Answer:
[tex]X_1 + X_2 \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]
Step-by-step explanation:
We are given the following in the question:
[tex]X_1[/tex] is a random normal variable with mean and variance
[tex]\mu_1\\\sigma_1^2[/tex]
[tex]X_1 \sim N(\mu_1,\sigma_1^2)[/tex]
[tex]X_2[/tex] is a random normal variable with mean and variance
[tex]\mu_2\\\sigma_2^2[/tex]
[tex]X_2 \sim N(\mu_2,\sigma_2^2)[/tex]
[tex]X_1, X_2[/tex] are independent events.
Let
[tex]Z =X_1 + X_2[/tex]
Then, Z will have a normal distribution with mean equal to the sum of the two means and its variance equal the sum of the two variances.
Thus, we can write:
[tex]\mu = \mu_1 + \mu_2\\\sigma^2 = \sigma_1^2 + \sigma_2^2\\Z \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]
Final answer:
The sum of two independent normal random variables X1 and X2 is also a normal random variable with a mean of μ1 + μ2 and variance of σ²₁ + σ²₂, denoted by N(μ1 + μ2, σ²₁ + σ²₂).
Explanation:
The distribution of the sum of two independent normal random variables, X1 + X2, is itself a normal random variable. Given that X1 is a normal random variable with mean μ1 and variance σ²₁, and X2 is a normal random variable with mean μ2 and variance σ²₂, the sum X1 + X2 will have a mean of μ1 + μ2 and variance of σ²₁ + σ²₂ because of the properties of the normal distribution and the independence of X1 and X2.
The resulting distribution can be denoted as N(μ1 + μ2, σ²₁ + σ²₂). This conclusion comes from the central limit theorem, which states that the sum of independent random variables tends towards a normal distribution as the number of variables increases, and the means and variances add up.
A firm that stocks lightbulbs gathers the following information: Demand = 19,500 units per year, Ordering cost = $25 per order, Holding cost = $4 per unit per year. The firm wants to calculate the: a) EOQ for the lightbulbs. b) Annual holding costs for the lightbulbs. c) Annual ordering costs for the lightbulbs.
Answer:
Step-by-step explanation:
Demand = 19,500 units per year (D)
Ordering cost = $25 per order (O)
Holding cost = $4 per unit per year (C)
a) [tex]EOQ=\sqrt{ \frac{2\times D\times O}{C}}[/tex]
[tex]EOQ=\sqrt{ \frac{2\times 19,500\times 25}{4}}[/tex]
= [tex]\sqrt{\frac{975000}{4}}[/tex]
= [tex]\sqrt{243,750}[/tex]
= 493.71044 ≈ 494
b) Annual holding cost = [tex]4\times(\frac{Q}{2})[/tex]
= [tex]4\times(\frac{494}{2})[/tex]
= 4 × 247
= 988
c) Annual ordering cost = [tex]O\times(\frac{D}{Q} )[/tex]
= [tex]25\times(\frac{19,500}{494} )[/tex]
= 25 × 39.47
= 986.75
AOQ = 494
Annual holding cost = 988
Annual ordering cost = 986.75
A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flows into the tank at a rate of Rin=80 L/minutes (min). The fluid mixes instantaneously and is pumped out at a specified rate Rout. Let y(t) denote the quantity of salt in the tank at time t. Assume that Rout=40L/min. (a) Set up and solve the differential equation for y(t). (b) What is the salt concentration when the tank overflows?
Final answer:
It would take roughly 22.22 hours to fill an 80,000 L swimming pool with a garden hose at a rate of 60 L/min. However, if diverting a moderate size river flowing at 5000 m³/s, it would fill the pool in approximately 0.016 seconds.
Explanation:
To estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min, we simply divide the total volume of the pool by the flow rate of the hose. For this calculation:
Time (min) = Total Volume (L) / Flow Rate (L/min)
Time = 80,000 L / 60 L/min = 1333.33 min
To convert minutes to hours, we divide by 60:
Time = 1333.33 min / 60 = 22.22 hours
For part (b), if you could divert a moderate size river, flowing at 5000 m³/s, into the pool, we need to convert this flow rate into liters per second (L/s) for consistency:
Flow Rate (L/s) = 5000 m³/s * 1000 L/m³
Flow Rate = 5000000 L/s
Now we can calculate the time it would take to fill the pool with this flow rate:
Time = 80,000 L / 5000000 L/s
Time = 0.016 seconds
Which shows a fractional and decimal equivalent of 9/20 ?
4/10=0.4
5/10=0.5
36/100=0.45
45/100=0.45
Answer:
45/100 = 0.45
Step-by-step explanation:
Divide 9 by 20.
9/20 = 0.45
Now write 36/100 as a decimal.
36/100 = 0.36
Write 45/100 as a decimal: 0.45
Answer: 45/100 = 0.45
A group of thirty people is selected at random. What is the probability that at least two of them will have the same birthday? (Round your answer to four decimal places.)
Answer:
0.706
Step-by-step explanation:
For uniformity, let's assuming none of the people selected is born on 29th of February. Therefore, the total possible birthday (sample space) for anybody selected is 365 days.
Probability of having at least two people having or not having the same birthday sum up to 1. It is easier to calculate the probability of people not having the same birthday.
The probability of at least two people having the same birthday = 1 - The probability of at least two people not having the same birthday
Pr (First Person selected is born in a day in a year) = [tex]\frac{365}{365}[/tex]
The next person selected is limited to 364 possible days
Pr (Second Person selected is born in a day in year) = [tex]\frac{364}{365}[/tex]
The next person selected is limited to 363 possible days
Pr (Third Person selected is born in a day in year) = [tex]\frac{363}{365}[/tex]
The next person selected is limited to 362 possible days
Pr (Fourth Person selected is born in a day in year) = [tex]\frac{362}{365}[/tex]
The next person selected is limited to 361 possible days
Pr (Fifth Person selected is born in a day in year) = [tex]\frac{361}{365}[/tex]
The next person selected is limited to 360 possible days
Pr (Sixth Person selected is born in a day in year) = [tex]\frac{360}{365}[/tex]
The next person selected is limited to 359 possible days
Pr (Seventh Person selected is born in a day in year) = [tex]\frac{359}{365}[/tex]
Looking at the pattern, the pattern continues by descending by 1 day from 365. The last person selected will have 336 possible days (365 – 29 days ) the other people before have used up 29 potential days.
