Answer:
P1: An ant releases a chemical when it dies, and its fellows then carry it away to the compost heap.
P2: A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
C: Apparently the communication is highly effective
Step-by-step explanation:
Premises are statements that are always assumed to be truth or statements that have been given and made to be true.
Premises will be labelled as P1, P2, P3..........
Conclusions are derived statements from premises. Its truthfulness is derived from the truthfulness of premises. Meaning that, when the premises is true, the conclusion is also true.
Conclusions will be labelled as C1, C2, C3........
In the question above, it's noted that the conclusion starts at "Apparently the" while the premises are statements before "Apparently the"
The argument's premises are as follows:
1. An ant releases a chemical when it dies.
2. Its fellow ants then carry the dead ant to the compost heap.
3. A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
The conclusion is that the communication among ants when one of them dies is highly effective.
In the given passage, there is an implicit argument about the effectiveness of communication among ants when one of them dies. Let's identify the premises and the conclusion:
Premises:
1. An ant releases a chemical when it dies.
2. Its fellow ants then carry the dead ant to the compost heap.
3. A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
Conclusion:
4. Apparently, the communication among ants when one of them dies is highly effective.
These premises and conclusion can be organized as follows:
Premise 1: An ant releases a chemical when it dies.
Premise 2: Its fellow ants then carry the dead ant to the compost heap.
Premise 3: A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
Conclusion: Apparently, the communication among ants when one of them dies is highly effective.
In this argument, premises 1, 2, and 3 provide evidence or observations related to ant behavior when a member of their colony dies. The conclusion, stated in premise 4, is drawn from these observations and suggests that the communication system among ants regarding the death of a fellow ant is effective.
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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
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
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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).
a marketing survey compiled data on the total number of televisions in households where k is a positie constant. What is the probability that a randomly chosen household has at least two televisions?
The probability that a randomly chosen household has at least two televisions can be calculated by dividing the number of households with at least two televisions by the total number of households.
Explanation:To find the probability that a randomly chosen household has at least two televisions, we need to use the data from the marketing survey. Let's assume there are 'n' households in total. The probability of a household having at least two televisions can be calculated by dividing the number of households with at least two televisions by the total number of households, which is 'n'.
Let's say there are 'x' households with at least two televisions. The probability can be expressed as:
P(at least 2 TVs) = x/n
For example, if there are 100 households in total and 30 of them have at least two televisions, then the probability would be P(at least 2 TVs) = 30/100 = 0.3, which is 30%.
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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.
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]
Suppose 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.
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.
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.
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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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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.
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.
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 ≥ 0The parents of three children, ages 1, 3, and 6, wish to set up a trust fund that will pay X to each child upon attainment of age 18, and Y to each child upon attainment of age 21. They will establish the trust fund with a single investment of Z. Find the equation of value for Z. ?
To calculate the initial trust fund deposit for three children of diverse ages, you create an equation reflecting the present value of the future payouts for each child at ages 18 and 21. The equation of value here models the required single investment with respect to the annual interest rate.
Explanation:The problem presented, concerning setting up a trust fund for three children aged 1, 3, and 6, involves future value of lump sum investments, compound interest, and time value of money. Using an equation of value approach, the equation for the single investment Z made today that will pay each child X at age 18 and Y at age 21 can be modeled as follows:
[tex]Z= (X/(1+r)^{(18-1)} + Y/(1+r)^{(21-1)}) + (X/(1+r)^{(18-3)} + Y/(1+r)^{(21-3)} ) + (X/(1+r)^{(18-6)} + Y/(1+r)^{(21-6)})[/tex]where r represents the annual interest rate. This equation reflects the present value of the future payouts for each child. The term (1+r)years until payout is used to calculate the present value of each future payment. The equation adds up these present values for each child to calculate the initial trust fund deposit (Z).
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If a person rolls a six dash sided dierolls a six-sided die and then flips a coinflips a coin, describe the sample space of possible outcomes using 1 comma 2 comma 3 comma 4 comma 5 comma 61, 2, 3, 4, 5, 6 for the diedie outcomes and Upper H comma Upper TH, T for the coincoin outcomes. (Make sure your answers reflect the order stated.) The sample space is Sequals=_________{nothing}.
