MATHEMATICS COLLEGE

A. Use your calculator to approximate ∫^ b_0 e^-0.00001x dx for b=10, 50, 100 and 1000.b. Based on your answers to part a, does ∫^[infinity]_0 e^-0.00001 dx appear to be convergent or divergent?
c. To what value does the integral actually converge?

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

We are to integrate the function

$e^-0.00001x$ from 0 to b for different ascending values of x.

$\int e^-0.00001x = -10^5 e^-0.00001x$

Now we substitute the limits

When b =10

I = integral value = $-10^5 e^-0.00001*10$

b =50, I = $-10^5(e^-0.00001*50-1)$

b =100, I = $-10^5( e^-0.00001*100-1)$

b =1000 I= $-10^5 (e^-0.00001*1000-1)$

b) As b increases exponent increases in negative, or denominator increases hence when b becomes large this will be a decreasing sequence hence converges

c) Converges to $-10^5 (0-1)$ =10^5

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) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station

Answers

Complete Question

The probability that a single radar station will detect an enemy plane is 0.65.

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane? (Round your answer to one decimal place.)

Answer:

$n \approx 4$

Step-by-step explanation:

From the question we are told that

The probability that a single radar station will detect an enemy plane is $p =0.65$

Gnerally the probability that an enemy plane flying over will be detected by at least one station is mathematically represented as

$P(X \ge 1 ) = 0.98$

=> $P(X \ge 1 ) = 1 - P(X < 1) = 0.98$

=> $P(X = 0) = 1 - 0.98$ Note $P(X < 1) = P(X = 0)$

=> $P(X = 0) = 1 - 0.98$

=> $P(X = 0) = 0.02$

Generally from binomial probability distribution function

$P(X = 0) = ^nC_0 * p^(0) * (1- p)^(n- 0)$

Here C represents combination hence we will be making use of of combination functionality in our calculators

Generally any number combination 0 is 1

$P(X = 1) = 1 * 1* (1- 0.65)^(n- 0) = 0.02$

=> $(1- 0.65)^(n- 1) = 0.02$

taking log of both sides

$log [(0.35)^(n- 1) ] = log (0.02)$

=> ${n- 1}log[0.35] = -1.699$

=> ${n- 1}* -0.4559 = -1.699$

=> $n= 3.7264 + 1$

=> $n= 4.7264$

=> $n \approx 4$

Gnerally the expected number of stations that will detect an enemy plane is

$E(X) = 7 * 0.65$

=> $E(X) \approx4.5$

The weather in a certain locale consists of alternating wet and dry spells. Suppose that the number of days in each rainy spell is a Poisson distribution with mean 2, and that a dry spell follows a geometric distribution with mean 7. Assume that the successive durations of rainy and dry spells are independent. What is the long-run fraction of time that it rains?

Answers

Answer:

2/9

Step-by-step explanation:

The Poisson’s distribution is a discrete probability distribution. A discrete probability distribution means that the events occur with a constant mean rate and independently of each other. It is used to signify the chance (probability) of a given number of events occurring in a fixed interval of time or space.

In the long run, fraction of time that it rains = E(Number of days in rainy spell) / {E(Number of days in a rainy spell) + E(Number of days in a dry spell)}

E(Number of days in rainy spell) = 2

E(Number of days in a dry spell) = 7

In the long run, fraction of time that it rains = 2/(2 + 7) = 2/9

Final answer:

Given the parameters of the rainy spell and dry spell, the long-run fraction of time that it rains can be calculated by dividing the mean of the rainy days by the sum of the average rainy and dry days. Hence, it rains roughly 22.22% of the time in the long-term.

Explanation:

The question is asking about the long-run fraction of time that it rains, based on a rainy spell following a Poisson distribution with a mean of 2 days, and a dry spell following a geometric distribution with an average of 7 days, with the sequences being independent.

We are being asked to calculate the proportion of time that it rains in the long-run, given these distribution parameters. The Poisson and geometric distributions are often used in this type of probability assessment.

