Answer:
(A) The minimum sample size required achieve the margin of error of 0.04 is 601.
(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.
Step-by-step explanation:
Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.
(A)
The margin of error, MOE = 0.04.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601[/tex]
Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.
(B)
The margin of error, MOE = 0.02.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401[/tex]
Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.
Roll two dice, one white and one red. Consider these events: A : The sum is 7 B : The white die is odd C : The red die has a larger number showing than the white D : The dice match (doubles) Which pair(s) of events are disjoint (events A and B are disjoint if A ∩ B = ∅ )? Which pair(s) are independent? Which pair(s) are neither disjoint nor independent?
Final answer:
Events B and D are disjoint; events B and C, A and C, B and D are independent; events A and B, A and C, and B and D are neither disjoint nor independent.
Explanation:
The pairs of events that are disjoint (have no intersection) are:
Events B and D are disjoint because if the white die is odd (event B), then it cannot match with the red die (event D).
Events A and D are disjoint because if the sum of the dice is 7 (event A), then the dice cannot match (event D).
The pairs of events that are independent are:
Events B and C are independent because the outcome of one die does not affect the outcome of the other die.
Events A and C are independent for the same reason.
The pairs of events that are neither disjoint nor independent are:
Events A and B are neither disjoint nor independent. If the white die is odd (event B), it is still possible for the sum of the dice to be 7 (event A).
Events A and C are also neither disjoint nor independent. If the sum of the dice is 7 (event A), it is still possible for the red die to have a larger number showing than the white die (event C).
Events B and D are neither disjoint nor independent. If the white die is odd (event B), it is still possible for the dice to match (event D).
In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is infected with the virus is 0.4, independently of other computers. A technician tests the computers, one after another, to see if they are infected.1. What is the probability that she has to test at least 5 computers to find the first (if any) defective one?2. Find the probability that on this day at least 5 computers are infected.
Answer:
(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.
(2) The probability at least 5 computers are infected is 0.949.
Step-by-step explanation:
The probability that a computer is defective is, p = 0.40.
(1)
Let X = number of computers to be tested before the 1st defect is found.
Then the random variable [tex]X\sim Geo(p)[/tex].
The probability function of a Geometric distribution for k failures before the 1st success is:
[tex]P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...[/tex]
Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:
P (X ≥ 5) = 1 - P (X < 5)
= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]
[tex]=1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078[/tex]
Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.
(2)
Let Y = number of computers infected.
The number of computers in the company is, n = 20.
Then the random variable [tex]Y\sim Bin(20,0.40)[/tex].
The probability function of a binomial distribution is:
[tex]P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...[/tex]
Compute the probability at least 5 computers are infected as follows:
P (Y ≥ 5) = 1 - P (Y < 5)
= 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)] [tex]=1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904[/tex]
Thus, the probability at least 5 computers are infected is 0.949.
A company sells its product for $59 per unit. Write an expression for the amount of money received (revenue R) from the sale of x units of the product.
Answer:
[tex]R = 59x[/tex]
Step-by-step explanation:
We are given the following in the question:
Cost per unit = $59
Let x units of product be sold.
We have to write an expression the amount of money received (revenue R) from the sale of x units of the product.
Revenue:
It is the total income and is obtained by multiplying the quantity of goods sold by the unit price of the goods.Revenue =
[tex]\text{Unit Cost}\times \text{Units sold}\\R = 59x[/tex]
is the required expression of revenue.
A fire company keeps two rescue vehicles. Because of the demand on the vehicles and the chance of mechanical failure, the probability that a specific vehicle is available when needed is 90%. The availability of one vehicle is independent of the availability of the other. Find the probability that (a) both vehicles are available at a given time, (b) neither vehicle is available at a given time, and (c) at least one vehicle is available at a given time.
Answer:
(a) P (Both vehicles are available at a given time) = 0.81
(b) P (Neither vehicles are available at a given time) = 0.01
(c) P (At least one vehicle is available at a given time) = 0.99
Step-by-step explanation:
Let A = Vehicle 1 is available when needed and B = Vehicle 2 is available when needed.
Given:
The availability of one vehicle is independent of the availability of the other, i.e. P (A ∩ B) = P (A) × P (B)
P (A) = P (B) = 0.90
(a)
Compute the probability that both vehicles are available at a given time as follows:
P (Both vehicles are available) = P (Vehicle 1 is available) ×
P (Vehicle 2 is available)
[tex]P(A\cap B)=P(A)\times P(B)[/tex]
[tex]=0.90\times0.90\\=0.81[/tex]
Thus, the probability that both vehicles are available at a given time is 0.81.
(b)
Compute the probability that neither vehicles are available at a given time as follows:
P (Neither vehicles are available) = [1 - P (Vehicle 1 is available)] ×
[1 - P (Vehicle 2 is available)]
[tex]P(A^{c}\cap B^{c})=[1-P(A)]\times [1-P(B)]\\[/tex]
[tex]=(1-0.90)\times (1-0.90)\\=0.10\times0.10\\=0.01[/tex]
Thus, the probability that neither vehicles are available at a given time is 0.01.
