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Damm [24]
3 years ago
14

Rachel has 30 pounds of a mixture of candy that sells for $1.00/lb. If lollipops sell for $0.95/lb and caramel candies sell for

$1.10/lb. How many pounds of lollipops are in the mixture?
Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0
L = pounds of lollipop
c = pounds of caramel candies

0.95l + 1.10c = 30 ($0.95/lb & $1.10/lb)
l + c = 30 (mixture sells for $1/lb)
c = 30 - l

0.95l + 1.10(30 - l) = 30 (Substitution)
0.95l + 33 - 1.10l = 30
-0.15l = -3 (move terms)
l = 20lbs

Hope this helps.
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The drama club members at Millie's middle school are selling tickets to the school play. Millie wants to determine the average
lukranit [14]

Answer:

Every third member of the club in her physical educations class

Step-by-step explanation:

The reasoning behind this is because if Millie surveys her club members who are hosting the school play, she wouldn't be able to get an accurate average number of tickets sold because the club member have a bias in terms of purchasing the tickets as they run the club. Meanwhile the physical education class has a more random assessment because every third person in the class may or may not have an interest in the play the same way the club would.

3 0
3 years ago
Read 2 more answers
Which statements are true about the ordered pair (10, 5) and the system of equations?
Alexeev081 [22]

Answer:

C

Step-by-step explanation:

We have the system of equations:

\left\{\begin{array}{ll}2x-5y=-5 \\ x+2y=11\end{array}

And an ordered pair (10, 5).

In order for an ordered pair to satisfy any system of equations, the ordered pair must satisfy both equations.

So, we can eliminate choices A and B. Satisfying only one of the equations does not satisfy the system of equations.

Let’s test the ordered pair. Substituting the values into the first equation, we acquire:

2(10)-5(5)\stackrel{?}{=}-5

Evaluate:

20-25\stackrel{?}{=}-5

Evaluate:

-5\stackrel{\checkmark}{=} -5

So, our ordered pair satisfies the first equation.

Now, we must test it for the second equation. Substituting gives:

(10)+2(5)\stackrel{?}{=} 11

Evaluate:

20\neq 11

So, the ordered pair does not satisfy the second equation.

Since it does not satisfy both of the equations, the ordered pair is not a solution to the system because it makes at least one of the equations false.

Therefore, our answer is C.

3 0
3 years ago
Evan has 2 pieces of wood: one is 90 inches and the other is 72 inches.he wants to cut both pieces of wood into smaller pieces s
Gre4nikov [31]

72/2=36

90/2=45

2+2=4 pieces of wood

6 0
2 years ago
For the figure shown on the right, find the value of the variable and the measures of the angles. PLZ HELP
pychu [463]
P=69 degrees
Q=71 degrees
R=40 degrees

Explanation

All triangles equal up to 180 degrees.
First add all numbers without x.
Then add all xs.

5x+20=180
Subtract 20 from 180

5x=160
Divide 160 by 5

x=40

8 0
2 years ago
A cement walkway of uniform width has been built around an in-ground rectangular pool. the area of the walkway is 1344 square fe
Elza [17]

Answer:

6 feet.

Step-by-step explanation:

Dimensions of the Pool =80 feet long by 20 feet wide

Area of the walkway =1344 square feet.

If the width of the walkway=w

  • Length of the Larger Rectangle =80+2w
  • Width of the Larger Rectangle =20+2w

Area of the Walkway =Area of the Larger Rectangle - Area of the Pool

1344=(80+2w)(20+2w)-(80*20)\\1344=80*20+160w+40w+4w^2-80*20\\4w^2+200w=1344\\4w^2+200w-1344=0\\We factorize\\4(w^2+50w-336)=0\\4(w^2-6w+56w-336)=0\\w(w-6)+56(w-6)=0\\(w-6)(w+56)=0\\w-6=0$ or $ w+56=0\\w=6$ ft or $ w=-56$ \\Therefore, the width of the walkway , w=6 feet.

6 0
3 years ago
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