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sineoko [7]
3 years ago
12

Mr. Jackson took the 8th grade class on

Mathematics
2 answers:
timofeeve [1]3 years ago
4 0

Hi there! :)

Answer:

\huge\boxed{\frac{17}{7} }

Given ratio of 51 boys to 21 girls:

Write as a ratio:

51 : 21

Divide both terms by GCF, or 3:

\frac{51}{3} : \frac{21}{3}

17 : 7 or \frac{17}{7}

Therefore, the ratio is 17 boys to 7 girls, or   \boxed {\frac{17}{7}}

Firdavs [7]3 years ago
3 0

Answer:

17/7

Step-by-step explanation:

the ratio would be 51/21 then to simplify divide both numbers by 3 to get 17/7

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1) Find an equation of the tangent line at each given point on the curve.x = t2 − 4, y = t2 − 2tat (0, 0)at (−3, −1)at (−3, 3)2)
Anastasy [175]

Answer:

1) at ( 0,0) : y = x/2.   at(-3-1) : y = -1.   at(-3,3) : y = 2x +9

2) DNE ( does not exist )

Step-by-step explanation:

The general equation of tangent line

y - y1 = m( x - x1 )

attached below is the detailed solution on how i derived the answers above

8 0
3 years ago
Solve the following application problem. Be sure to show your system of equations and the steps to solve the problem.
ololo11 [35]

Answer:

  • boat: 6 mph
  • current: 2 mph

Step-by-step explanation:

The relationship between time, speed, and distance is ...

  speed = distance/time

For boat speed b and current speed c, the speed downstream is ...

  b +c = (16 mi)/(2 h) = 8 mi/h

The speed upstream is ...

  b -c = (16 mi)/(4 h) = 4 mi/h

Adding the two equations eliminates the c term:

  2b = 12 mi/h

  b = 6 mi/h . . . . . divide by 2

Solving the second equation for c, we get ...

  c = b -4 mi/h = 6 mi/h -4 mi/h = 2 mi/h

The speed of the boat in still water is 6 mi/h; the current is 2 mi/h.

6 0
2 years ago
A school attendance clerk wants to determine if there is a relationship between the number of times a student arrives to school
amid [387]
The correct answer is B because the data points are too scattered

Hope this helped :)
5 0
3 years ago
(a) Find the three-digit number that increases by 30% when each of its
Elden [556K]

Step-by-step explanation:

So a three digit number can be expressed as: 100a + 10b + c where a is the third digit, b is the second digit, and c is the first digit. Or in other words a is the hundreds place, b is the tens place, and c is the ones place. When something "increases" by 30%, it is 130% it's original value, and to calculate how much that is, you simply convert 130% to a decimal by dividing by 100, which gives you 1.30. And since all the digits are increased by 1, you have the equation:

100(a+1) + 10(b+1) + 1(c+1) = 1.30(100a+10b+c)

Distribute the multiplication on the left side:

100a+100+10b+10+c+1=1.30(100a+10b+c)

Distribute the multiplication on the right side:

100a+100+10b+10+c+1=130a+13b+1.3c

Add like terms on the left side:

100a+10b+c+111=130a+13b+1.3c

Subtract 100a, 10b, and c from both sides

111=30a+3b+0.3c

So "technically" you can just plug in any two values, and then solve for the last value, but since you have a 3 digit number, you have the restriction of a < 10, b < 10, and c < 10, and also a, b, and c, should only be integers and all have the same sign or it wouldn't be a 3 digit number.

So let's start with a, since it has the highest coefficient, well you can fit 30 into 111, 3 times without going over so that's the first value

111 = 30(3) + 3b + 0.3c

111 = 90 + 3b + 0.3c

Now subtract the 90 from both sides

21=3b+0.3c

Well 3 can fit into 21, 7 times!

21 = 3(7) + 0.3c

21 = 21 + 0.3c

subtract 21 from both sides

0 = 0.3c

and now obviously c is 0, if you want you can divide both sides by 0.3 but it's a bit redundant

c = 0

This gives you the three values, a=3, b=7, c=0. which is the number 370. Now let's double check. Adding 1 to each digit would give you 481 and 481/370 = 1.3, so it is correct!

part b:

So to prove there is no three digit number, is to realize there is no solution, given the restriction or integers, greater than or equal to 0, and less than 10, and all of them must have the same sign.

So let's start with the same equation except this time instead of 1.3 it's 1.4

100(a+1) + 10(b+1) + 1(c+1) = 1.40(100a+10b+c)

Distribute on the left side;

100a+100+10b+10+c+1=1.40(100a+10b+c)

Distribute on the right side:

100a+100+10b+10+c+1=140a + 14b + 1.4c

Add like terms on left side:

100a + 10b + c + 111 = 140a + 14b + 1.4c

Subtract 100a, 10b, and c from both sides:

111 = 40a + 4b + 0.4c

Now to do the same process, let's start by finding how many times we can fit 40 into 111, and if you're wondering why we start with 40, it's because let's say for example I just say, I can fit another 40 into it, but I decide not to, and let b do that, well even if it's just 40, b will have be at least 10, which does not fit our restrictions, so you have to fit as many 40's into the number first then go the other numbers.

So only 2 40's can fit in 111 without going over the value

111 = 40(2) + 4b + 0.4c

Subtract 80 from both sides

31=4b+0.4c

4 can fit into 31, 7 times

31 = 4(7) + 0.4c

31 = 28 + 0.4c

subtract 28 from both sides

3 = 0.4c

divide both sides by 0.4

7.5 = c.

Since c is not an integer there is no 3 digit number that exists that increases by 40% whenever you increase it by 1.

7 0
2 years ago
Which of the following sets of ordered pairs represents a function?
Mila [183]

{(-1,3),(-1,4),(-1,5),(-1,6)} is the set from the given question which is a set of ordered pairs representing a function.

<h3>What is ordered pair?</h3>

An ordered pair (a, b) in mathematics is a group of two things. The pair's order of objects matters because the ordered pair (a, b) differs from the ordered pair (b, a) unless a = b. (By contrast, an unordered pair of a and b equals an unordered pair of b and a.)

Ordered pairs are also known as 2-tuples, or sequences (or, in computer science, occasionally, lists) of length 2. Sometimes referred to as 2-dimensional vectors, ordered pairs of scalars. Technically speaking, this is a misuse of the term because an ordered pair need not be a component of a vector space. An ordered pair's entries may be other ordered pairs, allowing for the recursive definition of ordered n-tuples (ordered lists of n objects).

Learn more about ordered pairs

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3 0
1 year ago
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