Exercise 3.23 introduces a husband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes

(a) What is the probability that their first child will have green eyes and the second will not?

(b) What is the probability that exactly one of their two children will have green eyes?

(c) If they have six children, what is the probability that exactly two will have green eyes?

(d) If they have six children, what is the probability that at least one will have green eyes?

Final answer:

The convection heat transfer coefficient of air at the upper surface of the plate is found to be 133.33 W/m·K by using the Fourier's law of heat conduction in conjunction with Newton's law of cooling and rearranging the expression to solve for the convection heat transfer coefficient.

Explanation:

To determine the convection heat transfer coefficient of air at the upper surface of the plate, we apply Fourier's law of heat conduction for a steady-state situation and Newton's law of cooling:

Fourier's law gives us:

Q = -kA(dT/dx)

Where:

Q is the rate of heat transfer (Watts),

k is the thermal conductivity of the material (W/m·K),

A is the surface area (m²),

(dT/dx) is the temperature gradient through the material (K/m).

We then use Newton's law of cooling:

Q = hA(T_surface - T_fluid)

Where:

h is the convection heat transfer coefficient (W/m·K),

T_surface is the surface temperature (K),

T_fluid is the fluid temperature (K).

We know the temperature difference across the plate (dT) is 10°C (60°C - 50°C) and assuming the heat transfer rate (Q) is the same throughout the plate due to steady-state conditions, we can rearrange and solve for h:

h = (k/d) (dT/(T_surface - T_fluid))

Given the values:

k = 80 W/m·K,

d = 15 cm = 0.15 m,

50°C (T_surface) - 10°C (T_fluid) = 40 K.

Substituting these into the equation gives us:

h = (80/0.15) (10/40) = 133.33 W/m·K

The convection heat transfer coefficient of air at the upper surface is [tex]\( 133.33 \) W/m^\²\cdotK.[/tex]

[tex]\[ q = h_1 (T_{s1} - T_{\infty}) = \frac{k}{L} (T_{s1} - T_{s2}) = h_2 (T_{s2} - T_{\infty}) \][/tex]

where:

- [tex]\( q \)[/tex] is the heat transfer rate per unit area[tex](W/m^\²)[/tex],

- [tex]\( h_1 \)[/tex] and [tex]\( h_2 \)[/tex] are the convection heat transfer coefficients at the upper and lower surfaces, respectively [tex](W/m^\²K)[/tex],

- [tex]\( T_{s1} \)[/tex] and [tex]\( T_{s2} \)[/tex] are the temperatures at the upper and lower surfaces of the plate, respectively (°C),

-[tex]\( T_{\infty} \)[/tex] is the ambient air temperature (°C),

- [tex]\( k \)[/tex] is the thermal conductivity of the plate [tex](W/m\cdot K)[/tex],

- [tex]\( L \)[/tex] is the thickness of the plate (m).

Given:

- [tex]\( T_{s1} = 50 \)\°C[/tex] (temperature at the upper surface),

- [tex]\( T_{s2} = 60 \)\°C[/tex] (temperature at the lower surface),

- [tex]\( T_{\infty} = 10 \)\°C[/tex] (air temperature),

- [tex]\( k = 80 \) W/m\cdotK[/tex] (thermal conductivity of the plate),

- [tex]\( L = 0.15 \)[/tex] m (thickness of the plate).

Since we want to find [tex]\( h_1 \)[/tex], we can equate the expressions for [tex]\( q \)[/tex] that involve [tex]\( h_1 \)[/tex] and [tex]\( k \)[/tex]:

[tex]\[ h_1 (T_{s1} - T_{\infty}) = \frac{k}{L} (T_{s1} - T_{s2}) \][/tex]

Now, solve for [tex]\( h_1 \)[/tex]:

[tex]\[ h_1 = \frac{k}{L} \cdot \frac{(T_{s1} - T_{s2})}{(T_{s1} - T_{\infty})} \][/tex]

Substitute the given values:

[tex]\[ h_1 = \frac{80 \text{ W/m\cdotK}}{0.15 \text{ m}} \cdot \frac{(50 \text{ \°C} - 60 \text{ \°C})}{(50 \text{ \°C} - 10 \text{ \°C})} \][/tex]