Answer:
S={1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T}
Step-by-step explanation:
The sample space for a six sided die is {1,2,3,4,5,6} as there are 6 possible outcomes and for flipping a coin is {H,T} as there are two possible outcomes.
If a person rolls a six sided die and then flips a coin then the sample space for this event will be written as
S={1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T}.
The number of elements in the sample space are
n(S)=12.
Answer:
Step-by-step explanation:If a person draws a playing card and checks its color and then spins a six dash space spinner, describe the sample space of possible outcomes using Upper B comma Upper R for the card outcomes and 1 comma 2 comma 3 comma 4 comma 5 comma 6 for the spinner outcomes. (Make sure your answers reflect the order stated.)
The sample space is Sequals{
012
12}.
(Use a comma to separate answers as needed.)
A faculty leader was meeting two students in Paris, one arriving by train from Amsterdam and the other arriving from Brussels at approximately the same time. Let A and B be the events that the trains are on time, respectively. If P(A) = 0.93, P(B) = 0.89 and P(A \ B) = 0.87, then find the probability that at least one train is on time.
Answer: P(AUB) = 0.93 + 0.89 - 0.87 = 0.95
Therefore, the probability that at least one train is on time is 0.95.
Step-by-step explanation:
The probability that at least one train is on time is the probability that either train A, B or both are on time.
P(A) = P(A only) + P(A∩B)
P(B) = P(B only) + P(A∩B)
P(AUB) = P(A only) + P(B only) + P(A∩B)
P(AUB) = P(A) + P(B) - P(A∩B) ......1
P(A) = 0.93
P(B) = 0.89
P(A∩B) = 0.87
Then we can substitute the given values into equation 1;
P(AUB) = 0.93 + 0.89 - 0.87 = 0.95
Therefore, the probability that at least one train is on time is 0.95.
Solve, graph, and give interval notation for the compound inequality:
−4x + 1 > 13 AND 4(x + 2) ≤ 4
Answer:
The answer to your question is below
Step-by-step explanation:
Inequality 1
-4x + 1 > 13
-4x > 13 - 1
-4x > 12
x < 12/-4
x < -3
Inequality 2
4(x + 2) ≤ 4
4x + 8 ≤ 4
4x ≤ 4 - 8
4x ≤ - 4
x ≤ -4/4
x ≤ -1
Interval notation (-∞, -3)
See the graph below
The solution to these inequalities in where both intervals crosses
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 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:
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°
How much work is required to lift a 1400-kg satellite to an altitude of 2⋅106 m above the surface of the Earth? The gravitational force is F=GMm/r2, where M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the satellite and the Earth's center. The radius of the Earth is 6.4⋅106 m, its mass is 6⋅1024 kg, and in these units the gravitational constant, G, is 6.67⋅10−11.
To lift a 1400-kg satellite to an altitude of 2×10⁶ meters above the Earth's surface, the work required is approximately 2.079×10¹¹ Joules. This is calculated based on changes in gravitational potential energy using given values for mass, radius, and the gravitational constant.
To calculate the work required to lift a 1400-kg satellite to an altitude of 2×10⁶ meters (2000 km) above the Earth's surface, we need to consider the gravitational potential energy change.
