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Musya8 [376]
2 years ago
7

Greg spent 3/4 of an hour each day for 2 days working on his project.Giselle spent 1/4 of an hour each day for 6 days working on

her project.Who spent more time in total working on the project?
GregB) GiselleC) They spent the same amount of time on the project.D) There is not enough information to solve the problem.
PLEASE HELP ASAP
Mathematics
2 answers:
DerKrebs [107]2 years ago
8 0

Answer:

C)They spent the same amount of time.

Step-by-step explanation:

Greg did 3/4 of an hour is 45mins for 2 days is x2 so Greg in total did 1 and 1/2 hours

Giselle did 1/4 of an hour which is 15mins for 6 days so x6 so Giselle did 1 and 1/2 hours too.

C)They spent the same amount of time on the project

Ilia_Sergeevich [38]2 years ago
5 0
Answer: They spent same amount of time on project
Explanation:
For greg--- (3/4)*2 = 6/4=3/2
for Giselle--- (1/4) *6 = 6/4=3/2
As such they spent the same amount of time on the project
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1- The Canada Urban Transit Association has reported that the average revenue per passenger trip during a given year was $1.55.
serg [7]

Answer:

0.5

0.9545

0.68268

0.4986501

Step-by-step explanation:

The Canada Urban Transit Association has reported that the average revenue per passenger trip during a given year was $1.55. If we assume a normal distribution and a standard deviation of 5 $0.20, what proportion of passenger trips produced a revenue of Source: American Public Transit Association, APTA 2009 Transit Fact Book, p. 35.

a. less than $1.55?

b. between $1.15 and $1.95? c. between $1.35 and $1.75? d. between $0.95 and $1.55?

Given that :

Mean (m) = 1.55

Standard deviation (s) = 0.20

a. less than $1.55?

P(x < 1.55)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (1.55 - 1.55) / 0.20 = 0

p(Z < 0) = 0.5 ( Z probability calculator)

b. between $1.15 and $1.95?

P(x < 1.15)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (1.15 - 1.55) / 0.20 = - 2

p(Z < - 2) = 0.02275 ( Z probability calculator)

P(x < 1.95)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (1.95 - 1.55) / 0.20 = 2

p(Z < - 2) = 0.97725 ( Z probability calculator)

0.97725 - 0.02275 = 0.9545

c. between $1.35 and $1.75?

P(x < 1.35)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (1.35 - 1.55) / 0.20 = - 1

p(Z < - 2) = 0.15866 ( Z probability calculator)

P(x < 1.75)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (1.75 - 1.55) / 0.20 = 1

p(Z < - 2) = 0.84134 ( Z probability calculator)

0.84134 - 0.15866 = 0.68268

d. between $0.95 and $1.55?

P(x < 0.95)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (0.95 - 1.55) / 0.20 = - 3

p(Z < - 3) = 0.0013499 ( Z probability calculator)

P(x < 1.55)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (1.55 - 1.55) / 0.20 = 0

p(Z < 0) = 0.5 ( Z probability calculator)

0.5 - 0.0013499 = 0.4986501

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