[tex]\[ h_1 = \frac{80}{0.15} \cdot \frac{-10}{40} \][/tex]

[tex]\[ h_1 = 533.33 \cdot \frac{-1}{4} \][/tex]

[tex]\[ h_1 = -133.33 \text{ W/m}^2\text{\cdotK} \][/tex]

The negative sign indicates that the heat transfer is from the plate to the air, which is expected. However, the convection heat transfer coefficient is typically reported as a positive value, so we take the absolute value:

[tex]h_1 = -133.33 {(\frac{W}{m}})^2} { \cdot K}[/tex]

Answer:

I also gave aadded the complete question for the expressions of the function f(x)

a) Exact value of the net signed area = 1/2 or 0.5

b) Exact average value of f = 53/30 = 1.7667

Step-by-step explanation:

The step by step explanation and calculation is attached below.

Answer:

36

Step-by-step explanation: thats what i got

Answer:the speed of postman from A to B is 6 km/hr

Step-by-step explanation:

Let x represent the speed of the postman when moving from A to B.

From point A, the postman delivered a letter to point B in half hour.

Distance = speed × time

It means that

Distance from A to B = 0.5 × x = 0.5x

In back way he reduced speed by 1km/h and gets back in 36 min. It means that his speed on returning back would be (x - 1)km/h

Converting 36 minutes to hour, it becomes

36/60 = 0.6 hours

Distance from B to A = 0.6(x - 1)

Since distance from A to B = distance from B to A, then

0.5x = 0.6(x - 1) = 0.6x - 0.6

0.6x - 0.5x = 0.6

0.1x = 0.6

x = 0.6/0.1 = 6

Answer:

Definitely

Step-by-step explanation:

In order to completely surround a circle, you need six circles to do that while here in the question it is mentioned that currently there are only five circles surrounding the circle, hence there is enough room to easily fit one more circle.

I just hope that you are satisfied with the answer, Best of Luck.

Answer: 0.025

Step-by-step explanation: we reject null hypothesis if p<0.05

Answer:

[tex]\frac{a(450-8a^2)}{4}[/tex]

is the volume in terms of side a

Step-by-step explanation:

Given that you I built a storage shed in the shape of a rectangular box on a square base. The material that I used for the base cost $4 per square foot, the material for the roof cost $2 per square foot, and the material for the sides cost $2.50 per square foot; and I spent $450 altogether on material for the shed.

Let a be the side of square and h be the height

Total cost of materials = cost for floor + cost for roof + cost for sides

= area of floor (4) + lateral area (2.50)+roof area (2)

[tex]=4a^2+4ah+4a^2\\=8a^2+4ah = 450[/tex][tex]h = \frac{450-8a^2}{4a}[/tex]

Now coming to volume

Volume = V = lbh = [tex]a^2 h\\= a^2*\frac{450-8a^2}{4a} \\=\frac{a(450-8a^2)}{4}[/tex]

Final answer:

To answer the student's combinatorics question, there are a total of 8008 potential delegations, 84 of which are all men and 7924 that include at least one woman.

Explanation:

The question involves combinatorics, which is a branch of mathematics. In particular, we are dealing with combinations since the order of selection does not matter for the delegation.

Number of possible delegations:

To find the total number of delegations possible, we must select 6 individuals out of 16 (7 women + 9 men) without regard to order. This is done using the combination formula:

C(n, k) = n! / (k!(n-k)!)

So, C(16, 6) = 16! / (6!(16-6)!) = 8008

Delegations with all men:

To find the number of delegations with all men, we select 6 men out of the 9 available. Using the combination formula again:

C(9, 6) = 9! / (6!(9-6)!) = 84

Delegations with at least one woman:

To find this, we subtract the number of all-male delegations from the total number of delegations:

8008 - 84 = 7924

So, there are 7924 delegations that include at least one woman.

Final answer:

To answer the student's question about the total number of possible delegations, all male delegations, and delegations with at least one woman, combinations are used. There are 8008 total possible delegations, 84 all-male delegations, and 7924 delegations with at least one woman.

Explanation:

To solve this problem, we will use combinatorics, specifically the concept of combinations as the order in which the delegation members are selected does not matter.

a) Total number of delegations possible

The total number of ways to choose 6 members from a board of 16 (7 women + 9 men) is calculated using the combination formula C(n, k) = n!/(k!(n-k)!), where n is the total number of items, and k is the number of items to choose.

