Answer:
12:6 or 6:12 cant remember but i thinks its the first one lol
Step-by-step explanation:
a) The probability that a new municipal bond issued by a city will receive an A rating is 0.625 or 62.5%.
b) 56% of municipal bonds are issued by cities.
c) The proportion of municipal bonds issued by suburbs is 0.325 or 32.5%.
Let's solve each part of the problem:
a. If a new municipal bond is to be issued by a city, what is the probability that it will receive an A rating?
Use conditional probability here.
Using conditional probability notation, we have:
P(A | City)
To calculate this, use the following formula:
P(A | City) = P(A and City) / P(City)
We are given:
- P(A) = 0.70 (probability of an A rating)
- P(B) = 0.20 (probability of a B rating)
- P(C) = 0.10 (probability of a C rating)
For bonds issued in cities:
- P(City | A) = 0.50 (probability that it's a city if it's rated A)
- P(City | B) = 0.60 (probability that it's a city if it's rated B)
- P(City | C) = 0.90 (probability that it's a city if it's rated C)
Now, let's calculate:
P(A and City) = P(A) * P(City | A)
P(City) = P(A) * P(City | A) + P(B) * P(City | B) + P(C) * P(City | C)
Substitute the values:
P(A and City) = 0.70 * 0.50
= 0.35
P(City) = (0.70 * 0.50) + (0.20 * 0.60) + (0.10 * 0.90)
= 0.35 + 0.12 + 0.09
= 0.56
Now, calculate the conditional probability:
P(A | City) = P(A and City) / P(City)
= 0.35 / 0.56
= 0.625
So, the probability is 0.625 or 62.5%.
b. What proportion of municipal bonds are issued by cities?
56% of municipal bonds are issued by cities.
c. What proportion of municipal bonds are issued by suburbs?
To find the proportion of municipal bonds issued by suburbs, use a similar approach:
P(Suburb) = P(A) * P(Suburb | A) + P(B) * P(Suburb | B) + P(C) * P(Suburb | C)
We are given:
- P(Suburb | A) = 0.40
- P(Suburb | B) = 0.20
- P(Suburb | C) = 0.05
Now, calculate:
P(Suburb) = (0.70 * 0.40) + (0.20 * 0.20) + (0.10 * 0.05)
= 0.28 + 0.04 + 0.005
= 0.325
So, the proportion of municipal bonds issued by suburbs is 0.325 or 32.5%.
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The probability that a municipal bond issued by a city will receive an A rating is 35%. The proportion of all municipal bonds issued by cities is 56%. The proportion of all municipal bonds issued by suburbs is 32.5%.
This question requires an understanding of probability and conditional probability.
a) To find the probability that a new municipal bond issued by a city will receive an A rating, we must first determine the likelihood that an A-rated municipal bond is issued by a city. Given that 50% of A-rated bonds are issued by cities and that 70% of all bonds receive an A rating, we can calculate this probability as (0.50)*(0.70) = 0.35, or 35%.
b) To find the proportion of municipal bonds issued by cities, we must add up the bonds issued by cities across all ratings. So, (0.70*0.50) + (0.20*0.60) + (0.10*0.90) = 0.35 + 0.12 + 0.09 = 0.56, or 56%.
c) To calculate the proportion of municipal bonds issued by suburbs, we do the same calculation as in part b) but for suburbs. So, (0.70*0.40) + (0.20*0.20) + (0.10*0.05) = 0.28 + 0.04 + 0.005 = 0.325, or 32.5%.
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Expression:
When x=5, the value of the expression is
Answer:
Answer:
The answer is 46 when x = 5
Step-by-step explanation:
8*x + 6
So x = 5
8*5 + 6
40 + 6
46
Answer:
b
Step-by-step explanation:
lim x → ∞ x^4 x^8 + 2
Combine exponents:
lim x → ∞ x^(4 +8) + 2
lim x → ∞ x^12 + 2
The limit at infinity of a polynomial, when the leading coefficient is positive is infinity.
80
120
48
144
Answer:
144
Step-by-step explanation:
first digit of the code: only 2 and 4 are the only even numbers availible, so only 2 possibilities for the first digit.
2nd digit of code: 2,5,7,9,1 so 6 possibilities for the 2nd digit.
3rd digit of code: 2,5,7,9,1 so 6 possibilities for the 3rd digit.
4th digit of code: only 7 and 9 are greater tyhan 6 so only 2 possibilities for the 4th digit.
SO...
2 x 6 x 6 x 2 = 144
Answer:
look below for question
Step-by-step explanation:
Which is true about the solution to the system of inequalities shown?
y > 3x + 1
y < 3x – 3
Only values that satisfy y > 3x + 1 are solutions.
Only values that satisfy y < 3x – 3 are solutions.
Values that satisfy either y > 3x + 1 or y < 3x – 3 are solutions.
There are no solutions.