Pr (Last Person selected is born in a day in year) = [tex]\frac{359}{365}[/tex]
The probability of selecting at least two people that will not have the same birthday, is the product of their 30 individual birthday probability.
[tex]\frac{365}{365} * \frac{364}{365} * \frac{363}{365} * \frac{362}{365} * \frac{361}{365} * \frac{360}{365} * \frac{359}{365} ... \frac{337}{365} * \frac{336}{365} = [/tex]
This is a huge a hug e calculation to simplify. For simplicity the numerator and the denominator will be calculated separately.
The denominator is product of 365 in 30 times.
The denominator = 365³⁰
To simplify the numerator, the factorials method is used. Using factorials method, 365! = 365 × 364 × 363 × 362 × 361 × 360 × 359 × 358 ... × 3 × 2 × 1. But we only need the product of the integers from 365 to 336, so we’ll divide the unwanted numbers by dividing 365! by 335!
The numerator = [tex]\frac{365!}{335!}[/tex]
Therefore,
[tex]\frac{365!}{335! * 365^{30}} = 0.294[/tex]
The probability that no one selected in the group has the same birthday is 0.294. The probability that at least two people will have the same birthday is the complement of the probability that no one in the group has the same birthday.
1 - 0.294 = 0.706
The probability that at least two of them will have the same birthday = 0.706
The correct answer is approximately 0.7064.
To solve this problem, it is easier to calculate the probability that no one shares a birthday and then subtract this from 1 to find the probability that at least two people share a birthday.
Let's denote the probability that no two people share a birthday as P(no shared birthday). There are 365 days in a year, so the first person can have any birthday.
The second person then has 364 out of 365 days to have a different birthday, the third person has 363 out of 365 days, and so on, until the thirtieth person has 336 out of 365 days to avoid sharing a birthday with anyone else.
Thus, the probability that no one shares a birthday is:
[tex]\[ P(\text{no shared birthday}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \ldots \times \frac{336}{365} \][/tex]
We can calculate this product to find P(no shared birthday). Then, to find the probability that at least two people share a birthday, we subtract this from 1:
[tex]\[ P(\text{at least one shared birthday}) = 1 - P(\text{no shared birthday}) \][/tex]
Calculating the product for P(no shared birthday) and subtracting from 1 gives us the probability that at least two people in the group of thirty will have the same birthday.
When rounded to four decimal places, this probability is approximately 0.7064.
The mean annual income for people in a certain city (in thousands of dollars) is 43, with a standard deviation of 29. A pollster draws a sample of 41 people to interview. What is the probability that the sample mean income is less than 42 (thousands of dollars)?
A) 0.4721 B) 0.4129 C) 0.3483 D) 0.5279
The probability that the average income of a 41-person sample would be less than $42,000, given a population with mean annual income of $43,000 and a standard deviation of $29,000, is 0.4129.
Explanation:This is a problem of probability concerning the Mean and Standard Deviation of the sampling distribution of the sample mean. We first have to find the Standard Error (SE) for the sampling distribution of the sample mean, which is given by SE = σ/√n , where σ is the standard deviation (29) and n is the sample size (41). Thus, SE = 29/√41 = 4.52.
Next, we find the Z score corresponding to the sample mean of 42. Z = (X - μ)/SE, where X is the sample mean (42), μ is the population mean (43), and SE is the standard error. So Z = (42 - 43)/4.52 = -0.22.
Looking up a Z score of -0.22 in a standard normal distribution table, we find a probability of 0.4129. Therefore, the answer is B) 0.4129.
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Final answer:
Using the Central Limit Theorem and provided statistics, the probability that the sample mean income is less than 42 thousand dollars is found by calculating a z-score and consulting standard normal distribution tables, resulting in a probability of 0.4129.
Explanation:
The task at hand is to calculate the probability that the sample mean income is less than 42 thousand dollars. Given the population mean (μ) of 43 thousand dollars and the standard deviation (σ) of 29 thousand dollars, with a sample size (n) of 41, we can use the Central Limit Theorem to determine that the sampling distribution of the sample mean will be normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
To find the probability, we first calculate the standard error (SE), which is:
SE = σ / √n = 29 / √41 ≈ 4.53
We then calculate the z-score for the sample mean of 42 thousand dollars as:
Z = (X - μ) / SE = (42 - 43) / 4.53 ≈ -0.22
Using standard normal distribution tables or a calculator, the probability of a z-score being less than -0.22 is approximately 0.4129. Thus, the answer is B) 0.4129.
Which triangle congruence postulate proves these two triangle are congruent?
A. AAS
B. HL
C. SAS
D. ASA
Please answer this
Answer:
A. AAS
Step-by-step explanation:
The vertical angles are congruent, the adjacent marked angles are congruent, and the marked sides on the far side of both angles are congruent. This geometry matches the conditions for Angle-Angle-Side (AAS) congruence.
________ and currency risks are to key country success factors as land costs and ________ are to key region success factors. Cultural issues; zoning restrictions Labor cost; proximity to customers Land costs; air and rail systems Exchange rates; environmental impact All of the above are accurate relationships.
Country: Cultural issues, currency risks. Region: Land costs, air and rail systems. (Total: 10 words)
In assessing country and region success factors, certain variables hold significant weight. Country success factors encompass aspects such as cultural issues and currency risks, which directly influence a nation's economic stability and investment attractiveness.
Cultural nuances affect business practices and consumer behaviors, while currency risks impact trade competitiveness and profitability.