(c)
Compute the probability that at least one vehicle is available at a given time as follows:
P (At least one vehicle is available) = 1 - P (None of the vehicles are available)
[tex]=1-[P(A^{c})\times P(B^{c})]\\=1-0.01.....(from\ part\ (b))\\ =0.99[/tex]
Thus, the probability that at least one vehicle is available at a given time is 0.99.
The Bradley family owns 410 acres of farmland in North Carolina on which they grow corn and tobacco. Each acre of corn costs $105 to plant, cultivate, and harvest; each acre of tobacco costs $210. The Bradleys have a budget of $52,500 for next year. The government limits the number of acres of tobacco that can be planted to 100. The profit from each acre of corn is $300; the profit from each acre of tobacco is $520. The Bradleys want to know how many acres of each crop to plant to maximize their profit. Formulate a linear programming model for this problem.
Answer:
Step-by-step explanation:
First let's identify decision variables:
X1 - acres of corn
X2 - acres of tobacco
Bradley needs to maximize the profit, MAX = 300X1 + 520X2
The Bradley family owns 410 acres, X1+X2≤410
Each acre of corn costs $105, each acre of tobacco costs $210
The Bradleys have a budget of $52,500
So 105X1 +210X2≤52,500
There is a restriction on planting the tobacco - 100acres
X2≤100
Also, since outcomes can be only positive, X1X2 ≥0
So, what we have:
MAX = 300X1 + 520X2
X1+X2≤410
105X1 +210X2≤52,500
X2≤100
X1X2 ≥0
To maximize profit, the Bradleys can formulate a linear programming model based on the costs and profits of each crop, subject to budget and government constraints.
Explanation:Linear programming model formulation:
Let x be the number of acres of corn and y be the number of acres of tobacco to be planted.Maximize profit: $300x + $520ySubject to constraints: $105x + $210y ≤ $52,500, y ≤ 100, x,y ≥ 0Suppose that the probability of a defective part is 0.03. Suppose that you have a shipment of 1000 parts. What is the probability that more than 10 parts will be defective? Answer to at least five decimal places. You may find it easier to use excel than to use a calculator for this one.
Answer:
The probability that more than 10 parts will be defective is 0.99989.
Step-by-step explanation:
Let X = a part in the shipment is defective.
The probability of a defective part is, P (Defect) = p = 0.03.
The size of the sample is: n = 1000.
Thus, the random variable [tex]X\sim Bin(1000, 0.03)[/tex].
But the sample size is very large.
The binomial distribution can be approximated by the Normal distribution if the following conditions are satisfied:
np ≥ 10n (1 - p) ≥ 10Check the conditions:
[tex]np=1000\times0.03=30>10\\n(1-p)=1000\times(1-0.03)=970>10[/tex]
Thus, the binomial distribution can be approximated by the Normal distribution.
The sample proportion (p) follows a normal distribution.
Mean: [tex]\mu_{p}=0.03[/tex]
Standard deviation: [tex]\sigma_{p}=\sqrt{\frac{p(1-p)}{n} } =\sqrt{\frac{0.03(1-0.03)}{1000} } =0.0054[/tex]
Compute the probability that there will be more than 10 defective parts in this shipment as follows:
The proportion of 10 defectives in 1000 parts is: [tex]p=\frac{10}{1000}=0.01[/tex]
The probability is:
[tex]P(p>0.01)=P(\frac{p-\mu_{p}}{\sigma_{p}}> \frac{0.01-0.03}{0.0054}) =P(Z>-3.704)=P(Z<3.704)[/tex]
Use the standard normal table for the probability.
[tex]P(p>0.01)=P(Z<3.704)=0.99989[/tex]
Thus, the probability that more than 10 parts will be defective is 0.99989.
Determine if the following is consistent Subscript[x, 1] - 2 Subscript[x, 2] + Subscript[x, 3] = 0 Subscript[ , ] 2Subscript[x, 2] - 8Subscript[x, 3] = 8 5Subscript[x, 1] - 5Subscript[x, 3] = 10
Answer:
The systems of equation is CONSISTENT
Step-by-step explanation:
The detailed steps using crammers rule to ascertain the CONSISTENCY is as shown in the attached file
POLYGONS AND CIRCLES
PLEASE HELP
Answer:
Step-by-step explanation:
The formula for finding the sum of the measure of the interior angles in a regular polygon is expressed as (n - 2) × 180.
Where
n represents the number of sides of the polygon.
5a) The polygon has 11 sides. Therefore, sum of angles is
(11 - 2) × 180 = 1620
The measure of each angle is
1620/11 = 148.3°
b) n = 24
Therefore, sum of angles is
(24 - 2) × 180 = 3960
The measure of each angle is
3960/24 = 165°
5a) n = 7
Therefore, sum of angles is
(7 - 2) × 180 = 900
The measure of each angle is
900/7 = 128.6°
5b) n = 10
Therefore, sum of angles is
(10 - 2) × 180 = 1440
The measure of each angle is
1440/10 = 144°
Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter \frac{1}{20}. Let X = the distance people are willing to commute in miles. What is m, μ, and σ? What is the probability that a person is willing to commute more than 25 miles?