Given:
Mass of the satellite (m): 1400 kgAltitude above Earth's surface (h): 2×10⁶ mRadius of the Earth (Rₑ): 6.4×10⁶ mMass of the Earth (M): 6×10²⁴ kgGravitational constant (G): 6.67×10⁻¹¹ N·m²/kg²The total distance from the center of the Earth to the satellite is:
r = Rₑ + h = 6.4×10⁶ m + 2×10⁶ m = 8.4×10⁶ m.The work required is equal to the change in gravitational potential energy:
The gravitational potential energy at a distance r from the Earth’s center is given by:
U = -GMm/rThe work done (W) to move the satellite from the Earth's surface to this altitude is the difference in potential energy:
W = GMm (1/Rₑ - 1/r)Substitute the given values:
W = (6.67×10⁻¹¹ N·m²/kg²)(6×10²⁴ kg)(1400 kg) [(1/6.4×10⁶ m) - (1/8.4×10⁶ m)]Calculate the values inside the brackets first:
(1/6.4×10⁶ - 1/8.4×10⁶) ≈ 1.5625×10⁻⁷ - 1.1905×10⁻⁷ ≈ 0.372×10⁻⁷Now, multiply:
W ≈ 6.67×10⁻¹¹ × 6×10²⁴ × 1400 × 0.372×10⁻⁷W ≈ 2.079×10¹¹ JoulesThe work required to lift the satellite to the desired altitude is approximately 2.079×10¹¹ Joules.
You go to the Huron Valley Humane Society so you can adopt a dog.
For each random variable below, determine whether it is categorical, quantitative discrete, or quantitative continuous.
a) The number of days (to the nearest day) the dog has been at Huron Valley Humane Society
b) Whether or not the dog has a microchip
c) The breed of the dog
d) How much the dog weighs (in pounds)
e) The amount of food (in cups) the dog eats
f) The number of people who have taken the dog out for a walk
g) Whether you decide to adopt the dog
Answer:
a) quantitative discrete data
b) categorical variable
c) categorical variable
d) quantitative continuous variable
e) quantitative discrete variable
f) categorical variable
Step-by-step explanation:
Categorical variable are the non parametric variables. Their value cannot be expressed in the form of numerical.Quantitative discrete are the variables whose values are expressed in whole numbers. These variable cannot take decimal values and hence cannot take all the values within interval.Quantitative continuous variable values can be expressed in the form of decimals and they can take any value within an interval.a) The number of days (to the nearest day) the dog has been at Huron Valley Humane Society
Since days will always have whole number numerical values it is a quantitative discrete data.
b) Whether or not the dog has a microchip
This will be answered with a yes or a no. Thus, it is a categorical variable. It does not have any numerical value.
c) The breed of the dog
This is a categorical variable. It does not have any numerical value. It will have non-parametric values.
d) How much the dog weighs (in pounds)
Since weight is always measured and not counted. Also weight can take decimal values, thus it is quantitative continuous variable.
e) The amount of food (in cups) the dog eats
Since this will have whole number values. Also, number of cups will be counted. Thus, it is a quantitative discrete variable.
f) The number of people who have taken the dog out for a walk
Since this will have whole number values. Also, number of people will be counted. Thus, it is a quantitative discrete variable.
g) Whether you decide to adopt the dog
This will be answered with a yes or a no. Thus, it is a categorical variable. It does not have any numerical value.
a,f are quantitave discrete b,c,g are categorical d,e,are quantitative continous.
a.he number of days (to the nearest day) the dog has been at Huron Valley Humane Society: This is quantitative discrete data because the number of days is counted in whole numbers.
b.Whether or not the dog has a microchip: This is categorical data because it categorizes the dog as either having or not having a microchip.
c.The breed of the dog: This is categorical data because breeds represent different categories.
d.How much the dog weighs (in pounds): This is quantitative continuous data because weight can be measured to a very fine degree.
e.The amount of food (in cups) the dog eats: This is quantitative continuous data because the amount can be measured to any level of precision.
f.The number of people who have taken the dog out for a walk: This is quantitative discrete data because it involves counting distinct individuals.
g.Whether you decide to adopt the dog: This is categorical data because it categorizes your decision into adopt or not adopt.
chang drank 10 fluid ounces of juice. how much is this in cups? write your answer as a whole number or a mixed number in simplest form.