Conversely, region success factors pivot around considerations like land costs and transportation infrastructure, particularly air and rail systems. Land costs dictate the feasibility of establishing operations or acquiring property within a region, shaping investment decisions.
Meanwhile, efficient air and rail systems facilitate logistical operations and market access, enhancing the competitiveness of businesses located within the region.
These factors collectively contribute to a region's appeal for businesses seeking to establish or expand operations.
Thus, understanding and appropriately navigating these variables are crucial for organizations aiming to optimize their strategic positioning at both the national and regional levels, ensuring long-term success and sustainability in an increasingly complex global landscape.
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 447 gram setting. It is believed that the machine is underfilling the bags. A 19 bag sample had a mean of 443 grams with a standard deviation of 21. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.
The decision rule for rejecting the null hypothesis H₀ is
if t-statistic < -2.093.
First , let's state the null and alternative hypotheses:
Null hypothesis (H₀): The population mean weight of banana chips is 447 grams (the machine works correctly).
Alternative hypothesis (H₁): The population mean weight of banana chips is less than 447 grams (the machine is underfilling).
α = 0.025
Since this is a one-tailed test (we only care if the weight is less than 447 grams), we need to find the t-value for a single-tailed test with α = 0.025 and degrees of freedom (df) = n - 1 = 19 - 1 = 18.
We can use a t-distribution table or a statistical calculator to find this value. In most calculators, you can use the TINV function with df = 18, tails = 1, and p = 0.025. The critical value (t_α) is approximately -2.093.
Reject the null hypothesis (H₀) if the calculated t-statistic (t_stat) is less than the critical value (t_α). In other words, if the sample mean weight is significantly lower than 447 grams, we have evidence to suggest the machine is underfilling.
Reject H₀ if t_stat < -2.093.
This will require sample data (the weights of the 19 bags). Once we have that, we can calculate the t-statistic using the formula:
t_stat = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size)).
The complete question is : A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 447 gram setting. It is believed that the machine is underfilling the bags. A 19 bag sample had a mean of 443 grams with a standard deviation of 21. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the comparison property to find the lower and upper bounds for I = \displaystyle \int_{-2}^{2} f(x) \, dxI = ∫ − 2 2 f ( x ) d x.
[tex]f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}[/tex]
[tex]f[/tex] has critical points where the derivative is 0:
[tex]2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1[/tex]
The second derivative is
[tex]f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}[/tex]
and [tex]f''(-1)=\frac2e>0[/tex], which indicates a local minimum at [tex]x=-1[/tex] with a value of [tex]f(-1)=\frac1e[/tex].
At the endpoints of [-2, 2], we have [tex]f(-2)=1[/tex] and [tex]f(2)=e^8[/tex], so that [tex]f[/tex] has an absolute minimum of [tex]\frac1e[/tex] and an absolute maximum of [tex]e^8[/tex] on [-2, 2].
So we have
[tex]\dfrac1e\le f(x)\le e^8[/tex]
[tex]\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx[/tex]
[tex]\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}[/tex]
(1 point) You are given the parametric equations x=2t3+3t2−12t,y=2t3+3t2+1. (a) List all of the points (x,y) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of x. If two or more points share the same value of x, list those points starting with the smallest value of y. If any blanks are unused, type an upper-case "N" in them.
Answer:
(0,1) and (13,2)
Step-by-step explanation:
We are given that parametric equations
[tex]x=2t^3+3t^2-12t[/tex]
[tex]y=2t^3+3t^2+1[/tex]
a.We have to find the points where tangent line is horizontal.
Differentiate x and y w.r.t.time
[tex]\frac{dx}{dt}=6t^2+6t-12[/tex]
[tex]\frac{dy}{dt}=6t^2+6t[/tex]
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
Substitute the values
[tex]\frac{dy}{dx}=\frac{6t^2+6t}{6t^2+6t-12}[/tex]
We know that when tangent line is horizontal then
[tex]\frac{dy}{dx}=0[/tex]
[tex]\frac{6t^2+6t}{6t^2+6t-12}=0[/tex]
[tex]6t^2+6t=0[/tex]
[tex]t^2+t=0[/tex]
[tex]t(t+1)=0[/tex]
[tex]t=0, t+1=0[/tex]
[tex]t+1=0\implies t=-1[/tex]
Substitute the t=0 then we get
[tex]x=2(0)^3+3(0)^2-12(0)=0[/tex]
[tex]y=2(0)^3+3(0)^2+1=1[/tex]
Substitute t=-1
[tex]x=2(-1)^3+3(-1)^2-12(-1)=-2+3+12=13[/tex]
[tex]y=2(-1)^3+3(-1)^2+1=-2+3+1=2[/tex]
The points (0,1) and (13,2) where tangent line is horizontal .
To find the points where the tangent line is horizontal, we need to find the values of t that make the derivative of y with respect to x equal to 0. The points where the tangent line is horizontal are (0,1) and (20,0).
Explanation:To find the points where the tangent line is horizontal, we need to find the values of t that make the derivative of y with respect to x equal to 0. Let's start by finding the derivatives of x and y with respect to t:
[d/dt(x)]=(6t^2)+(6t)-12
[d/dt(y)]=(6t^2)+(6t)
To find the values of t that make [d/dt(y)] equal to 0, we set (6t^2)+(6t)=0. By factoring out a 6t, we get t(t+1)=0. So t=0 or t=-1. We can substitute these values of t back into the parametric equations to find the corresponding points (x,y):
t=0: x=0, y=1t=-1: x=20, y=0Therefore, the points (x,y) where the tangent line is horizontal are (0,1) and (20,0).
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PLEASE HELP
Answer all three of the questions below.
1. Write an equivalent expression to 2 + 3 + 5 + 6 by combining like terms.
2. Find the sum of (8a – 2b - 4) and (3b − 5).
3.Write the expression in standard form: 4(2a) + 7(−4b) + (3 ∙ c ∙ 5).