Therefore, C(16, 6) = 16!/(6!*(16-6)!) = 8008 different delegations are possible.

b) Delegations with all men

To find the number of all-male delegations, we choose 6 men from a group of 9, which is C(9, 6) = 9!/(6!*(9-6)!) = 84 delegations.

c) Delegations with at least one woman

Instead of calculating this directly, we use the complement principle. We subtract the number of all-male delegations from the total number of delegations. Thus, the number of delegations with at least one woman is 8008 - 84 = 7924 delegations.

Answer:Amy used 4 41-cent stamps and 8 6-cent stamps.

Step-by-step explanation:

Let x represent the number of 41-cent stamps that Amy used. Let y represent the number of 6-cent stamps that Amy used.

41 cents = 41/100 = $0.41

6 cents = 6/100 = $0.06

She uses 41-cent stamps and 6-cent stamps to pay $2.12 in postage. It means that

0.41x + 0.06y = 2.12

Multiplying through by 100, it becomes

41x + 6y = 212

6y = 212 -41x

We would test for corresponding values of x and y that satisfies the equation and they must be whole numbers.

If x = 3,

6y = 212 - 41 × 3 = 89

y = 89/6 = 14.8333

If x = 4,

6y = 212 - 41 × 4 = 48

y = 48/6 = 8

To solve this problem, we establish two linear equations based on the given information. After logical assumption, we deduce Amy used five 41-cent stamps (X) and seven 6-cent stamps (Y).

This is a problem of linear equations. We need to figure out the number of 41-cent stamps (let's denote them as X) and 6-cent stamps (we'll call them Y) Amy used to total the postage price of $2.12 We can write the following two equations based on the given in the question:

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Answer:

The x-coordinate of the solution is 3.

The ordered pair that is the solution to the system lies in Quadrant I

The y-coordinate of the solution is 4

The graph in the attached figure

Step-by-step explanation:

we have

[tex]8x+6y=48[/tex] ----> equation A

[tex]2x-3y=-6[/tex] ----> equation B

To graph the system of equations , find the intercepts of each line

Line A

Find the y-intercept (value of y when the value of x is equal to zero)

For x=0

[tex]8(0)+6y=48\\y=8[/tex]

The y-intercept is the point is (0,8)

Find the x-intercept (value of x when the value of y is equal to zero)

For y=0

[tex]8x+6(0)=48\\x=6[/tex]

The x-intercept is the point is (6,0)

Plot the points (0,8) and (6,0), and connect them to graph the line A

Line B

Find the y-intercept (value of y when the value of x is equal to zero)

For x=0

[tex]2(0)-3y=-6\\y=2[/tex]

The y-intercept is the point is (0,2)

Find the x-intercept (value of x when the value of y is equal to zero)

For y=0

[tex]2x-3(0)=-6\\x=-3[/tex]

The x-intercept is the point is (-3,0)

Plot the points (0,2) and (-3,0), and connect them to graph the line B

Remember that the solution of the system of equations is the intersection point both graphs

The solution of the system is the point (3,4)

The graph in the attached figure

Verify each statement

case 1) The x-coordinate of the solution is −3

The statement is false

Because the x-coordinate of the solution is 3

case 2) The x-coordinate of the solution is 3.

The statement is true (see the explanation)

case 3) The ordered pair that is the solution to the system lies in Quadrant I

The statement is true

Because, the x-coordinate and the y-coordinate of the solution are positive values

case 4) The y-coordinate of the solution is 4

The statement is true (see the explanation)

case 5) The ordered pair that is the solution to the system lies in Quadrant II

The statement is false

Because, the ordered pair that is the solution to the system lies in Quadrant I

case 6) The y-coordinate of the solution is 3

The statement is false

Because, the y-coordinate of the solution is 4

Therefore, the particular solution is:

[tex]\[ y = \frac{1}{2}e^{-e^{-2y}} + \ln(4) - \frac{1}{2}e^{-\frac{1}{4}} \][/tex]

To find the particular solution of the given differential equation, we need to integrate both sides with respect to x. However, since the equation is not separable, we can use the method of integrating factors.