Answer:
m = [tex]\frac{1}{20}[/tex] μ = 20σ = 20The probability that a person is willing to commute more than 25 miles is 0.2865.
Step-by-step explanation:
Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.
A random variable X follows an exponential distribution with parameter m.
The decay parameter is, m.
The probability distribution function of an Exponential distribution is:
[tex]f(x)=me^{-mx}\ ;\ m>0, x>0[/tex]
Given: The decay parameter is, [tex]\frac{1}{20}[/tex]
X is defined as the distance people are willing to commute in miles.
The decay parameter is m = [tex]\frac{1}{20}[/tex]. The mean of the distribution is: [tex]\mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20[/tex]. The standard deviation is: [tex]\sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20[/tex]Compute the probability that a person is willing to commute more than 25 miles as follows:
[tex]P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865[/tex]
Thus, the probability that a person is willing to commute more than 25 miles is 0.2865.
Suppose two events A and B are two independent events with P(A) > P(B) and P(A ∪ B) = 0.626 and P(A ∩ B) = 0.144, determine the values of P(A) and P(B).
Answer:
P(A)= 0.606 and P(B)= 0.237
Step-by-step explanation:
Since A and B are independent
P(A ∩ B) = P(A) * P(B) → P(B) = P(A ∩ B) / P(A)
and also
P(A ∪ B)= P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = P(A) + P(A ∩ B) / P(A) - P(A ∩ B)
[P(A ∪ B) + P(A ∩ B) ]* P(A) = P(A)² + P(A ∩ B)
P(A)² - [P(A ∪ B) + P(A ∩ B) ]* P(A) + P(A ∩ B) = 0
P(A)² - [ 0.626+0.144] * P(A) + 0.144 =0
P(A)² - 0.77* P(A) + 0.144 =0
thus
P(A)₁= 0.606 or P(A)₂= 0.1647
for P(A)₁→ P(B)₁ = P(A ∩ B) / P(A)₁ = 0.144/0.606 = 0.237
thus P(A)₁ > P(B)₁ → correct
for P(A)₂→ P(B)₂ = P(A ∩ B) / P(A)₂ = 0.144/0.1647= 0.8743
thus P(A)₂ < P(B)₂ → incorrect
therefore
P(A)= 0.606 and P(B)= 0.237
Find an equation of the line passing through the pair of points (4 comma 5 )and (7 comma 11 ). Write the equation in the form Ax plus By equals Upper C.
Answer:
the line equation is 2*x - y = 3
Step-by-step explanation:
for the line equation in implicit form
A*x + B*y = C
then if the point (x=4,y=5) belong to the line
4*A + 5*B = C
and if the point (x=7,y=11) belong to the line
7*A + 11*B = C
then since we can choose C freely , we set C=1 for simplicity , then
4*A + 5*B = 1 → B= (1-4*A)/5
7*A + 11*B = 1
7*A + 11*(1-4*A)/5 = 1
7*A - 44/5*A + 11/5 = 1
-9/5*A = -6/5
A= 2/3
B= (1-4*A)/5 = (1-4*2/3)/5 = -1/3
therefore
2/3*x - 1/3*y = 1
2*x - y = 3
1. How do you multiply powers with the same base?
2. How do you divide powers with the same base?
3. How do you find the power of a power?
4. Simplify. (x^2*x^4)^3
____________
X^8
Answer:
[tex]\frac{(x^2\times x^4)^3}{x^8}[/tex] [tex]=x^{10}[/tex]
Step-by-step explanation:
1.
Power of the same base multiply by adding their exponent.
Example: [tex]a^m\times a^n = a^{(m+n)}[/tex]
2.
Power of the same base divide by subtracting their exponent.
Example:[tex]a^m \div a^n = a^{(m-n)}[/tex]
3.
Find power of a power we may multiple the exponent.
Example:[tex](a^m)^n = a^{mn}[/tex]
4.
[tex]\frac{(x^2\times x^4)^3}{x^8}[/tex]
[tex]=\frac{(x^6)^3}{x^8}[/tex] [ using multiplication rule]
[tex]=\frac{(x^{18})}{x^8}[/tex] [ using power of power rule]
[tex]=x^{18-8}[/tex] [ Using division rule]
[tex]=x^{10}[/tex]
Marine biologists have been studying the effects of acidification of the oceans on weights of male baluga whales in the Arctic Ocean. One of the studies involves a random sample of 16 baluga whales. The researchers want to create a 95% confidence interval to estimate the true mean weight of male baluga whales. Their data follow a normal distribution. The population standard deviation of weights of male baluga whales is LaTeX: \sigma = 125????=125 kg, and the researchers feel comfortable using this standard deviation for their confidence interval.
Use this information to answer Questions.
. Assuming the relevant requirements are met, calculate the margin of error in estimating the true mean weight of male baluga whales in the Artic Ocean.
15.31 kg
51.40 kg
61.25 kg
80.49 kg
Assuming the relevant requirements are met, what sample size would be required if the researchers wanted the margin of error to be 45 kg?
25
30
35
40
Are the requirements for the use of a confidence interval met? Explain.