Answer:
a)[tex]2 + 3 + 5 + 6 = 16[/tex]
b)[tex]8a + b-9[/tex]
c)[tex]8a - 28b + 15c[/tex]
Step-by-step explanation:
We are given the following in the question:
1. Equivalent expression
[tex]2 + 3 + 5 + 6[/tex]
Since all the terms are like terms, we can write,
[tex]2 + 3 + 5 + 6 = 16[/tex]
2. Sum of given expression
[tex](8a - 2b - 4) \text{ and } (3b - 5)\\ (8a - 2b - 4) + (3b -5)\\= 8a -2b -4 + 3b -5\\\text{Collecting the like terms}\\8a + (-2b+3b) + (-4-5)\\= 8a + b-9[/tex]
3. Standard form of the expression
[tex]4(2a) + 7(-4b) + (3 \times c \times 5)\\=(8a) + (-28b) + (15c)\\=8a - 28b + 15c[/tex]
which is the required standard form of the given expression.
There are 20 members of a basketball team.
(a) The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?
(b) From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. How many ways are there for her to select the starting line-up?
(c) From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?
There are 125,970 ways to select the players who will travel to an away game. There are 79,833,600 ways to select the starting line-up. With a restriction on the center position, there are 12 ways to select the starting line-up.
Explanation:(a) The coach must select 12 players to travel to an away game:
There are 20 members on the basketball team, and the coach needs to select 12 players to travel. This is a combination problem, since the order doesn't matter. The formula for calculating combinations is: nCr = n! / (r! * (n-r)!), where n is the total number of players and r is the number of players to be selected.
Using this formula, we can calculate the number of ways to select the traveling players: 20C12 = 20! / (12! * (20-12)!) = 125,970 ways.
(b) The coach must select her starting line-up:
There are 12 players who will travel, and the coach needs to select 5 players for the starting line-up. This is a permutation problem, since the order does matter. The formula for calculating permutations is: nPr = n! / (n-r)!, where n is the total number of players and r is the number of players to be selected.
Using this formula, we can calculate the number of ways to select the starting line-up: 12P5 = 12! / (12-5)! = 12! / 7! = 79,833,600 ways.
(c) The coach must select her starting line-up, with a restriction on the center position:
There are 12 players who will travel, and only 3 players who can play center. The remaining 4 positions have no restrictions. We can calculate the number of ways to select the starting line-up by first selecting the center player (3 ways), and then selecting the players for the other positions (4P4 = 4 ways).
Therefore, the total number of ways to select the starting line-up with the restriction on the center position is: 3 * 4 = 12 ways.
For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 505 and a standard deviation of 170. The grading process of this year's exam has just begun. The average score of the 35 exams graded so far is 530.What is the probability that a sample of 35 exams will have a mean score of 530 or more if the exam scores follow the same distribution as in the past?
Answer:
[tex]P(\bar X>530)=1-0.808=0.192[/tex]
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable that represent the scores, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=505,\sigma=170)[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(505,\frac{170}{\sqrt{35}})[/tex]
2) Calculate the probability
We want this probability:
[tex]P(\bar X>530)=1-P(\bar X<530)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X >530)=1-P(Z<\frac{530-505}{\frac{170}{\sqrt{35}}})=1-P(Z<0.87)[/tex]
[tex]P(\bar X>530)=1-0.808=0.192[/tex]
Using the central limit theorem, the probability of a sample mean of 530 or more is calculated considering the distribution of the sample mean, which is approximately normal due to the sample size.
Explanation:To calculate the probability that a sample of 35 exams will have a mean score of 530 or more, when the population mean is 505 with a standard deviation of 170, we use the concept of the sampling distribution of the sample mean. Since the sample size is 35, which is less than 10% of all psychology majors (assuming they number in the thousands), and the population distribution is normal or the sample size is large enough (which it is in this case), we can apply the Central Limit Theorem.
This theorem states that the sampling distribution of the sample mean will be approximately normally distributed with a mean equal to the population mean (μ = 505) and a standard deviation equal to σ/√n, where σ is the population standard deviation and n is the sample size (which is √35 in this case).
Based on the following information, what is the standard deviation of returns? State of Economy Probability of State of Economy Rate of Return if State Occurs Recession .22 − .112 Normal .25 .127 Boom .53 .237
The standard deviation of returns is 2.82% by calculating expected value, variance and standard deviation.
Given that:
State of Economy Probability Rate of Return if State Occurs
Recession 0.22 -0.9
Normal 0.47 0.105
Boom 0.31 0.215
To calculate the standard deviation of returns, by using the formula:
Standard Deviation = [tex]\sqrt{}[/tex][Σ (Probability * (Rate of Return - Expected Return[tex])^2[/tex])]
Step 1: calculate the Expected Return: Multiply each rate of return by its corresponding probability and sum the results:
E = Σ(Rate of Return* Probability )
Plugging the given data gives:
E = (0.22 * -0.9) + (0.47 * 0.105) + (0.31 * 0.215)
On multiplication gives
E = 0.0225 - 0.04935 + 0.06665
On solving gives:
E = 0.0398.
Step 2: Calculate the Variance: For each rate of return, subtract the expected return, square the result, multiply by its corresponding probability, and sum the results:
Variance ( V) = Σ (Probability * (Rate of Return - Expected Return[tex])^2[/tex]
Plugging the given data gives:
[tex]V = [(0.22 * (-0.9 - 0.0398)^2) + (0.47 * (0.105 - 0.0398)^2) + (0.31 * (0.215 - 0.0398)^2)][/tex]
On squaring and solving gives:
[tex]V = [0.0225 + 0.02741 + 0.0412][/tex]
On adding gives:
[tex]V = 0.09111.[/tex]
Step3: Find the square root of the variance to get the Standard Deviation:
[tex]\sigma = \sqrt V[/tex]
Plugging the given data gives:
[tex]\sigma = \sqrt{0.09111 }[/tex]
On taking square root gives:
[tex]\sigma \approx 0.282[/tex] or 2.82%.
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Question:
Based on the following information, what is the standard deviation of returns?
