My total owed with interest is $507.50. I borrowed $290 at 12.5% interest.How long was my loan for in years?
1 answer:
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Answer:
a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.
b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.
Step-by-step explanation:
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
To use the Poisson approximation for the binomial, we have that:
1 in every 2500 automobiles produced has a particular manufacturing defect.
This means that
a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.
This is when
. So
0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.
(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.
Using the approximation:
. So
0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.
Answer:
a. L{t} = 1/s² b. L{1} = 1/s
Step-by-step explanation:
Here is the complete question
The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0
Solution
a. L{t}
L{t} = ∫₀⁰⁰
Integrating by parts ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = and v =
and du/dt = dt/dt = 1
So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w
So, ∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + [
]₀⁰⁰
= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]
= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]
= -1/s[(0 - 0] - 1/s²[0 - 1]
= -1/s[(0] - 1/s²[- 1]
= 0 + 1/s²
= 1/s²
L{t} = 1/s²
b. L{1}
L{1} = ∫₀⁰⁰
= []₀⁰⁰
= -1/s[exp(-∞s) - exp(-0s)]
= -1/s[exp(-∞) - exp(-0)]
= -1/s[0 - 1]
= -1/s(-1)
= 1/s
L{1} = 1/s