Answer:
1 1/4 cups
Step-by-step explanation:
Their are 8 ounces in one cup and their are 2 ounces in 1/4's of a cup
CUP FL. OZ
1 8
3/4 6
2/3 5
1/2 4
1/3 3
1/4 2
1/8 1
1/16 0.5
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
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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What is the standard deviation of a random variable q with the following probability distribution? (Do not round intermediate calculations. Enter your answer in numbers not in percentage. Round your answer to 4 decimal places.) Value of q Probability 0 0.25 1 0.25 2 0.50
Answer:
The standard deviation of given probability distribution is 0.8292
Step-by-step explanation:
We are given the following in the question:
q: 0 1 2
Probability: 0.25 0.25 0.50
Formula:
[tex]E(q) = \displaystyle\sum q_ip(q_i)\\=0(0.25) + 1(0.25) + 2(0.50) = 1.25[/tex]
[tex]E(q^2)= \displaystyle\sum q_i^2p(q_i)\\=0^2(0.25) + 1^2(0.25) + 2^2(0.50) = 2.25[/tex]
Variance =
[tex]\sigma^2 = E(q^2) = (E(q))^2\\= 2.25- (1.25)^2\\=0.6875\\\sigma = \sqrt{0.6875} = 0.8292[/tex]
Thus, the standard deviation of given probability distribution is 0.8292
The standard deviation of a random variable can be calculated using the formula: √∑(xi-μ)2 ⋅ P(xi) . We find the mean (μ) by multiplying each value of q by its corresponding probability and summing them up. Next, we calculate the squared difference between each value of q and the mean, multiplied by their respective probabilities. Finally, we take the square root of this value to obtain the standard deviation.
Explanation:The standard deviation of a random variable can be calculated using the formula:
\sqrt{\sum (x_i - \mu)^2 \cdot P(x_i)}
First, we need to find the mean (μ) of the probability distribution. The mean can be calculated by multiplying each value of q by its corresponding probability and summing them up:
μ = (0 × 0.25) + (1 × 0.25) + (2 × 0.5) = 0.5
Next, we calculate the squared difference between each value of q and the mean, multiplied by their respective probabilities:
(0 - 0.5)^2 × 0.25 + (1 - 0.5)^2 × 0.25 + (2 - 0.5)^2 × 0.5 = 0.5
Finally, we take the square root of this value to obtain the standard deviation:
√0.5 ≈ 0.7071
Therefore, the standard deviation of the random variable q is approximately 0.7071.
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SOLUTION: if you know the names of the remaining 5 students in the spelling bee, what is the probability of randomly selecting an order and getting the order that is used in the spelling bee
There are
[tex]5! = 5\cdot 4 \cdot 3 \cdot 2 = 120[/tex]
possible arrangements of 5 students. So, if you pick a particular one, you'll have a probability of 1/120 to guess the correct one.
The probability of getting the order used in the spelling bee is 1/120 or approximately 0.0083.
Explanation:The probability of randomly selecting an order and getting the order used in the spelling bee is 1 divided by the number of possible orders. To find the number of possible orders, we use the concept of permutations. If there are 5 students participating in the spelling bee, there are 5 factorial (5!) possible orders. So the probability is 1/5! which is equal to 1/120 or approximately 0.0083.
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A farmer caught some wild pigs to use as founders for a domestic herd. The farmer wants all of her pigs to have one distinctive color. Most of the wild pigs are brown, but she catches a few with red hair and a few with black hair. The farmer plans to breed either the black-haired or the red-haired pigs to establish a pure-breeding herd (eliminating all other colors). Black hair (B) is dominant over brown (W). Black (B) and brown (W) are dominant over red (R). If the farmer wants her whole herd to reach its new color in as few generations as possible, should she breed for red-haired pigs, or should she breed for black-haired pigs? Why?
Answer: Breed for black-haired pigs
Step-by-step explanation:
Now according to the statement, the farmer catches wild pigs and mostly are brown.
Since black hair is dominant over brown and also black and brown are dominant over red, this implies that red is the least dominant color meaning if breeding occurs, there would be less chance of getting red colored pigs.
She wants to breed either black or red haired pigs and since black is more dominant so she should breed the black ones with the brown ones or with themselves so she would eventually get pure-breeding herd i.e. of black color.