State of Economy Probability Rate of Return if State Occurs
Recession 0.22 -0.9
Normal 0.47 0.105
Boom 0.31 0.215
To calculate the standard deviation of returns, follow these steps: 1) Calculate the expected return for each state by multiplying the probability of the state occurring by the rate of return. 2) Calculate the squared difference between each state's return and the expected return. 3) Multiply each squared difference by the probability of the state occurring. 4) Sum up all the values obtained in step 3. 5) Take the square root of the sum obtained in step 4.
Explanation:To calculate the standard deviation of returns, we need to follow these steps:
Calculate the expected return for each state of the economy by multiplying the probability of the state occurring by the rate of return for that state.Calculate the squared difference between each state's return and the expected return.Multiply each squared difference by the probability of the state occurring.Sum up all the values obtained in step 3.Take the square root of the sum obtained in step 4.Using the given information:
Expected return for Recession: (.22)*(-.112) = -0.02464Expected return for Normal: (.25)*(0.127) = 0.03175Expected return for Boom: (.53)*(0.237) = 0.12561Next, we calculate the squared difference between each state's return and the expected return:
For Recession: (-0.112 - (-0.02464))^2 = 0.005019024For Normal: (0.127 - 0.03175)^2 = 0.00718400625For Boom: (0.237 - 0.12561)^2 = 0.0122397638Then, we multiply each squared difference by the probability of the state occurring:
For Recession: 0.005019024 * 0.22 = 0.00110478528For Normal: 0.00718400625 * 0.25 = 0.0017960015625For Boom: 0.0122397638 * 0.53 = 0.006496061714Finally, we sum up all the values obtained:
0.00110478528 + 0.0017960015625 + 0.006496061714 = 0.0093968483025And take the square root of the sum:
Square root of 0.0093968483025 = 0.096919769Therefore, the standard deviation of returns is approximately 0.0969.
Find the quadratic function f (x )equals ax squared plus bx plus c for which f (1 )equals negative 2, f (negative 3 )equals 26, and f (3 )equals 32.
Answer:
f(x) =4x² + x - 7
Step-by-step explanation:
f(x) = ax² + bx + c
f(1) = a + b + c
f(-3) = 9a - 3b + c
f(3) = 9a + 3b + c
a + b + c = -2 -----(1)
c = -2-a-b -----(2)
9a - 3b + c = 26 ------(3)
9a + 3b + c = 32 -------(4)
(2)->(3)
9a-3b-2-a-b = 26
8a-4b = 28
4(2a-b) = 28
2a-b = 7 ----(5)
(2)->(4)
9a+3b-2-a-b = 32
8a+2b = 34
2(4a+b) = 34
4a+b = 17 ----(6)
(5)+(6)
6a = 24
a = 4
sub a=4 into(6)
4(4)+b = 17
b = 1
sub a=4,b=1 into (2)
c = -2-4-1
c = -7
You want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a simple random sample of 30 apartments advertised in the local newspaper is $1,450. The standard deviation of the population is $220. Find a 99% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community.
Answer: 99% confidence interval would be [tex](1346.37,1553.63)[/tex]
Step-by-step explanation:
Since we have given that
Sample size = 30
Sample mean = $1450
Standard deviation = $220
We need to find the 99% confidence interval.
So, z = 2.58
So, confidence interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1450\pm 2.58\times \dfrac{220}{\sqrt{30}}\\\\=1450\pm 103.63\\\\=(1450-103.63,1450+103.63)\\\\=(1346.37,1553.63)[/tex]
Hence, 99% confidence interval would be [tex](1346.37,1553.63)[/tex]
Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0=2x2+3x-8?
Quadratic formula: x =StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction
A. 1.39
B. 2.00
C. 2.89
D. 3.50
Answer:
Step-by-step explanation:
The given quadratic equation is
2x^2+3x-8 = 0
To find the roots of the equation. We will apply the general formula for quadratic equations
x = -b ± √b^2 - 4ac]/2a
from the equation,
a = 2
b = 3
c = -8
It becomes
x = [- 3 ± √3^2 - 4(2 × -8)]/2×2
x = - 3 ± √9 - 4(- 16)]/2×2
x = [- 3 ± √9 + 64]/2×2
x = [- 3 ± √73]/4
x = [- 3 ± 8.544]/4
x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4
x = 5.544/4 or - 11.544/4
x = 1.386 or x = - 2.886
The positive solution is 1.39 rounded up to the nearest hundredth
Answer:
its a. 1.39
Step-by-step explanation:
According to a survey of American households, the probability that the residents own two cars if annual household income is over $50,000 is 80%. Of the households surveyed, 60% had incomes over $50,000 and 70% had two cars. The probability that the residents of a household do not own two cars and have an income over $50,000 a year is __________?
Answer:
12%
Step-by-step explanation:
The probability that the residents of a household do not own two cars and have an income over $50,000 a year is 100% subtracted by the probability of owning two cars and having an income over $50,000(P(2∩50+) and the probability of not having an income over $50,000 (P(50-))
[tex]P = 1 - P(2\cap 50+) - P(50-)\\P = 1 - (0.6*0.7) - (1-0.6)\\P=1-0.48-0.40\\P=0.12[/tex]
Therefore, the probability that the residents of a household do not own two cars and have an income over $50,000 a year is 12%
Final answer:
To find the probability that a household has an income over $50,000 and does not own two cars, multiply the probability of a household not owning two cars (20%) by the probability of having an income over $50,000 (60%), which equals 12%.
Explanation:
The question is asking us to find the probability that the residents of a household do not own two cars and have an income over $50,000 a year. We know that the probability a household with income over $50,000 owns two cars is 80%, so the probability that such a household does not own two cars is 1 - 0.80 = 0.20 or 20%. Since 60% of households have an income over $50,000, we can calculate the probability that a household has an income over $50,000 and does not own two cars by multiplying these two probabilities: 0.60 * 0.20 = 0.12 or 12%.
