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nikdorinn [45]
3 years ago
9

A private college advertised that last year their freshman students, on average, had a score of 1120 on the college entrance exa

m. Assuming that average refers to the mean, which of the following claims must be true based on this information?
a. Last year, the number of their freshman students who had a score of more than 1120 on the exam was equal to the number of their freshman students who had a score of less than 1120 on the exam.
b. Last year at least one of their freshman students had a score of less than 1240 on the exam.
c. Last year some of their freshman students had a score of exactly 1120 on the exam.
d. Last year some of their freshman students had a score of at least 1120 on the exam.
e. Next year at least one of their freshman students will have a score of at least 1120 on the exam.
f. None of the above statements is true.
Mathematics
1 answer:
Ahat [919]3 years ago
3 0

Answer:. d. Last year some of their freshman students had a score of at least 1120 on the exam.

Step-by-step explanation:

If the average score is 1120, then some students scored more than 1120. Hence, those students scored at least 1120

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How many are 16 x 16 ?​
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Please answer this correctly
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4 0
2 years ago
5. Write down the nth term of each of the following G.Ps
yawa3891 [41]

Answer:

1i) nth term = 12(-b/4)^(n-1)

ii) nth term = 3(-1/9)^(n-1)

2i) Sixth term= (-3b^5)/256

ii) Eight term = -1/1594323

Question:

The complete question as found on brainly (question-10746111):

Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.

i) 12,-3b,......sixth term

ii) 3,-1/3,..........;8th term?

Step-by-step explanation:

1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.

i) 12,-3b,......;sixth term

1st term = a= 12

r = common ratio = 2nd term /1st term

r = -3b/12 = -b/4

Then we would find the nth term

See attachment for details

ii) 3,-1/3,..........;8th term

1st term = a= 3

r = common ratio = 2nd term /1st term

r = (-⅓)/3 = (-⅓)(⅓) = -1/9

Then we would find the nth term

See attachment for details

2) we are to determine the sixth term in (i) and eight term in (ii)

See attachment for details

Sixth term= (-3b^5)/256

Eight term = -1/1594323

7 0
3 years ago
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