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BabaBlast [244]
3 years ago
8

nine people are to be distributed among three committees of two, three, four members and a chairperson is to be selected for eac

h. How many ways can this be done? a) select the chair of the two-person committee b) select the chair of the three- person committee c) select the chair of the four- person committee
Mathematics
1 answer:
Sonja [21]3 years ago
8 0

b is the answer to your promblem

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A collection of coins consists of nickels, dimes, and quarters. There are three fewer quarters than nickels and six more dimes t
weeeeeb [17]

Answer:

You have 15 dimes. You have 9 quarters. You have 12 nickels

Step-by-step explanation:

lets set some variables:

let "n" = the number of nickels

let "d" = the number of dimes

let "q" = the number of quarters

So, the total amount of money you have should be: $4.35 = 0.25q + 0.10d + 0.05n

Now let's look at the relationships between the coins:

"There are three fewer quarters than nickels": n - 3 = q

"six more dimes than quarters": q + 6 = d

So now you have three equations with three variables, all you need to do is solve.

\left \{ {{4.35 = 0.25q + 0.10d + 0.05n} \atop {n - 3 = q}} \atop {q + 6 = d}}\right.

first, you can substitute "n-3" for "q" (according to the 2nd equation) in the 1st and 3rd equation, you get:

\left \{ {{4.35=0.25(n-3)+0.10d+0.05n} \atop {(n-3)+6=d}} \right.

You now only have two equations and two variables.

Simplify:

\left \{ {{4.35=0.25n-0.75+0.10d+0.05n} \atop {n+3=d}} \right.

\left \{ {{4.35=0.30n-0.75+0.10d} \atop {n+3=d}} \right.

Now substitute "n+3" for "d" (according to the 2nd equation) in the 1st equation:

4.35=0.30n-0.75+0.10(n+3)

simplify:

4.35=0.30n-0.75+0.10n+0.30

4.35=0.40n-0.45

4.35+0.45=0.40n

4.80=0.40n

n=12

You have 12 nickels. Now sub "n" back into your equations to find the number of dimes and quarters:

n - 3 = q

12 - 3 = q

q = 9

You have 9 quarters.

q + 6 = d

9 + 6 = d

d = 15

You have 15 dimes.

3 0
3 years ago
If
Rudiy27
Looking at this problem in terms of geometry makes it easier than trying to think of it algebraically.

If you want the largest possible x+y, it's equivalent to finding a rectangle with width x and length y that has the largest perimeter.

If you want the smallest possible x+y, it's equivalent to finding the rectangle with the smallest perimeter.

However, the area x*y must be constant and = 100.

We know that a square has the smallest perimeter to area ratio. This means that the smallest perimeter rectangle with area 100 is a square with side length 10. For this square, x+y = 20.

We also know that the further the rectangle stretches, the larger its perimeter to area ratio becomes. This means that a rectangle with side lengths 100 and 1 with an area of 100 has the largest perimeter. For this rectangle, x+y = 101.

So, the difference between the max and min values of x+y = 101 - 20 = 81.
6 0
3 years ago
kai calculates that he spends 15% of a school day in science class. If he spends 75 minutes in science class, how many minutes d
jasenka [17]
Im not sure but I would think you would divide 75 by .15. That would get you 500. So 500 minutes would be your answer.
5 0
3 years ago
Determine whether the function f(x) = -6x^2 - 5 is a quadratic function<br> A) Yes<br> B) No
Yuki888 [10]
The answer is no!!!!!!!
6 0
3 years ago
How many whole numbers have a sum of 40
Olenka [21]
In a positive integers there are twenty whole numbers I hope this help
7 0
3 years ago
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