First, let's rewrite the equation in the form:

[tex]\[\frac{dy}{dx} - e^{-2y}(x-4) = 0\][/tex]

To find the integrating factor, we consider the term multiplying dy/dx, which is [tex]\(-e^{-2y}\).[/tex]

The integrating factor, denoted by [tex]\( \mu \)[/tex], is given by the exponential of the integral of [tex]\(-e^{-2y}\):[/tex]

[tex]\[ \mu = e^{\int -e^{-2y} dx} \][/tex]

[tex]\[ = e^{\frac{1}{2}e^{-2y}} \][/tex]

Multiplying both sides of the differential equation by the integrating factor [tex]\( \mu \)[/tex], we get:

[tex]\[ e^{\frac{1}{2}e^{-2y}}\frac{dy}{dx} - (x-4)e^{-\frac{1}{2}e^{-2y}} = 0 \][/tex]

This can be written as the derivative of a product:

[tex]\[ \frac{d}{dx}\left( e^{\frac{1}{2}e^{-2y}}y \right) = (x-4)e^{-\frac{1}{2}e^{-2y}} \][/tex]

Now, integrating both sides with respect to x, we get:

[tex]\[ e^{\frac{1}{2}e^{-2y}}y = \int (x-4)e^{-\frac{1}{2}e^{-2y}} dx + C \][/tex]

[tex]\[ e^{\frac{1}{2}e^{-2y}}y = \int (x-4)e^{-\frac{1}{2}e^{-2y}} dx + C \][/tex]

[tex]\[ e^{\frac{1}{2}e^{-2y}}y = \int (x-4)e^{-\frac{1}{2}e^{-2y}} dx + C \][/tex]

At this point, it seems difficult to directly integrate the right-hand side. So, let's substitute [tex]\( u = e^{-\frac{1}{2}e^{-2y}} \), then \( du = -\frac{1}{2}e^{-2y}e^{-\frac{1}{2}e^{-2y}} dy \).[/tex]

After making this substitution, the equation becomes:

[tex]\[ y = \int (x-4) du + C \][/tex]

[tex]\[ y = \frac{1}{2}u^2 + C \][/tex]

[tex]\[ y = \frac{1}{2}e^{-e^{-2y}} + C \][/tex]

To solve for  C , we use the initial condition [tex]\( y(4) = \ln(4) \):[/tex]

[tex]\[ \ln(4) = \frac{1}{2}e^{-e^{-2\ln(4)}} + C \][/tex]

[tex]\[ \ln(4) = \frac{1}{2}e^{-\frac{1}{4}} + C \][/tex]

[tex]\[ C = \ln(4) - \frac{1}{2}e^{-\frac{1}{4}} \][/tex]

Therefore, the particular solution is:

[tex]\[ y = \frac{1}{2}e^{-e^{-2y}} + \ln(4) - \frac{1}{2}e^{-\frac{1}{4}} \][/tex]

The Correct question is:

Find the particular solution of the differential equation

dydx=(x−4)e^(−2y) satisfying the initial condition y(4)=ln(4).

Answer: y=

The particular solution of the differential equation [tex]\(\frac{dy}{dx} = (x - 5)e^{-2y}\)[/tex] satisfying the initial condition [tex]\(y(5) = \ln(5)\)[/tex] is given by the implicit equation [tex]\(e^{2y} - xe^{2y} + 2y = 2\ln(5) + 5\)[/tex].

To find the particular solution, we start by separating the variables in the differential equation:

[tex]\[\frac{dy}{dx} = (x - 5)e^{-2y}\][/tex]

[tex]\[e^{2y} dy = (x - 5) dx\][/tex]

Now, we integrate both sides:

[tex]\[\int e^{2y} dy = \int (x - 5) dx\][/tex]

[tex]\[\frac{1}{2}e^{2y} = \frac{1}{2}x^2 - 5x + C\][/tex]

To find the constant of integration [tex]\(C\)[/tex], we use the initial condition [tex]\(y(5) = \ln(5)\)[/tex]:

[tex]\[\frac{1}{2}e^{2\ln(5)} = \frac{1}{2}(5)^2 - 5(5) + C\][/tex]

[tex]\[\frac{1}{2}e^{\ln(25)} = \frac{1}{2}(25) - 25 + C\][/tex]