A survey of 61 comma 646 people included several questions about office relationships. Of the respondents, 26.5% reported that bosses scream at employees. Use a 0.05 significance level to test the claim that more than 1 divided by 4 of people say that bosses scream at employees. How is the conclusion affected after learning that the survey is an online survey in which Internet users chose whether to respond? Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
Answer:
H ₀ : p ≤ 0.25
H ₁: p > 0.25
Test statics value z = 12.91
Decision Rule:
Reject the value of null hypothesis
Conclusion:
If survey is online then there is sufficient evidence to claim . If the sample is voluntary response sample then conclusion is not valid.
Step-by-step explanation:
step by step explanation is shown in attachment.
One hundred draws are made at random with replacement from a box with ninety-nine tickets marked "0" and one ticket marked "1." True or false, and explain:
(a) The sum will be around 1, give or take 1
(b) There is about a 68% chance that the sum will be in the range 0 to 2.
(a) False. The sum will not be around 1, give or take 1. (b) False. The probability that the sum will be in the range 0 to 2 is not about 68%.
(a).
Each draw has a 99/100 probability of resulting in a sum of 0 and a 1/100 probability of resulting in a sum of 1. Since there are 100 draws made with replacement, the expected value of the sum is given by the average of the individual probabilities multiplied by the number of draws:Expected sum = (99/100 x 0) + (1/100 x 1)
= 1/100
= 0.01
Therefore, the expected value of the sum is 0.01, which is much closer to 0 than 1. While it is possible to observe a sum of 1 or close to 1 in a particular set of draws, it is not likely to be the average or typical outcome.
(b) False. The probability that the sum will be in the range 0 to 2 is not about 68%.
To determine the probability, to consider the possible outcomes that result in a sum between 0 and 2.
The possible outcomes are:
Sum = 0: This can only happen if all 100 draws result in 0. The probability of this occurring is [tex](99/100)^{100} \approx 0.366[/tex].Sum = 1: This can only happen if exactly one draw results in 1, and the remaining 99 draws result in 0. The probability of this occurring is [tex]100 \times (1/100) \times (99/100)^{99} \approx 0.369[/tex].Sum = 2: This can only happen if exactly two draws result in 1, and the remaining 98 draws result in 0. The probability of this occurring is [tex](100 choose 2) \times (1/100)^2 \times (99/100)^{98} \approx 0.185.[/tex]Therefore, the total probability of the sum being in the range 0 to 2 is approximately:
[tex]0.366 + 0.369 + 0.185 = 0.92,[/tex]
which is far greater than 68%.
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Two accounting professors decided to compare the variance of their grading procedures. To accomplish this, they each graded the same 10 exams, with the following results: Mean Grade Standard Deviation Professor 1 79.3 22.4 Professor 2 82.1 12.0 At the 10% level of significance, what is the decision regarding the null hypothesis? A) Reject the null hypothesis and conclude the variances are different. B) Fail to reject the null hypothesis and conclude no significant difference in the variances. C) Reject the null hypothesis and conclude the variances are the same. D) Fail to reject the null hypothesis and conclude the variances are the same.
Without the exact F-statistic and p-value, it is not possible to definitively choose between rejecting or failing to reject the null hypothesis. The decision would depend on whether the calculated p-value is less than the level of significance (0.10).
Explanation:To determine whether there is a significant difference in the variance of grading procedures between two professors, we need to conduct an F-test for variance. The null hypothesis (H0) would be that there is no significant difference in the variances of grading, while the alternative hypothesis (H1) would be that there is a significant difference in the variances.
Using the data provided:
Professor 1: Mean grade = 79.3, Standard Deviation = 22.4Professor 2: Mean grade = 82.1, Standard Deviation = 12.0The F-statistic is calculated by dividing the variance of the set with the larger variance (squared standard deviation) by the variance of the set with the smaller variance, which means:
F = (22.4)^2 / (12.0)^2
However, the exact F value is not provided in the data, nor is the p-value. To make the decision, we would compare the p-value to the level of significance (
0.10). If the p-value is less than 0.10, we would reject the null hypothesis, indicating that there is a significant difference between the variances. If the p-value is greater than 0.10, we would fail to reject the null hypothesis, indicating there is no significant difference in variances.
Based on the options provided in the question, without the exact p-value or F-statistic, we cannot provide the correct answer. Thus, the answer would depend on the calculated p-value from the F-statistic. The options, therefore, are between A (rejecting the H0) or B (failing to reject the H0), and an exact answer requires additional information from an F-test.
In a recent poll of 750 randomly selected adults, 589 said that it is morally wrong to not report all income on tax returns. Use a 0.05 significance level to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
Answer:
z= 5.08
The p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .
Step-by-step explanation:
1) Data given and notation
n=750 represent the random sample taken
X=589 represent the adults that said that it is morally wrong to not report all income on tax returns
[tex]\hat p=\frac{589}{750}=0.785[/tex] estimated proportion of adults that said that it is morally wrong to not report all income on tax returns
[tex]p_o=0.7[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
pv represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:
Null hypothesis:[tex]p=0.7[/tex]
Alternative hypothesis:[tex]p \neq 0.7[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>5.08)=0.000000377[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .
A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a random sample of size n = 36 and find a sample mean of ¯ x = 98 . What is a 95% confidence interval for the mean of x ?
Answer: The 95% confidence interval for the mean of x is (94.08, 101.92) .
Step-by-step explanation:
We are given that ,
A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.
i.e. [tex]\sigma= 12[/tex]
Also, it is given that , Sample mean [tex]\overline{x}=98[/tex] having sample size : n= 36
For 95% confidence ,
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
By using the z-value table , the two-tailed critical value for 95% Confidence interval :
[tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]
We know that the confidence interval for unknown population mean[tex](\mu)[/tex] is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]\sigma[/tex] = Population standard deviation
[tex]z_{\alpha/2}[/tex] = Critical z-value.
Substitute all the given values, then the required confidence interval will be :
[tex]98\pm (1.96)\dfrac{12}{\sqrt{36}}[/tex]
[tex]=98\pm (1.96)\dfrac{12}{6}[/tex]
[tex]=98\pm (1.96)(2)[/tex]
[tex]=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)[/tex]
Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .
