aTime to Study Frequency Relative Frequency (%)Early Morning0% ELate MorningEarly AfternoonLate AfternoonEveningNightTotal(O10ETT1086591856SUBMIT QUESTION0 %0 %0 %0 %0 %100%SAVE PROGRESS> The responses are reported in the following table (EM= Early Morning, LM = LateMorning, EA Early Afternoon, LA = Late Afternoon, EV = Evening, NT = Night):EV NTLALMEMEVLMEMEMEVLANTNTEAEVNTLM EM NTEALM NTNT LANTEVNTEMSAMSUNGEMNTMRAMM M LA NNTEVNTEA EVHAMMERESEMESENTEV EANT NTSUBMIT QUESTIONEVLMNTLMLALMLMEA NT NT LASAVE PROGRESSSU

Answered: Time to Study Frequency Relative…

Time to Study Frequency Relative Frequency (%) Early Morning 0% = Late Morning Early Afternoon Late Afternoon Evening Night Total 10 8 6 5 9 18 56 0 % 0% 0% 0% 0% = 100%

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

a
Time to Study Frequency Relative Frequency (%)
Early Morning
0% E
Late Morning
Early Afternoon
Late Afternoon
Evening
Night
Total
(
O
10
ETT
10
8
6
5
9
18
56
SUBMIT QUESTION
0 %
0 %
0 %
0 %
0 %
100%
SAVE PROGRESS
>

The responses are reported in the following table (EM= Early Morning, LM = Late
Morning, EA Early Afternoon, LA = Late Afternoon, EV = Evening, NT = Night):
EV NT
LA
LM
EM
EV
LM
EM
EM
EV
LA
NT
NT
EA
EV
NT
LM EM NT
EA
LM NT
NT LA
NT
EV
NT
EM
SAMSUNG
EM
NT
MRAMM M LA N
NT
EV
NT
EA EV
HAMMERE
SEMESE
NT
EV EA
NT NT
SUBMIT QUESTION
EV
LM
NT
LM
LA
LM
LM
EA NT NT LA
SAVE PROGRESS
SU

Expert Solution

Step 1

$Given data is LA EV LM EM NT EV NT LM LM LA EA EM EV EV EA NT EM NT LM NT EA EV EM LM EV NT NT LA NT EM EM LA LM EA NT EM NT NT EM LM EM EV EV NT NT EV EA LM EM NT NT NT EA NT NT LA EM = Early morning LM = Late morning EA = Early Afternoon LA = Late Afternoon EV = Evening NT = Night$

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1
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Ages Number of students 15-18 4 19-22 6 23-26 9 27-30 3 31-34 4 35-38 6 Find the relative frequency for the class with lower class limit 31Relative Frequency = %Give your answer as a percent, rounded to two decimal places
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Question content area top Part 1 Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.6 degrees. Low Temperature (◦F) 40−44 45−49 50−54 55−59 60−64 Frequency 3 6 13 7 2 Part 1 The mean of the frequency distribution is enter your response here degrees. (Round to the nearest tenth as needed.) Part 2 Which of the following best describes the relationship between the computed mean and the actual mean? A. The computed mean is close to the actual mean because the difference between the means is less than 5% of the actual mean. B. The computed mean is close to the actual mean because the difference between the means is more than 5% of the actual mean. C. The computed mean is not close to the actual mean because the difference between the means is less than 5% of the actual mean. D. The computed mean is not close to the…
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(36 complete) HW Score: 83.64%, - 2.1.22 Daily Low ("F) EQuestion H Frequency 35-39 1 40-44 Daily Low Temperature (°F) Cumulative 45-49 Frequency 4 Less than 40 50-54 15 55-59 Less than 45 9. 60-64 Less than 50 65-69 1 Less than 55 Construct the cumulative frequency distribution. Less than 60 Less than 65 Less than 70 Enter your answer in the edit fields and then click Check Answer. All parts showing Clear All Check Answer 41 1,452 48 tv MAR MacBook Pro F9 F10 F8 888 F7 F6 F5 F3 F4 esc FI F2 %23 %24