In one week, Mohammed can knit 5 sweaters or bake 240 cookies. In one week Aisha can knit 15 sweaters or bake 480 cookies. Mohammed's opportunity cost knitting one sweater is: A. 480 cookies. B. 240 cookies. C. 48 sweaters. D. 1/48 of a cookie E. 48 cookies.

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:

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.        

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

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:

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.

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°

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.

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:

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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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]

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.

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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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

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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:

Therefore, the volume of the water in the hemispherical bowl is (2/3)πa³ - (1/3)π(h²)(3a - h).

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.

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.

(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.

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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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:

Check 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.

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.)

Answer: you need to add each side .45+45+90=180

Step-by-step explanation:

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.

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.

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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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

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.

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.

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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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:

Revenue =

[tex]\text{Unit Cost}\times \text{Units sold}\\R = 59x[/tex]

is the required expression of revenue.

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.

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.

Linear programming model formulation:

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.

For more such questions on argument's premises:

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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

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

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]

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..

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.

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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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

Answer:

The 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.

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.