To tackle this, we need to divide the mean of the rainy days by the sum of the average rainy and dry days. Thus, the long-run fraction of time it rains is given by $2/(2+7) = 2/9.$

So, in the long run, it rains roughly 22.22% (or 2/9) of the time.

Learn more about Probability Distributions here:

brainly.com/question/14210034

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The equation 2|3x−1|=10 has two solutions, one positive and one negative. Solve the equation to find the solutions.

Answers

The two solutions of the equation 2|3x - 1| = 10 are x = 2 and

x = -4/3.

We have the following equation -

2|3x -1| = 10

We have to solve the equation to find the solutions.

What is modulus function ?

The modulus function is as follows -

for x > 0 , |x| = x

for x < 0 , |x| = - x

According to the question, we have -

2|3x - 1| = 10

|3x -1| = 5

Now, using the modulus property -

3x - 1 = 5 and 3x - 1 = -5

3x = 6 and 3x = -4

x = 2 and x = -4/3

Hence, the two solutions of the equation 2|3x - 1| = 10 are x = 2 and

x = -4/3.

To learn more about Modulus function, visit the link below-

brainly.com/question/13103168

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WILL GIVE BRAINLIEST!!! 50POINTS. PLEASE EXPLAIN!What is the recursive rule for this geometric sequence? 2, 1/2, 1/8, 1/32, ... Enter your answers in the boxes.
an=___⋅an−1, a1=___

Answers

$\bf \stackrel{a_1}{2}~~,~~\stackrel{2\cdot (1)/(4)}{\cfrac{1}{2}}~~,~~\stackrel{(1)/(2)\cdot (1)/(4)}{\cfrac{1}{8}}~~,~~\stackrel{(1)/(8)\cdot (1)/(4)}{\cfrac{1}{32}}\n\n-------------------------------\n\na_n=\cfrac{1}{4}\cdot a_(n-1)\qquad \qquad a_1=2$

Answer:

an= 1/4 · an-1 a1= 2

Step-by-step explanation:

Got it correct on the test.

In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.02 margin of error and use a confidence level of 99%. Complete parts (a) through (c) below.a. Assume that nothing is known about the percentage to be estimated.n = ________b. Assume prior studies have shown that about 55% of full-time students earn bachelor's degrees in four years or less.n = _______c. Does the added knowledge in part (b) have much of an effect on the sample size?

Answers

Answer:

(a) The sample size required is 2401.

(b) The sample size required is 2377.

Step-by-step explanation:

The confidence interval for population proportion p is:

$CI=\hat p\pm z_(\alpha/2)\sqrt{(\hatp(1-\hat p))/(n)}$

The margin of error in this interval is:

$MOE=z_(\alpha/2)\sqrt{(\hatp(1-\hat p))/(n)}$

The information provided is:

MOE = 0.02

$z_(\alpha/2)=z_(0.05/2)=z_(0.025)=1.96$

(a)

Assume that the proportion value is 0.50.

Compute the value of n as follows:

$MOE=z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}\n0.02=1.96* \sqrt{(0.50(1-0.50))/(n)}\nn=(1.96^(2)*0.50(1-0.50))/(0.02^(2))\n=2401$

Thus, the sample size required is 2401.

(b)

Given that the proportion value is 0.55.

Compute the value of n as follows:

$MOE=z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}\n0.02=1.96* \sqrt{(0.55(1-0.55))/(n)}\nn=(1.96^(2)*0.55(1-0.55))/(0.02^(2))\n=2376.99\n\approx2377$

Thus, the sample size required is 2377.

(c)

On increasing the proportion value the sample size decreased.

Ten measurements of impact energy on specimens of A238 steel at 60 ºC are as follows: 64.1, 64.7, 64.5, 64.6, 64.5, 64.3, 64.6, 64.8, 64.2, and 64.3 J. a. Use the Student’s t distribution to find a 95% confidence interval for the impact energy of A238 steel at 60 ºC. b. Use the Student’s t distribution to find a 98% confidence interval for the impact energy of A238 steel at 60 ºC.