The amounts due on a mobile phone bill in Ireland are normally distributed with a mean of € 53 and a standard deviation of € 15. If a monthly phone bill is chosen at random, find the probability that we get a phone bill between 47 and 74 euros.
Answer:
0.5746
Step-by-step explanation:
Lets call X monthly bill chosen. X is a random variable with distribution N(μ=53,σ =15). We first standarize X, lets call W = (X-53)/15, random variable obtained from X by substracting μ and dividing by σ. W is a random variable with distribution N(0,1) and the cummulative function of W, Φ, is tabulated. You can find the values of Φ on the attached file. Using this information we obtain
[tex] P(47 < X < 74) = P(\frac{47-53}{15} < \frac{X-53}{15} < \frac{74-53}{15}) = P(-0.4 < W < 1.4) = \phi(1.4) - \phi(-0.4) [/tex]
A standard normal random variable has a symmetric density function, as a result Φ(-0.4) = 1- Φ(0.4) = 1- 0.6554 = 0.3446. Also, Φ(1.4) = 0.9192. We conclude that Φ(1.4)-Φ(-0.4) = 0.9192-0.3446 = 0.5746, that is the probability that the phone bill is between 47 and 74 euros.
A postmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of x¯=1.19 fluid ounces and the sample standard deviation is s=0.015 fluid ounces. Find a 95% two-sided confidence interval on the mean volume of syrup dispensed. Assume population is approximately normally distributed.
Answer: (1.1838, 1.1962)
Step-by-step explanation:
The formula we use to calculate the confidence interval for population mean ( if population standard deviation is not given)is given by :-
[tex]\overline{x}\pm t*\dfrac{s}{\sqrt{n}}[/tex],
where n= sample size
s= sample standard deviation.
[tex]\overline{x}[/tex]= sample mean
t* = Two-tailed critical t-value.
Given : n= 25
Degree of freedom : df = n-1 =24
Significance level [tex]=\alpha=1-0.95=0.95[/tex]
Now from students' t-distribution table , check the t-value for significance level [tex]\alpha/2=0.025[/tex] and df=24:
t*=2.0639
[tex]\overline{x}=1.19[/tex] fluid ounces
s= 0.015 fluid ounces
We assume that the population is approximately normally distributed
Now, the 95% two-sided confidence interval on the mean volume of syrup dispensed :-
[tex]1.19\pm (2.0639)\dfrac{0.015}{\sqrt{25}}\\\\=1.19\pm (2.0639)(\dfrac{0.015}{5})\\\\=1.19\pm0.007728=(1.19-0.0061917,\ 1.19+0.0061917)\\\\=(1.1838083,\ 1.1961917)\approx(1.1838,\ 1.1962)[/tex]
∴ The required confidence interval = (1.1838, 1.1962)
To find the 95% two-sided confidence interval for the mean volume of syrup dispensed, use the t-distribution with the given sample mean, standard deviation, and size. The resulting interval is (1.183808, 1.196192) fluid ounces.
To calculate the 95% two-sided confidence interval for the mean volume of syrup dispensed, we can use the t-distribution, as the population standard deviation is unknown, and the sample size is less than 30.
Step-by-Step Explanation:
First, identify the sample mean (ar{x}), which is 1.19 fluid ounces, the sample standard deviation (s), which is 0.015 fluid ounces, and the sample size (n), which is 25.
Next, find the t-score that corresponds to a 95% confidence level and 24 degrees of freedom (n-1). You can use a t-distribution table or calculator for this, which typically gives you a t-score around 2.064.
Now calculate the margin of error (ME) using the formula ME = t*(s/sqrt{n}).
Finally, construct the confidence interval by adding and subtracting the margin of error from the sample mean: (1.19 - ME, 1.19 + ME).
Plugging in the values, we get ME = 2.064*(0.015/sqrt{25}) = 2.064*0.003 = 0.006192. So, the 95% confidence interval is (1.19 - 0.006192, 1.19 + 0.006192) = (1.183808, 1.196192).
solve for −5(x+1)=−3(2x−2)
Answer:
x=11
Step-by-step explanation:
The local orchestra has been invited to play at a festival. There are 111 members of the orchestra and 6 are licensed to drive large multi-passenger vehicles. Busses hold 25 people but are much more expensive to rent. Passenger vans hold 12 people. make a system of equations to find the smallest number of busses the orchestra can rent. Let x= the number of busses and y= the number of vans make an equation representing the number of vehicles needed. __________ make an equation representing the total number of seats in vehicles for the orchestra members. __________ Solve the system of equations. How many busses does the orchestra need to rent? ____________ How many 12-passenger vans does the orchestra need to rent? ____________
- Make an equation representing the number of vehicles needed.
We have six drivers so
x + y ≤ 6
That's not really an equation; it's an inequality. We want to use all our drivers so we can use the small vans, so
x + y = 6
- Make an equation representing the total number of seats in vehicles for the orchestra members.
s = 25x + 12y
That's how many seats total; it has to be at least 111 so again an inequality,
25x + 12y ≥ 111
We solve it like a system of equations.
x + y = 6
y = 6 - x
111 = 25x + 12y = 25x + 12(6-x)
111 = 25x + 72 - 12x
111 - 72 = 13 x
39 = 13 x
x = 3
Look at that, it worked out exactly. It didn't have to.
y = 6 - x = 3
Answer: 3 buses, 3 vans
Using a System of equations, the number of buses and vans required are 3 respectively.
Using the system of equations :
Total number of buses :
b + v ≤ 6 - - - - (1)Total number of passengers :
25b + 12v ≥ 111 - - - - (2)From (1)
b = 6 - v - - - - - (3)Substitute (3) into (2)
25(6-v) + 12v = 111
150 - 25v + 12v = 111
-13v = 111 - 150
-13v = - 39
v = 39/13
v = 3
From (3)
b = 6 - 3
b = 3
Hence, the number of vans and buses required are 3 and 3 respectively.