Yes. The distribution of sample means is normal because the data are normal.
Yes. The distribution of sample means is normal because the sample size is large.
No. The distribution of sample means is not normal because the sample size is small.
No. The fact that the data are normal does not imply that the distribution of sample means is normal.
Answer: a) margin of error = 61.25, b) sample size when margin of error is 45 = 30
Step-by-step explanation:
The formulae to get the margin of error of a confidence interval is given as
Margin of error = critical value * (σ/√n)
Where σ = population standard deviation = 125
n = sample size = 16
Critical value =Zα/2 = 1.96 ( this is so because we are performing a 95% confidence level test then level of significance (α) will be 5% and since our test is of two values, it will be 2 tailed).
Margin of error = 1.96 * (125/√16)
Margin of error = 1.96 * 125/4
Margin of error = 1.96 * 31.25
Margin of error = 61.25
Question b)
Margin of error = 45
Critical value =Zα/2 = 1.96
Population standard deviation = σ = 125
Sample size =n =??
By recalling the formulae
Margin of error = critical value * (σ/√n)
45 = 1.96 * (125/√n)
45 = (1.96 * 125)/√n
45 = 245/√n
45 * √n = 245
√n = 245/ 45
√n = 5.444
n = (5.444)²
n= 29.64 which is approximately 30.
Final answer:
The margin of error calculation for estimating the mean weight of male baluga whales, determining the necessary sample size, and confirming the requirements for a confidence interval.
Explanation:
The margin of error in estimating the true mean weight of male baluga whales in the Arctic Ocean can be calculated using the formula:
Margin of Error = Z * (Standard Deviation / √sample size)
Given Z value for 95% confidence interval is 1.96,
Margin of Error = 1.96 * (125 / √16) = 61.25 kg.
To find the required sample size for a margin of error of 45 kg:
Set up the formula: Margin of Error = Z * (Standard Deviation / √sample size)Plug in the values: 45 = 1.96 * (125 / √sample size)Solve for the sample size: sample size ≈ 35The requirements for the use of a confidence interval are considered met when the distribution of sample means is normal.
In this case, the distribution of sample means is normal because the sample size is large, ensuring the validity of utilizing a confidence interval for estimation.
3.20 × 105 gallons of tar (SG = 1.20) is stored in a 20.0-ft tall storage tank. What is the total mass of the liquid in the tank?
Answer:
Mass of the liquid in the tank will be [tex]1.45\times 10^8kg[/tex]
Step-by-step explanation:
We have given volume of tar [tex]V=3.20\times 10^5gallon[/tex]
1 gallon = 3.785 liter
So [tex]3.20\times 10^5gallon=3.20\times 10^{5}\times 3.785=12.112\times 10^5liter[/tex]
Specific gravity = 1.2
Density of water [tex]=1000kg/m^3[/tex]
We know that specific gravity [tex]=\frac{density\ of\ liquid}{density\ of\ water}[/tex]
[tex]1.2=\frac{density\ of\ liquid}{density\ of\ water}[/tex]
Density of liquid [tex]=1200kg/m^3[/tex]
So mass of liquid = volume × density
= [tex]12.112\times 10^5\times 1200=1.45\times 10^8kg[/tex]
So mass of the liquid in the tank will be [tex]1.45\times 10^8kg[/tex]
Customers arrive at a checkout counter at times 1,4,5,10,20,22,23,28,29,35. The time it takes fortheir checkout are the amounts 4,4,4,4,3,3,5,1,5. Compute for each customer the wait time fromwhen the customer arrives until the customer begins to get served. Use the formula given in classand in Dai&Park’s text.
Answer:
wait time in order = 0,1,4,3,0,1,3,3,3,2
Step-by-step explanation:
We need to make a table for easy computation of wait time for each customer.
Customer Arrival Time taken Checkout Wait
Serial no. time for checkout time time
1 1 4 1+4= 5 0
2 4 4 4+5= 9 5-4= 1
3 5 4 9+4= 13 9-5= 4
4 10 4 13+4= 17 13-10=3
5 20 3 20+3= 23 0
6 22 3 23+3= 26 23-22=1
7 23 5 26+5= 31 26-23=3
8 28 1 31+1= 32 31-28=3
9 29 5 32+5= 37 32-29=3
10 35 37-35=2
if the equation of m is given by y= ax + c, which of the following represents the perpendicular slope to m?
A. - 1/a
B. -a
C. 1/a
D. c
Answer: A. - 1/a
Step-by-step explanation:
Two lines are said to be perpendicular if the product of their slope equals -1 . That is , if the slope of the first line is [tex]m_{1}[/tex] and the slope of the second line is [tex]m_{2}[/tex] , if they are perpendicular , then :
[tex]m_{1}[/tex] x [tex]m_{2}[/tex] = -1.
Therefore : the perpendicular slope to m is [tex]\frac{-1}{a}[/tex]
A hemispherical bowl of radius a contains water to a depth h. Find the volume of the water in the bowl. b. Water runs into a sunken concrete hemispherical bowl of radius 5 m at the rate of 0.2 m cubed divided by sec. How fast is the water level in the bowl rising when the water is 4 m deep?