[tex]\[\frac{1}{2}(25) = \frac{1}{2}(25) - 25 + C\][/tex]

[tex]\[C = 25\][/tex]

Substituting [tex]\(C\)[/tex] back into the equation, we get:

[tex]\[\frac{1}{2}e^{2y} = \frac{1}{2}x^2 - 5x + 25\][/tex]

Multiplying through by 2 to clear the fraction:

[tex]\[e^{2y} = x^2 - 10x + 50\][/tex]

Now, we add [tex]\(2y\)[/tex] to both sides to isolate [tex]\(e^{2y}\)[/tex]:

[tex]\[e^{2y} + 2y = x^2 - 10x + 50 + 2y\][/tex]

Since [tex]\(e^{2y} - xe^{2y} + 2y = e^{2y} + 2y - xe^{2y}\)[/tex], we can rewrite the equation as:

[tex]\[e^{2y} - xe^{2y} + 2y = 50 - 10x + 2y\][/tex]

Using the initial condition [tex]\(y(5) = \ln(5)\)[/tex] again, we have:

[tex]\[e^{2\ln(5)} - 5e^{2\ln(5)} + 2\ln(5) = 50 - 10(5) + 2\ln(5)\][/tex]

[tex]\[25 - 5(25) + 2\ln(5) = 50 - 50 + 2\ln(5)\][/tex]

[tex]\[25 - 125 + 2\ln(5) = 2\ln(5)\][/tex]

[tex]\[-100 + 2\ln(5) = 2\ln(5)\][/tex]

This confirms that the constant [tex]\(C\)[/tex] is correct. Therefore, the particular solution of the differential equation satisfying the initial condition is:

[tex]\[e^{2y} - xe^{2y} + 2y = 2\ln(5) + 5\][/tex]

Answer:

a) [tex] P(X \leq 100) = 1- e^{-0.01342*100} =0.7387[/tex]

[tex] P(X \leq 200) = 1- e^{-0.01342*200} =0.9317[/tex]

[tex] P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930[/tex]

b) [tex] P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498[/tex]

c) [tex] m = \frac{ln(0.5)}{-0.01342}=51.65[/tex]

d) [tex] a = \frac{ln(0.05)}{-0.01342}=223.23[/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 we have that X is represented by the following distribution:

[tex] X\sim Exp (\lambda=0.01342)[/tex]

Is important to remember that th cumulative distribution for X is given by:

[tex] F(X) =P(X \leq x) = 1-e^{-\lambda x}[/tex]

Part a

For this case we want this probability:

[tex] P(X \leq 100)[/tex]

And using the cumulative distribution function we have this:

[tex] P(X \leq 100) = 1- e^{-0.01342*100} =0.7387[/tex]

[tex] P(X \leq 200) = 1- e^{-0.01342*200} =0.9317[/tex]

[tex] P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930[/tex]

Part b

Since we want the probability that the man exceeds the mean by more than 2 deviations

For this case the mean is given by:

[tex] \mu = \frac{1}{\lambda}=\frac{1}{0.01342}= 74.516[/tex]

And by properties the deviation is the same value [tex] \sigma = 74.516[/tex]

So then 2 deviations correspond to 2*74.516=149.03

And the want this probability:

[tex] P(X > 74.516+149.03) = P(X>223.547)[/tex]

And we can find this probability using the complement rule:

[tex] P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498[/tex]

Part c

For the median we need to find a value of m such that:

[tex] P(X \leq m) = 0.5[/tex]

If we use the cumulative distribution function we got:

[tex] 1-e^{-0.01342 m} =0.5[/tex]

And if we solve for m we got this:

[tex] 0.5 = e^{-0.01342 m}[/tex]

If we apply natural log on both sides we got:

[tex] ln(0.5) = -0.01342 m[/tex]

[tex] m = \frac{ln(0.5)}{-0.01342}=51.65[/tex]

Part d

For this case we have this equation:

[tex] P(X\leq a) = 0.95[/tex]

If we apply the cumulative distribution function we got:

[tex] 1-e^{-0.01342*a} =0.95[/tex]

If w solve for a we can do this:

[tex] 0.05= e^{-0.01342 a}[/tex]

Using natural log on btoh sides we got:

[tex] ln(0.05) = -0.01342 a[/tex]

[tex] a = \frac{ln(0.05)}{-0.01342}=223.23[/tex]

The question involves applying the exponential distribution formula to calculate certain probabilities and expectations involving the distances travelled by kangaroo rats. You need to calculate these by using the formula, P(X≤x) = 1 - e^(-λx), and using that the standard deviation of an exponential distribution is the reciprocal of the parameter. Median and the distance that only 5% will exceed can be calculated by setting the P(X≤a) = 0.50 and 0.95, respectively.

