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Sergeeva-Olga [200]
2 years ago
5

An arts academy requires there to be 6 teachers for every 144 students and 5 tutors for every 40 students. How many students doe

s the academy have per teacher? Per tutor? How many tutors does the academy need if it has 72 students
Mathematics
2 answers:
sleet_krkn [62]2 years ago
7 0

Answer:

3 teachers and 9 tutors

Step-by-step explanation:

Greeley [361]2 years ago
7 0

Answer:

There are 24 students per teacher and 8 students per tutor. If the academy has 72 students, then 9 tutors would be needed.

Step-by-step explanation:

We will start with the question, how many students does the academy have per teacher. To find this, we need to set up the equation. 144 will be the number of students, 6 will be the number of teachers, and x will be the number of students per teacher.

144/6=x

24=x

So, there are 24 students for every teacher.

We could also do this as a ratio. We would do the same thing. We would divide by six to get our answer.

144:6

Make the six one.

24:1

This shows that there are 24 students for every teacher.

Now to find out how many per tutor.

We will do the same thing as before, minus the ratio, though you could still use the ratio. Remember, 40 is the number of students, 5 is the number of tutors, and x is the number of students per tutor.

40\5=x

8=x

So there are 8 students per tutor.

Now we will use this knowledge to solve for how many tutors are needed if the academy has 72 students.

We will set up our equation. 72 will be the students, we will use eight since that is how many students are needed per tutor, and x will be the number of tutors needed.

72/8=x

9=x

So this means that if the academy has 72 students, then it will need 9 tutors.

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monitta

Answer:

P * 300000 = 30

.01%

Step-by-step explanation:

Let P equal the percent.  Of means multiply and is means equals

We need to make sure the units are the same.  I will convert km to cm

3 km *1000m/1km  * 100cm/1m =300000 cm

Now we can write the equation

P * 300000 = 30

Divide each side by 300000

P 300000/300000 = 30/300000

P =.0001

Changing this from decimal to percent

P = .0001*100 = .01 %

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3 years ago
What are the domain and range
natta225 [31]

For this case, the first thing we must do is observe that the graph has two asymptotes.

A horizontal asymptote or equivalently:

y = 0

A vertical asymptote or equivalently:

x = 0

Therefore, the domain and the range of the function is the same:

All reals excluding zero.

Answer:

The domain and range is the mime:

All reals excluding zero.

3 0
3 years ago
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Answer:

ok

Step-by-step explanation:

7 0
3 years ago
A scuba diver diving with an 80 cubic foot tank of air, needs to resurface when his air in the
Burka [1]

At sea level atmospheric pressure is 1 bar absolute (1 standard atmosphere =101 kPa=1.013 bars). The weight of the atmosphere exerts a pressure which will support a column of water 10 m high; 10 m under water the pressure on a diver is 200 kPa. The volume of gas in an early diving bell full of air at sea level is halved at 10 m according to Boyle’s law; at 20 m pressure is 300 kPa absolute and the gas is compressed into one third the volume.

Dry air is composed of roughly 21% oxygen, 78% nitrogen, and 1% other gases. According to Dalton’s law the partial pressure of oxygen at any depth will be 21% of the total pressure exerted by the air and the partial pressure of nitrogen will be 78% of total pressure.

Gases dissolve in the liquid with which they are in contact. Nitrogen is fat soluble and at sea level we have several litres dissolved in our bodies. If the partial pressure of nitrogen is doubled (by breathing air at 10 m depth) for long enough for equilibration to take place we will contain twice as many dissolved nitrogen molecules as at sea level.

6 0
3 years ago
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
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