The water level in the bowl is rising at a rate of approximately 0.00269 meters per second when the water is 4 meters deep.
Let's address each part of the problem:
a. To find the volume of water in a hemispherical bowl of radius "a" to a depth "h," we can use the formula for the volume of a spherical cap. The volume of a spherical cap is given by:
V = (1/3)πh^2(3a - h)
In this case, "a" is the radius of the hemisphere, and "h" is the depth of the water.
So, the volume of water in the bowl is:
V = (1/3)πh^2(3a - h)
b. To find how fast the water level in the bowl is rising, we can use related rates. Let's denote the radius of the water-filled hemisphere as "R" (which is equal to "a" since it's the same hemisphere), and the depth of the water as "h."
Given that water is running into the bowl at a rate of 0.2 cubic meters per second, we can express the change in volume with respect to time:
dV/dt = 0.2 m^3/sec
We want to find dh/dt, the rate at which the water level is rising when the water is 4 meters deep.
We have the formula for the volume of water in the hemisphere from part (a):
V = (1/3)πh^2(3a - h)
Differentiate both sides of this equation with respect to time (t):
dV/dt = (1/3)π(2h)(dh/dt)(3a - h) - (1/3)πh^2(d(3a - h)/dt)
Now, plug in the values we know:
dV/dt = 0.2 m^3/sec
h = 4 m
a = 5 m
Now, solve for dh/dt:
0.2 = (1/3)π(24)(dh/dt)(35 - 4) - (1/3)π(4^2)(d(3*5 - 4)/dt)
0.2 = (8/3)π(dh/dt)(15 - 4) - (16/3)π(d(15 - 4)/dt)
0.2 = (8/3)π(11)(dh/dt) - (16/3)π(d(11)/dt)
0.2 = (88/3)π(dh/dt) - (16/3)π(0)
Now, solve for dh/dt:
(88/3)π(dh/dt) = 0.2
dh/dt = 0.2 / [(88/3)π]
Now, calculate dh/dt:
dh/dt ≈ 0.00269 meters per second
for such more question on hemispherical bowl
https://brainly.com/question/31998695
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Final answer:
To find the volume of water in a hemispherical bowl, subtract the volume of the upper portion of the hemisphere from the volume of the entire hemisphere. Therefore, the volume of the water in the hemispherical bowl is (2/3)πa³ - (1/3)π(h²)(3a - h).
Explanation:
To find the volume of water in a hemispherical bowl, we need to consider the shape of the bowl. The volume of a hemisphere is given by the formula V = (2/3)πr³, where r is the radius of the hemisphere. However, we only need to find the volume of the water in the bowl, not the entire bowl. To do this, we can subtract the volume of the upper portion of the hemisphere that is not filled with water.
Let's break down the steps:
Find the volume of the entire hemisphere using the formula V = (2/3)πa³, where a is the radius of the bowl.Find the volume of the upper portion of the hemisphere using the formula V' = (1/3)π(h²)(3a - h), where h is the depth of the water in the bowl.The volume of the water in the bowl is given by V - V'.Therefore, the volume of the water in the hemispherical bowl is (2/3)πa³ - (1/3)π(h²)(3a - h).
Find the mean, median, and mode of the following data. If necessary, round to one more decimal place than the largest number of decimal places given in the data.MLB Batting Averages0.3270.2950.3180.3100.2850.2800.3140.3230.3100.2950.2830.2790.2770.3130.3270.3060.3270.3170.2920.275
Answer:
Mean = 0.3026
Median = 0.308
Mode = 0.327
Step-by-step explanation:
We are given the following data set:
0.327, 0.295, 0.318, 0.310, 0.285, 0.280, 0.314, 0.323, 0.310, 0.295, 0.283, 0.279, 0.277, 0.313, 0.327, 0.306, 0.327, 0.317, 0.292, 0.275
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{6.053}{20} = 0.3026[/tex]
Sorted data: 0.275, 0.277, 0.279, 0.280, 0.283, 0.285, 0.292, 0.295, 0.295, 0.306, 0.310, 0.310, 0.313, 0.314, 0.317, 0.318, 0.323, 0.327, 0.327, 0.327
[tex]Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}[/tex]
[tex]\text{Median} = \dfrac{10^{th} + 11^{th}}{2} = \dfrac{0.306 + 0.310}{2} = 0.308[/tex]
Mode is the mos frequent observation in the data.
Mode = 0.327
It appeared 3 times.
The mean of the MLB Batting Averages is around 0.306, the median is 0.308, and the mode is 0.327.
Explanation:In the field of mathematics, especially in statistics, mean, median and mode are measures of central tendency that provide an overview of the data set. Given the MLB Batting Averages, we start by arranging the data in ascending order. After that, we calculate the mean by adding up all the values and dividing by the number of values. The median is the middle value in an ordered dataset, and the mode is the value that appears most often in a dataset.