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Juan only paid $10 for his new letter. He got 25% discount what was the original price of the sweater?
Answer:
$12.50
Step-by-step explanation:
so Juan got his sweater at a 25% discount. So we woukd find 25% of 10 and then add it to $10
Answer:
Step-by-step explanation:
Juan only paid $10 for his new Sweater. He got a 25% discount
Let the original price of the sweater be $x. If he gets s 35% discount, it means that the price which she paid was lower than the original price.
Discount of 25% on the original price = 25/100 × x = 0.25x
The amount of money that Juan paid is the original price minus the discount of 25%. It becomes
x - 0.25x = 0.75x
Amount paid = $10. Therefore
0.75x = 10
x = 10/0.75 = $13.33
Write the quadratic equation whose roots are -2 and 1, and whose leading coefficient is 4 .
Answer: the equation is
4x^2 + 4x - 12
Step-by-step explanation:
A quadratic equation is an equation in which the highest power of the unknown is 2.
The general form of a quadratic equation is expressed as
ax^2 + bx + c
Where
a is the leading coefficient
c is a constant
Assuming we want to write the quadratic equation in x, from the information given, the roots which are given are -2 and 1 and the leading coefficient is 4.
Therefore, the linear factors of the quadratic equation will be (x+2) and (x-1)
the equation becomes
(x+2)(x-1)
= x^2 - x +2x - 3
= x^2 + x - 3
Given a leading coefficient of 4, we will multiply the quadratic expression by 4. It becomes
4(x^2 + x - 3)
= 4x^2 + 4x - 12
Final answer:
The quadratic equation with roots -2 and 1 and a leading coefficient of 4 is 4x^2 - 12x + 8 = 0.
Explanation:
To write a quadratic equation with roots -2 and 1 and a leading coefficient of 4, we need to use the fact that if α and β are roots of a quadratic equation, then the equation can be expressed in the form:
a(x - α)(x - β) = 0
Here, α is -2 and β is 1, and the leading coefficient, a, is 4. Therefore, we can write:
4(x + 2)(x - 1) = 0
Multiplying it out, we have:
4(x2 - 1x - 2x + 2) = 0
Combining like terms, we get:
4x2 - 12x + 8 = 0, which is the desired quadratic equation.
For an average person, the rate of change of weight W (in pounds) with respect to height h (in inches) is given approximately by the following formula.
dW/dh=0.0018h^2
Find W(h) if W(80)equals=287.2 pounds. Also, find the weight of a person who is 5 feet, 8 inches tall.
W(h)=_______
A Person who is 5 feet, 8 inches tall weighs about _______ lbs
Answer:
[tex]W(h)=0.0006h^3-20[/tex].
A Person who is 5 feet, 8 inches tall weighs about 168.7 lbs.
Step-by-step explanation:
We know the rate of change of weight W (in pounds) with respect to height h (in inches)
[tex]\frac{dW}{dh}=0.0018h^2[/tex]
This is a separable equation. A separable equation is a first-order differential equation in which the expression for [tex]\frac{dy}{dx}[/tex] can be factored as a function of x times a function of y. In other words, it can be written in the form
[tex]\frac{dy}{dx}=g(x)f(y)[/tex]
To find W(h), we write the equation in terms of differentials and integrate both sides:[tex]\frac{dW}{dh}=0.0018h^2\\\\dW=(0.0018h^2)dh\\\\\int dW=\int (0.0018h^2)dh\\\\W=0.0006h^3+C[/tex]
To find the value of C, we use W(80) = 287.2 lbs
[tex]287.2=0.0006(80)^3+C\\0.0006\left(80\right)^3+C=287.2\\307.2+C=287.2\\307.2+C-307.2=287.2-307.2\\C=-20[/tex]
Thus,
[tex]W(h)=0.0006h^3-20[/tex]
To find the weight of a person who is 5 feet, 8 inches tall you must:Convert the 5 feet into inches
[tex]5 \:ft \:\frac{12 \:in}{1\:ft} = 60 \:in[/tex]
Add 60 in and 8 in, to find the total height of the person
h = 68 in
Substitute h = 68 in into [tex]W(h)=0.0006h^3-20[/tex] to find the weight:[tex]W(68)=0.0006(68)^3-20=168.7[/tex]
A Person who is 5 feet, 8 inches tall weighs about 168.7 lbs.
To find the weight function W(h) and the weight of a person who is 5 feet, 8 inches tall, we need to integrate the given rate of change formula and substitute the values into the equation.
Explanation:To find the weight function W(h), we need to integrate the given rate of change formula dW/dh = 0.0018[tex]h^2[/tex].
Integrating both sides, we get:
∫ dW = ∫ 0.0018[tex]h^2[/tex] dh
Integrating, we have:
W(h) = 0.0018 * (1/3) * [tex]h^3[/tex] + C
To find the value of C, we use the given information W(80) = 287.2 pounds.
Substituting the values, we have:
287.2 = 0.0018 * (1/3) * [tex]80^3[/tex] + C
Simplifying the equation, we solve for C and find:
C = 287.2 - 0.0018 * (1/3) * [tex]80^3[/tex]
Now, we can write the weight function W(h) as:
W(h) = 0.0018 * (1/3) * [tex]h^3[/tex] + (287.2 - 0.0018 * (1/3) * [tex]80^3[/tex])
To find the weight of a person who is 5 feet, 8 inches tall, we need to convert the height to inches.
5 feet is equal to 60 inches, and 8 inches is 8 inches. So, the total height is 60 + 8 = 68 inches.
Substituting the value into the weight function, we have:
W(68) = 0.0018 * (1/3) * [tex]68^3[/tex] + (287.2 - 0.0018 * (1/3) * [tex]80^3[/tex])
Simplifying the equation, we find the weight of a person who is 5 feet, 8 inches tall.
Learn more about Weight function and finding weight here:https://brainly.com/question/38986050
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