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kow [346]
3 years ago
9

The ratio of chaperones to students on a field trip to New York city is 2to11.If there are 297 students on the trip how many Cha

perones are there
Mathematics
2 answers:
Rus_ich [418]3 years ago
8 0

Answer:

54 chaperones

Step-by-step explanation:

The basic idea behind this kind of question is that you want to keep the proportions equal. When there are 11 students, there are 2 chaperones. If there are now 297 students, how many chaperones must there be keep the proportions equal?

When answering this sort of question, you should construct an equation in which the basic ratio is equal to the proportions you're interested in. In this case:

2 chaperones / 11 students = X chaperones / 297 students

To find the value of X, you then cross-multiply and solve for X.

Cross-multiplying means that you multiply the numerator (top) on one side by the denominator (bottom) on the other, and vice versa.

Here,

2 chaperones x 297 students = 11 students x X chaperones

You then solve for X.

First, multiply the left side:

594 = 11X

Now, divide 594 / 11 to solve for X.

X = 54.

Because X stood for the number of chaperones, the answer is 54 chaperones.

boyakko [2]3 years ago
5 0

Answer:

54

Step-by-step explanation:

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slavikrds [6]

Answer:

Step-by-step explanation:

Decrease = 100 - 49 = 51

Percentage of decrease =

=\frac{51}{100}*100

= 51%

6 0
3 years ago
Write the value of the numerator 374 as the product of prime factors
ycow [4]
The prime factors of 374 are 2,11and,17
8 0
3 years ago
How would I find a? What formula would I use?
xenn [34]

Answer:

  You can use either of the following to find "a":

  • Pythagorean theorem
  • Law of Cosines

Step-by-step explanation:

It looks like you have an isosceles trapezoid with one base 12.6 ft and a height of 15 ft.

I find it reasonably convenient to find the length of x using the sine of the 70° angle:

  x = (15 ft)/sin(70°)

  x ≈ 15.96 ft

That is not what you asked, but this value is sufficiently different from what is marked on your diagram, that I thought it might be helpful.

__

Consider the diagram below. The relation between DE and AE can be written as ...

  DE/AE = tan(70°)

  AE = DE/tan(70°) = DE·tan(20°)

  AE = 15·tan(20°) ≈ 5.459554

Then the length EC is ...

  EC = AC - AE

  EC = 6.3 - DE·tan(20°) ≈ 0.840446

Now, we can find DC using the Pythagorean theorem:

  DC² = DE² + EC²

  DC = √(15² +0.840446²) ≈ 15.023527

  a ≈ 15.02 ft

_____

You can also make use of the Law of Cosines and the lengths x=AD and AC to find "a". (Do not round intermediate values from calculations.)

  DC² = AD² + AC² - 2·AD·AC·cos(A)

  a² = x² +6.3² -2·6.3x·cos(70°) ≈ 225.70635

  a = √225.70635 ≈ 15.0235 . . . feet

3 0
3 years ago
6a*2-a-15 factoring and soulutions
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6a(2)-a-15
=12a+-a+-15
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4 0
3 years ago
Someone please help me quickly i will give brainlyist
scZoUnD [109]

(16,5) it should be

16 input

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if you went to 5 ont he y-axis and keep going till it hits the line the x seems to be 16.

im deeply sorry if this is wrong.

4 0
3 years ago
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