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ryzh [129]
3 years ago
9

an arts academy requires there to be 3 teachers for every 51 students and 5 tutors for every 45 students. how many students does

the academy have per teacher? per tutor? how many tutors does the academy need if it has 90 students? The academy has ___ students per tutor
Mathematics
1 answer:
padilas [110]3 years ago
6 0

Answer:

17 students per teacher, 9 students per tutor. You need 10 tutors if the school had 90 students.

Step-by-step explanation:

51 students divided by 3 teachers = 17 students per teacher.

45 students divided by 5 tutors = 9 students per teacher.

If there are 9 students per tutor 9 * 10 = 90 so ten tutors

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3 years ago
Dose someone know how to figure this out can someone pleas help its a test
riadik2000 [5.3K]

Complete question is;

The levels of mercury in two different bodies of water are rising. In one body of water the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Which equation can be used to find y, the year in which both bodies of water have the same amount of mercury?

A) 0.05 – 0.1y = 0.12 – 0.06y

B) 0.05y + 0.1 = 0.12y + 0.06

C) 0.05 + 0.1y = 0.12 + 0.06y

D) 0.05y – 0.1 = 0.12y – 0.06

Answer:

Option C: 0.05 + 0.1y = 0.12 + 0.06y

Step-by-step explanation:

In the first body, the initial measure of mercury is 0.05 parts per billion (ppb) while it's rising at a rate of 0.1 ppb each year. We are told to use y for the number of years.

Thus, amount of mercury for y years in this first body is;

A1 = 0.05 + 0.1y

Now, for the second body, we are told that;the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Thus, amount of mercury for y years in this second body is;

A2 = 0.12 + 0.06y

Since we want to find the year in which both bodies of water have the same amount of mercury. Thus, it means;

A1 = A2

Thus;

0.05 + 0.1y = 0.12 + 0.06y

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