The heights, in inches, of the starting five players on a college basketball team are 6868, 7373, 7777, 7575, and 8484. Considering the players as a sample, the mean and standard deviation of the heights are 75.475.4 inches and 5.95.9 inches, respectively. When the players are regarded as a population, the mean and standard deviation of the heights are 75.475.4 inches and 5.25.2 inches, respectively. Explain why, numerically, the sample mean of 75.475.4 inches is the same as the population mean but the sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches
Answer:
The sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches because of their formulas for calculating it.
Step-by-step explanation:
We are given the heights, in inches, of the starting five players on a college basketball team ;
68, 73, 77, 75 and 84
Now whether we treat this data as sample data or population data, the mean height would remain same in both case because the formula for calculating mean is given by ;
Mean = Sum of all data values ÷ No. of observations
Mean = ( 68 + 73 + 77 + 75 + 84 ) ÷ 5 = 75.4 inches
So, numerically, the sample mean of 75.4 inches is the same as the population mean.
Now, coming to standard deviation there will be difference in both sample and population standard deviation and that difference occurs due to their formulas;
Formula for sample standard deviation = [tex]\frac{\sum (X_i - Xbar)^{2} }{n-1}[/tex]
where, [tex]X_i[/tex] = each data value
X bar = Mean of data
n = no. of observations
Sample standard deviation = [tex]\frac{ (68 - 75.4)^{2} +(73 - 75.4)^{2}+(77- 75.4)^{2}+(75- 75.4)^{2}+(84 - 75.4)^{2} }{5-1}[/tex]
= 5.9 inches
Whereas, Population standard deviation = [tex]\frac{\sum (X_i - Xbar)^{2} }{n}[/tex]
= [tex]\frac{ (68 - 75.4)^{2} +(73 - 75.4)^{2}+(77- 75.4)^{2}+(75- 75.4)^{2}+(84 - 75.4)^{2} }{5}[/tex] = 5.2 inches .
So, that's why sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches only because of formula.
According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe people's bad habits as they sneeze.a. What is the probability that among 10 randomly observed individuals exactly 4 do not cover their mouth when sneezing?b. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth when sneezing?c. Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? why?
Answer and Step-by-step explanation:
From the question statement we get know that it is Binomial distribution because there are only two possible outcomes so we need to use Binomial Probability Distribution for this question.
Formula for the Binomial Probability Distribution:
P(X)= p^x q^(n-x)
Where,
C_x^n=n!/(n-x)!x! (i.e. combination)x= total number of successes p=probability of success (p=1-q) q=probability of failure (q=1-p) n=number of trials P(X)= probability of total number of successesAnswer and explanation for each part of the question are as follow:
a.What is the probability that among 10 randomly observed individuals exactly 4 do not cover their mouth when sneezing?
Solution:
Given that
n=10
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=4 (number of successes i.e. individuals not covering their mouths)
C_x^n=n!/(n-x)!x!=10!/(10-4)!4!=210
P(X)=C_x^n p^x q^(n-x)=210×〖(0.267)〗^4×〖0.733〗^(10-4)
P(X)=210×0.00508×0.155
P(X)=0.165465
b. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth when sneezing?
Solution:
Given that
n=10
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=3 (number of successes i.e. individuals not covering their mouths)
C_x^n=n!/(n-x)!x!=10!/(10-3)!3!=120
P(X)=C_x^n p^x q^(n-x)=120×(0.267)^3×〖0.733〗^(10-3)
P(X)=120×0.01903×0.1136
P(X)=0.25962
c. Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? why?
Solution:
Given that
n=18
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=9 (x is the number of successes “number of individuals not covering their mouths when sneezing”, if less than half cover their mouth then more than half will not cover), so let x=9
C_x^n=n!/(n-x)!x!=18!/(18-9)!9!=48620
P(X)=48620×(0.267)^9×〖0.733〗^(18-9)
P(X)=48620×0.00000689×0.0610
P(X)=0.020
Yes, I am surprised that probability of less than 9 individuals covering their mouth when sneezing is 0.020. Which is extremely is small.
How many equations can you write that will equal 100
Step-by-step explanation:
Infinitely many equations can be written that will be equal to 100.
x + y = 100
2x - y = 100
and many more..
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]