(x+5)=-3+18
15
3x-2(x+5)=15
3x-2x-10=15
x-10=15
+10 +10
--------------
x=25
Procedure:
1) calculate the number of diferent teams of four members that can be formed (with the ten persons)
2) calculate the number of teams tha meet the specification (two girls and two boys)
3) Divide the positive events by the total number of events: this is the result of 2) by the result in 1)
Solution
1) the number of teams of four members that can be formed are:
10*9*8*7 / (4*3*2*1) = 210
2) Number of different teams with 2 boys and 2 girls = ways of chosing 2 boys * ways of chosing 2 girls
Ways of chosing 2 boys = 6*5/2 = 15
Ways of chosing 2 girls = 4*3/2 = 6
Number of different teams with 2 boys and 2 girls = 15 * 6 = 90
3) probability of choosing one of the 90 teams formed by 2 boys and 2 girls:
90/210 = 3/7
Andrew can buy maximum 3 meals this weekend.
From given question,
Andrew must spend less than 53$ on meals during the weekend.
He has already spent 21$ on meals costing 8$ average.
Let x, the number of meals
So, we get an inequality,
8x + 21 < 53
We need to find the number of meals he can buy this weekend.
From above inequality,
⇒ 8x + 21 < 53
⇒ 8x < 53 - 21
⇒ 8x < 32
⇒ x < 4
This means, from 1 to 3 meals.
Therefore, Andrew can buy maximum 3 meals this weekend.
Learn more about an inequality here:
brainly.com/question/19003099
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