Kieran’s wallet contains 7 bills: three $5 bills, two $10 bills, one $20 bill, and one $100 bill. if kieran pulls out two bills
2 answers:
Answer:
Step-by-step explanation:
Total number of bills in the wallet = 7
Total number of $5 bills in the wallet = 3
Total number of $10 bills in the wallet = 2
Total number of $20 bill in the wallet = 1
Total number of $100 bill in the wallet = 1
We need to find the probability that he selects exactly two $5 bills.
Number of favorable outcomes = 3
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Answer:
The population of the small community, 5 years into the future, after the initial 20-year period = 4268.
Step-by-step explanation:
t | 0 | 5 | 10 | 15 | 20
p | 100 | 200 | 450 | 950 | 2000
The exponential function will look like
p = aeᵏᵗ
where a and k are constants.
Take the natural logarithms of both sides
In p = In aeᵏᵗ
In p = In a + In eᵏᵗ
In p = In a + kt
In p = kt + In a.
We then use linear regression to fit the data of In p against t to obtain k and In a.
t | 0 | 5 | 10 | 15 | 20
p | 100 | 200 | 450 | 950 | 2000
In p | 4.605 | 5.298 | 6.109 | 6.856 | 7.601
In p = kt + In a.
y = mx + b
m = k and b = In a
Performing a linear regression analysis on the now-linear relationship between In p and t and also plotting a graph of the variables.
The regression equation obtained is
y = 0.151x + 4.584
The first attached image shows the equations necessary for the estimation of the linear regression parameters.
The second attached image shows the use of regression calculator and the plot of the function In p versus t.
Comparing
y = 0.151x + 4.584
With
In p = kt + In a.
y = In p
k = 0.151
x = t
In a = 4.584
a = 97.905
The exponential function relating p and t,
p = aeᵏᵗ now becomes
p = 97.905 e⁰•¹⁵¹ᵗ
So, to predict the population 5 years into the future, that is 5 years after the 20 year period.
we need p at t=25 years.
0.151 × 25 = 3.775
p(t=25) = 97.905 e³•⁷⁷⁵ = 4268.41 = 4268.
Hope this Helps!!!
Answer:
(A) 0.15625
(B) 0.1875
(C) Can't be computed
Step-by-step explanation:
We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.
Let X = Amount of time taken by student to complete a statistics quiz
So, X ~ U(32 , 64)
The PDF of uniform distribution is given by;
f(X) = , a < X < b where a = 32 and b = 64
The CDF of Uniform distribution is P(X <= x) =
(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)
P(X > 59) = 1 - P(X <= 59) = 1 - = 1 -
=
= 0.15625
(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)
P(X <= 43) = =
= 0.34375
P(X < 37) = =
= 0.15625
P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875
(C) Probability that student complete the quiz in exactly 44.74 minutes
= P(X = 44.74)
The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.