Mean: (0.327+0.295+0.318+0.310+0.285+0.280+0.314+0.323+0.310+0.295+0.283+0.279+0.277+0.313+0.327+0.306+0.327+0.317+0.292+0.275)/20 ≈ 0.306 Median: The middle numbers are 0.310 and 0.306, so the median is (0.310 + 0.306) / 2 = 0.308 Mode: The number 0.327 appears three times, more than any other number, so the mode is 0.327. Learn more about Central Tendency here:
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Determine whether the statement is true or false. Justify your answer.
If A and B are independent events with nonzero probabilities, then A can occur when B occurs.
Answer:Yes (A can occur when B occurs)
Step-by-step explanation: Independent events are events that are not determined by the occurrence of another. In this case EVENT A CAN OCCUR AS EVENT B OCCURS.
Justification: Landing on tails after tossing a coin and landing on a Five (5) after rolling the DIE Landing on Six (6) after rolling a DIE and landing on tails after tossing a coin.
Independent probabilities can occur together,they don't depend on each other.
A triangle has measures that are 45 45 90 The hypothenuse of the triangle is 10 What is the perimeter of the triangle
Answer: you need to add each side .45+45+90=180
Step-by-step explanation:
Determine if the statement is true or false, and justify your answer. If u4 is not a linear combination of {u1, u2, u3}, then {u1, u2, u3, u4} is linearly independent. True. If {u1, u2, u3, u4} is linearly dependent, then u4
Answer:The statement is true
Step-by-step explanation:
This is an implication, then if we by starting from the premise we can get to the conclusion, then the implication is true. First let us remember that a given set of n vectors, say [tex]\{u_1,\ldots, u_n\}[/tex] is l.i (linearly independent), by definition if the only solution to [tex]\alpha_1\cdot u_1 + \cdots + \alpha_n\cdot u_n = \textbf{0}, is (\alpha_1, \ldots, \alpha_n)= \textbf{0}\in \mathcal {R}^n, \mathcal {R}[/tex] the set of real numbers, that is the only linear combination equal to the null vector is the null combination, all the scalars must be zero. And since it is a definition it is if and only if.
Then if we say that the premise is true, then u4 is not a linear combination of {u1,u2,u3}, this means there do not exist [tex]\alpha_1, \alpha_2, \alpha_3 \in \mathcal{R}[/tex] not all zero, such that, [tex]\alpha_1\cdot u_1+ \alpha_2\cdot u_2+ \alpha_3\cdot u_3=u_4[/tex] which is equivalent to say that for every [tex]\alpha_1, \alpha_2, \alpha_3,\alpha_4 \in \mathcal{R}[/tex] with
[tex]\alpha_1\cdot u_1 + \alpha_2\cdot u_2+ \alpha_3\cdot u_3+ \alpha_4\cdot u_4 = \textbf{0}[/tex] implies [tex]\alpha_1= \alpha_2= \alpha_3= \alpha_4=0[/tex], then the given set is linearly independent as by definition and just as it is stated in the conclusion, therefore the affirmation is true.
The time between breakdowns of an alarm system is exponentially distributed with mean 10 days. What is the probability that there are no breakdowns on a given day?
Answer:
[tex] P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048[/tex]
Step-by-step explanation:
Previous concepts
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]
Solution to the problem
For this case the time between breakdowns representing our random variable T is exponentially distirbuted [tex] T \sim Exp (\mu = 10)[/tex]
So on this case we can find the value of [tex]\lambda[/tex] like this:
[tex] \lambda = \frac{1}{\mu} = \frac{1}{10}[/tex]
So then our density function would be given by:
[tex]P(T)=\lambda e^{-\frac{t}{10}}[/tex]
The exponential distribution is useful when we want to describe the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time between two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:
[tex]P(T>t)= e^{-\lambda t}[/tex]
And on this case we are looking for this probability:
[tex] P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048[/tex]
A patio was to be laid in a design with one tile in the
first row, two tiles in the second row, three
tiles in the third row, and so on. Mr. Tong had 60
tiles to use. How many tiles should be placed
in the bottom row to use the most tiles?
Answer:
10
Step-by-step explanation:
The number of tiles in the design is 1 + 2 + 3 + ...
We can model this as an arithmetic series, where the first term is 1 and the common difference is 1. The sum of the first n terms of an arithmetic series is:
S = n/2 (2a₁ + d (n − 1))
Given that a₁ = 1 and d = 1:
S = n/2 (2(1) + n − 1)
S = n/2 (n + 1)
Since S ≤ 60:
n/2 (n + 1) ≤ 60
n (n + 1) ≤ 120
n must be an integer, so from trial and error:
n ≤ 10
Mr. Tong should use 10 tiles in the final row to use the most tiles possible.
Newlyweds Bryce & lauren need to rent a truck to move their belongings to their new apartment. They can rent a truck of the size they need from U-Haul for $29.95 per day plus 28 cents per mile or from Budget Truck rentals for $34.95 per day plus 25 cents per mile. After how many miles (to the nearest mile) would the budget rental be a better deal than the U-Haul one?
Answer:
After 167 miles the Budget rental deal would be better than the U-haul deal
Step-by-step explanation:
Let x be the number of miles.