This question surrounds the concepts within probability theory and specifically the exponential distribution.

Firstly, understand that an exponential distribution can be described by the formula: P(X≤x) = 1 - e^(-λx). Given λ = 0.01342, you can solve for Part (a), calculate the probabilities for distances at most 100m, 200m, and between 100 and 200m. Plug the distances into the formula and calculate.

For part (b), you need to know that the standard deviation of an exponential distribution is the reciprocal of the parameter (1/λ). Calculate the mean and standard deviation and use these values to find the required probability.

For Part (c), set P(X≤ a) = 0.50 and solve for 'a', this will give you the median.

For Part (d), set P(X≤ a) = 0.95 and solve for 'a', this will give you the distance that only 5% of animals will exceed.

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Answer:

a) There is a 24.14% probability that both marbles are red.

b) There is a 5.56% probability of first selecting a blue marble, then a green marble.

Step-by-step explanation:

a) Two marbles are selected with replacement. Find the probability that both marbles are red.

Initially, there are 30 marbles, of which 15 are red. So there is a 15/30 probability that the first marble selected is red.

After a red marble is selected, there are 29 marbles, of which 14 are red. So there is a 14/29 probability that the second marble selected is red.

The probability that both marbles are red is:

[tex]P = \frac{15}{30}*\frac{14}{29} = 0.2414[/tex]

There is a 24.14% probability that both marbles are red.

b) Two marbles are selected without replacement. Find the probability of first selecting a blue marble, then a green marble

There are 30 marbles, of which 10 are blue and 5 are green.

So, there is a 10/30 probability of selecting a blue marble and a 5/30 probability of selecting a red marble.

The probability of selecting a blue marble and then a green marble is:

[tex]P = \frac{10}{30}*\frac{5}{30} = 0.0556[/tex]

There is a 5.56% probability of first selecting a blue marble, then a green marble.

Answer: (a). prob = 0.1543

(b). Prob = 2⁵⁰/ \left[\begin{array}{ccc}100\\50\end{array}\right].

Step-by-step explanation:

The information in the question tells us that the United States contains two senators from each of the 50 states, this means there is a total of a 100 senators in the senate.

(a). this question tells us to find the probability that it will contain one of the the two senators from a certain specified state.

Explanation:

Consider selecting a committee of 8 senators at random,

the total number of ways to select 8 senators from the Senate of 100 senators is \left[\begin{array}{ccc}100\\8\end{array}\right].

Next is to consider an event Q that at least one of the senators from the specified state is added to the senate,

Let us call the two senators F and G.

where F is the number of possible ways to select 7 members from 98 senetors; \left[\begin{array}{ccc}98\\7\end{array}\right].

Also, G is the number of possible ways to select 6 members from 98 senators; \left[\begin{array}{ccc}98\\6\end{array}\right].

The probability that committee of 8 senators will contain at least one of the two senators from a certain specified state is given thus;

Prob (Q) = \left[\begin{array}{ccc}2\\1\end{array}\right]\left[\begin{array}{ccc}98\\7\end{array}\right]  / \left[\begin{array}{ccc}100\\8\end{array}\right]   + \left[\begin{array}{ccc}2\\2\end{array}\right] \left[\begin{array}{ccc}98\\6\end{array}\right] / \left[\begin{array}{ccc}100\\8\end{array}\right]

Prob (Q) = 0.1487 + 0.0057 = 0.1543

Prob (Q) = 0.1543

(b).  the questions tells us to find the probability it will contain one senator each from each of the states.

consider that a group of 50 senators is selected in random, the total number of ways to select 50 senators from the senate of 100 senators is

\left[\begin{array}{ccc}100\\50\end{array}\right].

