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kiruha [24]
3 years ago
7

Mary had a number of cookies. after eating one, she gave half the remainder to her sister. after eating another cookie, she gave

half of what was left to her brother. mary now had only five cookies left. how many cookies did she start with
Mathematics
2 answers:
Oduvanchick [21]3 years ago
7 0
For this question, you have to work backwards, so if she ate a cookie, working backwards, you would have to add, not subtract or divide.

Mary had 5 cookies after she gave half away. So 5 is half of something. 5 x 2 = 10. She had 10 cookies before she gave half to her brother.

Before she had 10, she ate one. So you add one to the ten.(Remember, working backwards). 10+1=11.

She had 11 cookies after she gave half. So 11 x 2 = 22.
She had 22 cookies before she gave half away.
She ate one before she had 22. 22+1=23.
We are finished. Mary started with 23 cookies.

Hope you understand! :)
bonufazy [111]3 years ago
6 0
7 cookies she ate 1 in total an she gave away 1 in total and she has 5 left
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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.
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Answer:

0.5

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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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