Considering the U-haul plan the cost can be expressed as:
[tex]u=29.95+0.28x[/tex]
Consider the Budget rental plan the cost can be expressed as:
[tex]b=34.95+0.25x[/tex]
For the Budget rental plan to be better than the U-haul plan the relation between the two cost equation will be:
[tex]b<u[/tex]
Substitute the equations of u and b and solve for x:
[tex]b<u\\34.95+0.25x<29.95+0.28x\\34.95-29.95<0.28x-0.25x\\5<0.03x\\0.03x>5\\x>166.67\approx167[/tex]
So after 167 miles the Budget rental deal would be better than the U-haul deal.
After approximately 167 miles, Budget Truck rental becomes a cheaper option than U-Haul for Bryce and Lauren's move.
Explanation:To find the point at which Budget Truck rentals becomes cheaper than U-Haul, we must set up the cost equation for each company and solve for the number of miles (m) that makes them equal. For U-Haul, the cost (C) is given by C = 29.95 + 0.28m, and for Budget it’s C = 34.95 + 0.25m.
Setting these two equations equal, we get: 29.95 + 0.28m = 34.95 + 0.25m.
We then solve for m: 0.03m = 5. Subtracting 29.95 from both sides, then dividing by 0.03 gives: m ≈ 167 miles.
So, after 167 miles, Budget Truck becomes the cheaper option.
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An important tool in archeological research is radiocarbon dating, developed by the American chemist Willard F. Libby.3 This is a means of determining the age of certain wood and plant remains, and hence of animal or human bones or artifacts found buried at the same levels. Radiocarbon dating is based on the fact that some wood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. This isotope is accumulated during the lifetime of the plant and begins to decay at its death. Since the half-life of carbon-14 is long (approximately 5730 years),4 measurable amounts of carbon-14 remain after many thousands of years. If even a tiny fraction of the original amount of carbon-14 is still present, then by appropriate laboratory measurements the proportion of the original amount of carbon-14 that remains can be accurately determined. In other words, if Q(t) is the amount of carbon-14 at time t and Q0 is the original amount, then the ratio Q(t)/Q0 can be determined, as long as this quantity is not too small. Present measurement techniques permit the use of this method for time periods of 50,000 years or more.
(a) Assuming that Q satisfies the differential equation Q' = -rQ, determine the decay constant r for carbon-14. (b) Find an expression for Q(t) at any time t, if Q(0) = Qo. (c) Suppose that certain remains are discovered in which the current residual amount of carbon-14 is 20% of the original amount. Determine the age of these remains.
Answer:
a) r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121/year
b) Q(t) = Q₀ (e^-rt)
c) Are of the 20% remnant of Carbon-14 = 13301.14 years.
Step-by-step explanation:
Q' = -rQ
Q' = dQ/dt
dQ/dt = -rQ
dQ/Q = -rdt
Integrating the left hand side from Q₀ to Q₀/2 and the right hand side from 0 to t1/2 (half life, t1/2 = 5730 years)
In ((Q₀/2)/Q₀) = -r(t1/2)
In (1/2) = -r(t1/2)
In 2 = r(t1/2)
r = (In 2)/(t1/2) = (In 2)/5730 = 0.000121 /year
b) Q' = -rQ
Q' = dQ/dt
dQ/dt = -rQ
dQ/Q = -rdt
Integrating the left hand side from Q₀ to Q(t) and the right hand side from 0 to t.
In (Q(t)/Q₀) = -rt
Q(t)/Q₀ = e^(-rt)
Q(t) = Q₀ (e^-rt)
c) Q(t) = Q₀ (e^-rt)
Q(t) = 0.2Q₀, t = ? and r = 0.000121/year
0.2Q₀ = Q₀ (e^-rt)
0.2 = e^-rt
In 0.2 = -rt
-1.6094 = - 0.000121 × t
t = 1.6094/0.000121 = 13301.14 years.
Hope this Helps!
Radiocarbon dating is a method used in archeological research to determine the age of artifacts based on the ratio of carbon-14 to the original amount. This technique is limited to time periods of 50,000 years or more.
Explanation:Radiocarbon dating is an important tool in archeological research that allows scientists to determine the age of wood, plant remains, and other artifacts. This method is based on the fact that some wood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. By measuring the ratio of carbon-14 to the original amount, scientists can accurately determine the age of these remains. Radiocarbon dating is limited to time periods of 50,000 years or more.
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Following is the data of new car color preferences for U.S. buyers. New Car Color Preferences for U.S. Buyers Color Percent Blue 12 Green 7 Natural 12 Red 13 Silver/Grey 24 White 16 Black 13 Other 3 Total 100
Answer:
The chart A is correct
Pareto Chart
Step-by-step explanation:
Given chart is missing (Attached)
Find:
- Which chart represents the correct data.
- What other chart can be used to express the given data
Solution:
- Use the given values for each color and compare with the three charts A,B and C given.
For Blue = A (12) , B(12) , C(11)
For Green = A(7) , B(13) , C(12)
- Hence, The chart A is correct.
- Any other chart which can correctly express the information given should be a chart that uses bars or frequency to expresses the percentages. Pareto Chart expresses both bars and line chart(curve) to express the frequency of the data.
