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Minchanka [31]
3 years ago
9

The bar graph below shows the attendance at school dances. The ticket per price for each dance is $5 per student. Which two danc

es together earned 37.5% of the total ticket sales. A. Fall dance,Spring Fling. B, Valentines day ball, Spring Fling. C. Fall dance,end of year bash. D. Winter snow ball, Valentines day ball

Mathematics
1 answer:
topjm [15]3 years ago
6 0

It's B because all the students that attended added together is 320 and 37.5 of 320 is 120

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Find the slope-intercept equation of the line
Natali [406]

Answer:

y=8.

Step-by-step explanation:

all the details are in the attachment.

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3 years ago
Marshall's office had already recycled 2 kilograms this year before starting the new recycling plan, and the new plan will have
Alona [7]

Answer: w x 4 + 2 = p

Step-by-step explanation:

W = Week

P = Paper

4 kg a week would be 4 x w

you also must add the 2 kilograms at the start of the year

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3 years ago
The number of problems that the math team successfully solved at this year's competition increased from last year by a factor of
Maksim231197 [3]
Earlier today I set you ten questions from this year’s International Singapore Maths Competition. Here are the questions and the answers. On the whole you did very well - smarter than a 10-year-old Singaporean! (With the caveat that they didn’t have multiple choice answers to choose from, and they are only ten). The only questions where your most popular answer the wrong one were 6 and 8. (C in Q6, and B in Q8). Thanks for taking part - now look through your workings...

For Year 5 pupils:

1. Mary cut off 2/5 of a piece of string. Later, she cut off another 14 m. The ratio of the length of string remaining to the total length cut off is 1 : 3. What is the length of the remaining string?

A. 5 m
B. 7 m
C. 10 m
D. 14 m
Solution is C. [73 per cent of readers got it right]

Oh Mary! This is how I would have solved it, using equations. Let L be the original length of the string, and R be what is remaining once you have cut the string twice. We know that R = (L x 3/5) – 14m, and that ((L x 2/5) +14) /R = 3, or 2L/5 + 14 = 3R. By substituting the first equation in the second we have 2L/5 + 14 = 9L/5 –42. Which rearranges to:7L/5 = 56, or L = 40. So R = 10m.

Interestingly, the Singapore method of solution is different. It requires us to think more visually about the string: We cut 2/5 of it. Then 14m, and are left with a piece that is a third of the size of what was cut. In other words, we are left with 1/4 of the original length. In order to compare the fractions 2/5 and then 1/4, lets change them to the lowest common denominator, which is 20. So, we cut off 8/20, subtract 14m and are left with 5/20. Let’s now draw the string divided into twentieths:

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The 14m must be 7/20 of the string, which mean each twentieth is 2m. The remaining piece of string is 5/20, i.e 10m

2. The areas of the faces of a rectangular box are 84 cm2, 70 cm2 and 30 cm2. What is the volume of the box?

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Photograph: ISMC
A. 300 cm3
B. 420 cm3
C. 490 cm3
D. 504 cm3
Solution is B. [85 per cent of readers got it right]

First we need to work out the possible side lengths, by seeing which two numbers multiply to be the area of each face. The 84 face could be 1 x 84, 2 x 42, 3 x 28, 4 x 21, 6 x 14 or 7 x 12. The 70 face could be 1 x 70, 2 x 35, 5 x 14 or 7 x 10. The 30 face could be 1 x 30, 2 x 15, 3 x 10 or 5 x 6.

The common factors between 84 and 70 are 1, 2, 7 and 14.
The common factors between 84 and 30 are 1, 2, 3 and 6.
The only way to make 84 with one each of these common factors are 14 from the top line and 6 from the bottom. So the edge bordering the 84 and 70 faces has length 14, and the edge bordering the 84 and 30 edges has length 6. Which means the height must be 30/6, or 70/14 = 5. Thus the volume is 14 x 6 x 5 = 420cm.

3. There are four numbers. If we leave out any one number, the average of the remaining three numbers will be 45, 60, 65 or 70. What is the average of all four numbers?

A. 50
B. 55
C. 60
D. 65
Solution is C. [82 per cent of readers got it right]

If the four numbers are A, B, C and D, then we know that

A + B + C = 45 × 3
A + B + D = 60 × 3
A + C + D = 65 × 3
B + C + D = 70 × 3
Now add them up to get 3A + 3B + 3C + 3D = (45 + 60 + 65 + 70) × 3

Which is A + B + C + D = (45 + 60 + 65 + 70) = 240. So their average is 240/4 = 60

6 0
3 years ago
How do u do b??????????
horrorfan [7]

the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.

the input goes to the 1st machine and the output of that goes to the 2nd machine.

a)

if she uses and input of 6 on the 2nd one, the result will be 6² - 6  = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.

b)

clearly we can never get an output  of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.

let's check something, we need a -5 on the 2nd, so

\bf \underset{final~out put}{\stackrel{y}{-5}}=x^2-6\implies 1=x^2\implies \sqrt{1}=x\implies 1=x

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

\bf \underset{first~out put}{\stackrel{y}{1}}=\sqrt{x-5}\implies 1^2=x-5\implies 1=x-5\implies 6=x

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

\bf \stackrel{first~machine}{y=\sqrt{\boxed{6}-5}}\implies y=\sqrt{1}\implies y=1 \\\\\\ \stackrel{second~machine}{y = \boxed{1}^2-6}\implies y = 1-6\implies y = -5

5 0
3 years ago
Find and interpret the mean absolute deviation of the data
Zigmanuir [339]

Answer:

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Step-by-step explanation:

5 0
3 years ago
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