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lara [203]
3 years ago
15

1) A group of 100 students was surveyed about their interest in a new international studies program. interest was measured in te

rms of high, medium, or low. in the study,30 students responded high interest, 40 students responded medium interest, and 30 students responded low interest. what is the relative frequency of students with high interest? 2) Monthly commissions class frequencies 600 up to 800 3 800 up to 1000 7 1000 up to 1200 11 1200 up to 1400 22 1400 up to 1600 40 1600 up to 1800 24 1800 up to 2000 9 2000 up to 2200 4 What is the relative frequency of those sales persona that earned 1600 or more 30.8?
Mathematics
1 answer:
liubo4ka [24]3 years ago
7 0

Answer:

1) 0.30

2) 0.308

Step-by-step explanation:

1)

Interest   Frequency

High          30

Medium     40

Low            30

Total frequency=30+40+30=100.

The relative frequency of student with high interest can be computed by dividing the frequency of high interest to the total frequency.

R.f of high interest=30/100=0.30

2)

Monthly Commission Frequencies

600-800                  3

800-1000                  7

1000-1200                 11

1200-1400                 22

1400-1600                40

1600-1800                24

1800-2000                 9

2000-2200                 4

Total frequency=3+7+11+22+40+24+9+4=120

The relative frequency of earning 1600 or more computed by dividing the frequency of earning 1600 or more to the total frequency.

R.f of earning more than 1600=24+9+4/120=37/120=0.308

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Given r(x) = x square +2x -1 and s(x) = x-1 what is the value of r(s(3))
Amanda [17]

Answer:

r(s(3))=7

Step-by-step explanation:

we have

r(x)=x^{2} +2x-1

s(x)=x-1

we know that

The function composition is equal to

(r ∘ s )(x) = r(s(x))

In the function r(x) substitute the variable x by the function s(x)

so

r(s(x))=(x-1)^{2} +2(x-1)-1

For x=3

substitute the value of x=3 in the function composition

r(s(3))=(3-1)^{2} +2(3-1)-1

r(s(3))=(2)^{2} +2(2)-1

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3 0
3 years ago
Problem 2:
kifflom [539]

Answer:

The answers to the questions about problem 2 are:

  1. The fraction of the length after Priya stops two times is 3/4.
  2. The fraction of the length after Priya stops four times is 15/16.
  3. Priya will never reach the end of the hallway.

Step-by-step explanation:

The explanation about each answer is below:

1. In the first stop, Priya has walked half of the total length, and there would still be half the hallway, however, when she stops the second time, she has walked the half of the half, I mean:

\frac{1}{2}/2=\frac{1}{4}

If you add the two values you obtain the total distance traveled by Priya:

\frac{1}{2}+\frac{1}{4}= \frac{3}{4}

And the distance traveled in two stops is 3/4 of the total length of the hallway.

2. With four stops the system is the same, we know with two stops Priya travels 3/4 of the hallway, now with the third stop, she will travel half of the remaining distance:

\frac{1}{4}/2=\frac{1}{8}

And in the fourth stop she will travel half of the remaining, I mean:

\frac{1}{8}/2=\frac{1}{16}

Now, we add all the values of the distances obtained:

\frac{3}{4}+\frac{1}{8}+\frac{1}{16}= \frac{15}{16}

So, the distance traveled by Priya in the fourth stop is 15/16 of the total length of the hallway.

3. How we know, the numbers are infinite, in the same forms the distances, by this reason, how the problem says that Priya walks just half of the distance, she will never reach the end because despite she has very near of the end, she will continue walking just half and ever smaller distances.

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