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swat32
3 years ago
6

Two sections of a class took the same quiz. Section A had 15 students who had a mean score of 80, and Section B had 20 students

who had a mean score of 90. Overall, what was the approximate mean score for all of the students on the quiz?
A. 85.7
B. 85.0
C. none of these
D. 84.3
E. It cannot be determined
Mathematics
1 answer:
Nezavi [6.7K]3 years ago
7 0

Answer: A. 85.7

Step-by-step explanation:

Given : Two sections of a class took the same quiz.

Section A had 15 students who had a mean score of 80, and Section B had 20 students who had a mean score of 90.

We know that  , \text{Mean}=\dfrac{\text{Sum of observations}}{\text{No. of observations}}

Then , for section A :

\text{Mean score }=\dfrac{\text{Sum of scores in sec A}}{\text{No. of students}}

\Rightarrow\ 80=\dfrac{\text{Sum of scores in sec A}}{15}\\\\\Rightarrow\ \text{Sum of scores in sec A}=80\times15=1200

Similarly in Section B, \text{Sum of scores in sec B}=90\times20=1800

Total scores = Sum of scores in sec A+Sum of scores in sec B

=1200+1800=3000

Total students = Students in sec A +Students in sec B

=15+20=35

Now , the mean score for all of the students on the quiz =\dfrac{\text{Total score}}{\text{Total students}}

=\dfrac{3000}{35}=85.7142857143\approx85.7

Hence, the approximate mean score for all of the students on the quiz = 85.7

Thus , the correct answer is option A. 85.7.

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JM = 9x + 6

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Answer:

# after 5 seconds, the ball strikes the ground

# The ball reaches maximum height at t = 1.5 seconds

# The max height is 196 feet

Step-by-step explanation:

The equation is:

d(t)=-16t^2+48t+160

t is the time

d(t) is the distance traveled

Initial height is 160 feet and initial velocity is 48 ft/sec

If we want to find the time it takes the ball to hit the ground, we let d(t) equal to 0 and find t. Shown below:

-16t^2+48t+160=0\\t^2-3t-10=0\\(t-5)(t+2)=0\\t=5,-2

We disregard t = -2 since time can't be negative. We take t = 5

Thus, after 5 seconds, the ball strikes the ground.

The equation is a quadratic of the form: ax^2+bx+c

Matching equations, we can say:

a = -16

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c = 160

The time when ball reaches max height is given as:

t=-\frac{b}{2a}

Substituting, we find:

t=-\frac{b}{2a}\\t=-\frac{48}{2(-16)}\\t=1.5

The ball reaches maximum height at t = 1.5 seconds

The max height can be found by putting t = 1.5 into the original equation. Shown below:

d(t)=-16t^2+48t+160\\d(1.5)=-16(1.5)^2+48(1.5)+160\\=196

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