Find the zero of the following function f. Do not use a calculator. f (x )equals 1.5 x plus 3 (x minus 3 )plus 5.5 (x plus 7 )
Answer:
(-2.95,0)
x = -2.95 is the zero of the function.
Step-by-step explanation:
We are given the function:
[tex]f(x) = 1.5x + 3(x-3) + 5.5(x+7)[/tex]
We have to find the zero of the function.
Zero of function:
It is the value where the function have a value zero.[tex](a,0)[/tex] such that [tex]f(a) = 0[/tex]We can find zero of the function in the following manner:
[tex]f(x) = 1.5x + 3(x-3) + 5.5(x+7) = 0\\1.5x + 3x -9 + 5.5x + 38.5 = 0\\(1.5x + 3x + 5.5x) + (38.5-9) = 0\\10x + 29.5 = 0\\10x = -29.5\\x = -2.95[/tex]
Thus, x = -2.95 is the zero of the function.
A slice of pizza whose edges form a 32degrees angle with an outer crust edge 4 inches long was found in a gym locker. What was the diameter of the original pizza?g
Answer:
The diameter was [tex]d=\frac{45}{\pi }[/tex] in.
Step-by-step explanation:
The arc of a circle is given by
[tex]s=r\theta[/tex]
where
s = arc length
r = radius of the circle
θ = measure of the central angle in radians.
From the information given
s = 4 in
θ = 32º
To find the diameter of the original pizza, we use the formula of the diameter of a circle
[tex]d=2r[/tex]
First, we need to convert the angle to radians
[tex]\theta=32\º \cdot \frac{\pi}{180\º} =\frac{8\pi }{45}[/tex]
Next, solve for r from the arc formula
[tex]r=\frac{s}{\theta} =\frac{4}{\frac{8\pi }{45}} =\frac{45}{2\pi }[/tex]
Then, we use the diameter of a circle formula
[tex]d=2r=2(\frac{45}{2\pi })=\frac{45}{\pi }[/tex]
Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year.
Answer:
[tex]n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689[/tex]
So the answer for this case would be n=689 rounded up to the nearest integer
Step-by-step explanation:
Assuming this complete question: "Suppose that the minimum and maximum ages for typical textbooks currently used in college courses are 0 and 8 years. Use the range rule of thumb to estimate the standard deviation.
Estimate the minimum and maximum ages for typical textbooks currently used in college courses, then use the range rule of thumb to estimate the standard deviation. Next, find the size of the sample required to estimate the mean age (in years) of textbooks currently used in college courses. Use a 90% confidence level and assume that the sample mean will be in error by no more than 0.25 year."
Solution for the problem
First we need ti find the estimation for the standard deviation using the Rule of thumb, with the following formula:
[tex] s \approx \frac{R}{4}[/tex]
Where R is the range defined as :
[tex] R = Max - Min = 8-0 = 8[/tex]
So then the deviation would be approximately:
[tex] s \approx 4[/tex]
Important concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error (ME) is the range of values below and above the sample statistic in a confidence interval.
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 margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
And on this case we have that [tex]ME =\pm 0.25[/tex] and we are interested in order to find the value of n, if we solve n from equation (1) we got:
[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (2)
We can assume that the estimator for the population deviation from the rule of thumb is [tex]\hat \sigma = s= 4[/tex]
The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.64[/tex], replacing into formula (2) we got:
[tex]n=(\frac{1.64(4)}{0.25})^2 =688.537 \approx 689[/tex]
So the answer for this case would be n=689 rounded up to the nearest integer
To estimate textbook ages, an assumed range of 1 to 20 years might give an estimated standard deviation of 4.75 years. Using the formula for sample size, about 275 textbooks would need to be sampled to estimate the mean age to within 0.25 years with 90% confidence.
Explanation:To estimate the minimum and maximum ages for typical textbooks currently used in college courses, we need more specific information. However, if we assume a range of 1 to 20 years, this would give us a range of 19 years. We can use the range rule of thumb for estimating the standard deviation, which is the range divided by 4. So the estimated standard deviation of the textbook ages is 19 / 4 = 4.75 years.
To find the required sample size to estimate the mean age of textbooks, we use the equation n = (Z*σ/E)^2 where Z is the z-score corresponding to the desired confidence level, σ is the standard deviation, and E is the maximum tolerable error. For a 90% confidence level, the z-score is 1.645. Thus with σ = 4.75 and E = 0.25. Filling these values into our equation we get: n = (1.645*4.75/0.25)^2 = 274.68. Rounded up, we need a sample of size 275 textbooks to estimate the mean age to within 0.25 years with 90% confidence.
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7. Martin sells cars. He earns $100 per day, plus
any commission on his sales. His daily salary s in
dollars depends on the amount of commission c.
Write an equation to represent his daily salary.
Answer:the equation representing his daily salary is
s = 100 + c
Step-by-step explanation:
Let s represent the Salary that Martins earns per day.
Let c represent the commission that Martins earns on his sales on that day.
He earns $100 per day, plus
any commission on his sales. Since
his daily salary in dollars depends on the amount of commission, then an equation to represent his daily salary would be
s = 100 + c