Of 60 students in a class 2/3 are girls, and 2/5 of the class are taking music lessons. What is the maximum number of girls that are not taking music lessons?

Answers

Answer 1

Answer: 24 girls. Because 60/3=20 20*2=40. So than u do 40/5=8 8*2=16 40-16=24.

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Find the value of k so that the line through the given points has slope m.(k+1, k-1), (k-k); m=k+1

Answers

Answer:

-3/2

Step-by-step explanation:

formula of slope (y2-y1/x2-x1)

Can anyone explain how to solve this 5(6-2)^2-5

Answers

$5\cdot(6-2)^2-5=5\cdot4^2-5=5\cdot16-5=80-5=75$

Expand and simplify the following expression: $(a-2b){3}$

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$(x-y)^3=x^3-3x^2y+3xy^2-y^3\n--------------------------\n(a-2b)^3=a^3-3a^2\cdot2b+3a\cdot(2b)^2-(2b)^3\n\n=a^3-6a^2b+12ab^2-8b^3$

Atlas Chemical Company wanted to keep its recruiting costs for the position of Chief Chemist to less than $100,000. The company spent $15,500 on advertising and $32,700 on interviewing expenses. They will pay 8% commission on selling the selected individual's home. What is the maximum selling price that Atlas would pay 8% commission on and stay within budget?

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647,500 is my answer :33

How do I round these numbers?

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Do your sum which is $329.69 subtract $59.63 then if the unit is 4 or less round your answer to the nearest ten eg, 103 would be rounded to 100. If the unit is 5 and above you would round your answer up eg, 437 would be rounded to 440.

Search results for "A shipment of 18 cars, some weighing 3,000 pounds, and the others weighing 5,000 pounds each. Together the shipment has a total weight of 30 tons. (60,000 lbs). Find the number of each kind of car."

Answers

The best way to solve a problem like this is to set up two equations. First assign a variable to each thing you are trying to find. In this case, it's two different kinds of cars. Let's call the cars that weigh 3,000 pounds x, and the ones that weigh 5,000 y. The two equations you should write are:

x+y=18 (because the problem tells you there were 18 cars in total)
3000x+5000y=60000 (because that is the total weight in the problem)

Next, you need to solve for one of the variables. I will solve for x first by subtracting y from both sides of the first equation.

x=18-y

Then you have to plug that into the other equation to get:

3000(18-y)+5000y=60000

Simplify and solve for y:

54000-3000y+5000y=60000
54000+2000y=60000
2000y=6000
y=3

Now that you know what y equals, you can put it into the equation we solved for x:

x=18-3
x=15

So there are 15 cars that weigh 3000 pounds and 3 that weigh 5000.

Answer:

x+y=18 (because the problem tells you there were 18 cars in total)

3000x+5000y=60000 (because that is the total weight in the problem)

Next, you need to solve for one of the variables. I will solve for x first by subtracting y from both sides of the first equation.

x=18-y

Then you have to plug that into the other equation to get:

3000(18-y)+5000y=60000

Simplify and solve for y:

54000-3000y+5000y=60000

54000+2000y=60000

2000y=6000

y=3