Two ocean beaches are being affected by erosion. The table shows the width, in feet, of each beach measured at high tide where 1
995 is represented by year 0: Year number Western Beach width (in feet) Dunes Beach width (in feet) 0 100 20 5 90 45 10 80 70 11 78 75 12 76 80 15 70 95 Describe the patterns shown by the erosion data measurements shown for each of the beaches in the table. Between which years will the beaches have approximately the same width? Assuming these rates remain constant, what can you do to get a better approximation of when the two beaches will have the same width?
1 answer:
Answer:
1. Western erodes 2 ft/yr; Dunes builds up at 5 ft/yr
2. Sometime in 2006
3. Solve simultaneous equations
Step-by-step explanation:
1. Erosion patterns
(a) Western Beach
In 15 yr, Western Beach erodes from 100 ft to 70 ft.
The rate of erosion is 30 ft/15 yr = 2 ft/yr.
(b) Dunes Beach
In 15 yr, Dunes Beach builds up from 20 ft to 95 ft.
The rate of buildup is 75 ft/15 yr = 5 ft/yr.
2. Beaches with equal width
From the table, it appears that the beaches will have the same width sometime in year 11 (2006).
3. Best approximation
The graph below also shows that it happens part way through year 11 (2006).
We could get an even better solution by calculating the equations of the two lines and solving the simultaneous equations.
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Answer:
a) 0.1829
b) 0.6823
c) 0.0413
Step-by-step explanation:
We are given the following information:
We treat adult having little confidence in the newspaper as a success.
P(Adult have little confidence) = 62% = 0.62
Then the number of adults follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 10
a) exactly 5
0.1829 is the probability that exactly 5 out of 10 U.S.adults have very little confidence in newspapers.
b) atleast six
0.6823 is the probability that atleast 6 out of 10 U.S. adults have very little confidence in newspapers.
c) less than four
0.0413 is the probability that less than 4 out of 10 U.S. adults have very little confidence in newspapers.
Recall that
There are three cases to consider:
(1) When
, we have 
and 
, so
(2) When
and 
, we get 
and , so
(3) When
, we have and 
, so
So
There are three cases to consider:
(1) When
(2) When
(3) When
So
Answer:
Step-by-step explanation:
Answer: The proportion of students spending at least 2 hours on social media equals 0.7257 .
Step-by-step explanation:
Given : The typical college freshman spends an average of μ=150 minutes per day, with a standard deviation of σ=50 minutes, on social media.
The distribution of time on social media is known to be Normal.
Let x be the number of minutes spent on social media.
Then, the probability that students spending at least 2 hours (2 hours = 120 minutes as 1 hour = 60 minutes) on social media would be:
Hence, the proportion of students spending at least 2 hours on social media equals 0.7257 .