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alexira [117]
3 years ago
14

Assume that advanced students average 93% on an achievement test and regular students averaged 75%. If 100 advanced students and

300 regular students took the test, what would you expect the average to be? How many of each kind of student would be needed to get a group of 90 students who average 87% on the test? Please include work/explaination, I along with a few of my classmates are confused by this problem. We are in a chapter about systems of equations.
Mathematics
1 answer:
Lemur [1.5K]3 years ago
7 0
Average (mean) = (sum of all the data) / (# of data)

sum of all the data = (average)(# of data)

Thus for 100 students with an average of 93,
sum of all data = (93)(100) = 9300

and for 300 students with an average of 75,
sum of all data = (75)(300) = 22500

Therefore you would expect the overall average to be
(9300 + 22500) / (100 + 300) = 79.5 %

Now if there are x # of advanced students and y # of regular students, then

x + y = 90 (total # of students) and 93x + 75y = 87(x + y) (overall average)

The second equation can be simplified to x - 2y = 0

Subtracting the two equations yields

x = 60 and y = 90

Therefore you would need 60 advanced and 30 regular students.
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valina [46]

Answer:

24 students got off at the middle school.

Step-by-step explanation:

15 students

+

12 students

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6 students

+

3 students

=

24 students (remaining students).

8 0
4 years ago
53 - 21 + 62 + (25 ÷ 5)
Alisiya [41]

Answer:

48

Step-by-step explanation:

So your equation is: 5y3 - 21 + 6y2 + (25 ÷ 5) and y = 2.

Evaluate for y=2

5(23)−21+6(22)+

25

5

5(23)−21+6(22)+

25

5

=48.

So the answer is 48.

4 0
3 years ago
(2 √-2)(5 √(3)<br> fully simplify please help and no links
mart [117]

Answer:

{ \tt{(2 \sqrt{ - 2})(5 \sqrt{3})  }} \\  \\  ={ \tt({2i \sqrt{2}) (5 \sqrt{3}) }} \\  \\  = { \tt{10i \sqrt{6} }}

5 0
2 years ago
What is the rate of change of the volume of the cylinder at that instance (in cubic kilometers per second)?
Kay [80]

Answer:

hello your question is incomplete attached below is the complete question

answer : -2400 π

Step-by-step explanation:

dr/dt = -12 km/s

h = 2.5 km

r = 40 km at some instant

hence the rate of change of Volume of cylinder at that instance

= dv / dt = d(\pi r^h ) / dt  

              = \pi h\frac{d(r^2)}{dt}  = -\pi h *2r *dr/dt

hence ; dv / dt = -2400 π

3 0
3 years ago
Let f(x)=x^2f ( x ) = x 2. Find the Riemann sum for ff on the interval [0,2][ 0 , 2 ], using 4 subintervals of equal width and t
sladkih [1.3K]

Answer:

A_L=1.75

Step-by-step explanation:

We are given:

f(x)=x^2

interval = [a,b] = [0,2]

Since n = 4 ⇒ \Delta x = \frac{b-a}{n} = \frac{2-0}{4}=\frac{1}{2}

Riemann sum is area under the function given. And it is asked to find Riemann sum for the left endpoint.

A_L= \sum\limits^{n}_{i=1}\Delta xf(x_i) = \frac{1}{2}(0^2+(\frac{1}{2})^2+1^2+(\frac{3}{2})^2)=\frac{7}{4}=1.75

Note:

If it will be asked to find right endpoint too,

A_R=\sum\limits^{n}_{i=1}\Delta xf(x_i) =\frac{1}{2}((\frac{1}{2})^2+1^2+(\frac{3}{2})^2+2^2)=\frac{15}{4}=3.75

The average of left and right endpoint Riemann sums will give approximate result of the area under f(x)=x^2 and it can be compared with the result of integral of the same function in the interval given.

So, (A_R+A_L)/2 = (1.75+3.75)/2=2.25

\int^2_0x^2dx=x^3/3|^2_0=8/3=2.67

Result are close but not same, since one is approximate and one is exact; however, by increasing sample rates (subintervals), closer result to the exact value can be found.

3 0
3 years ago
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