Now let us consider that an event Q, that a total of 50 groups contains exactly one senator from each state. for any state, there are two ways for exactly one senator to be in a group of 50.

∴ the probability becomes;

Pr (Q) = 2⁵⁰/ \left[\begin{array}{ccc}100\\50\end{array}\right].

cheers, i hope this helps.

This response provides a methodology for calculating probabilities related to the selection of senators from the U.S. Senate. It explains that the probability that a randomly selected eight-person committee includes a senator from a particular state is simply one minus the ratio of groups of eight not from that state to the total groups of eight. Furthermore, the probability of randomly selecting a fifty-person group that includes one senator from each state is 1.

The subject of this question is the calculation of probability in the context of selections from the United States Senate. Probability describes the likelihood of an event occurring and is often represented as a fraction or decimal, where 1 (or 100%) represents certainty, and 0 (or 0%) indicates impossibility.

(a) To find the probability that a committee of eight senators includes at least one senator from a particular state, first consider the total number of possible committees: C(100,8), or combinations of 100 things taken 8 at a time. The easiest way to ensure at least one senator from a particular state is to envision the opposite scenario: the committee consists of senators from any state but that one, calculated as C(98,8). The desired probability is then 1 minus this ratio, or 1 - C(98,8) / C(100,8).

(b) The probability that a group of 50 senators selected at random will contain one senator from each state is more straightforward, as there is only one way to draw one senator from each of the 50 states out of a pool of 100 senators. So, the probability for such an event to occur is C(100,50) / C(100,50) = 1.

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Answer:

b. False.

See explanation below.

Step-by-step explanation:

b. False.

When we have a significant result that means [tex] P_v < \alpha[/tex] where [tex] P_v[/tex] represent the p value for the test and [tex]\alpha[/tex] the significance level assumed at the begin of the hypothesis test.

For this case we have a null hypothesis [tex]H0[/tex] and an alternative hypothesis [tex]H_1[/tex] for a parameter of interest let's say [tex]\theta[/tex], and using the test we conclude thar we reject the null hypothesis, so on this case we need to have that [tex] p_v <\alpha[/tex], so then that means that we have a significant difference.

And when we have this situation we can't say that the difference between the sample statistic and the hypothesized parameter is just due to chance, since we are obtaining singificant results that are showing difference between the two values on statistical terms

The radius of the circumscribed circle is [tex]\( 3.2\sqrt{3} \)[/tex] centimeters.
The length of a side of the hexagon is [tex]\( 3.2\sqrt{3} \)[/tex] centimeters.
The perimeter of the hexagon is [tex]\( 19.2\sqrt{3} \)[/tex] centimeters.

Let's solve each part of the problem step by step:

a. **Find the radius of the circumscribed circle:**

The apothem of a regular polygon and the radius of the circumscribed circle are related by the formula:

[tex]\[ \text{Radius} = \frac{\text{Apothem}}{\cos(180^\circ / \text{number of sides})} \][/tex]

For a regular hexagon (6 sides), the formula becomes:

[tex]\[ \text{Radius} = \frac{4.8}{\cos(180^\circ / 6)} \][/tex]

[tex]\[ \text{Radius} = \frac{4.8}{\cos(30^\circ)} \][/tex]

[tex]\[ \text{Radius} = \frac{4.8}{\frac{\sqrt{3}}{2}} \][/tex]

[tex]\[ \text{Radius} = \frac{4.8 \times 2}{\sqrt{3}} \][/tex]

[tex]\[ \text{Radius} = \frac{9.6}{\sqrt{3}} \][/tex]

[tex]\[ \text{Radius} = \frac{9.6\sqrt{3}}{3} \][/tex]

[tex]\[ \text{Radius} = 3.2\sqrt{3} \][/tex]

b. **What is the length of a side of the hexagon?**

The length of a side of a regular hexagon can be found using the formula:

[tex]\[ \text{Side Length} = 2 \times \text{Apothem} \times \tan(180^\circ / \text{number of sides}) \]\\[/tex]

For a regular hexagon with apothem 4.8 cm:

[tex]\[ \text{Side Length} = 2 \times 4.8 \times \tan(30^\circ) \][/tex]

[tex]\[ \text{Side Length} = 2 \times 4.8 \times \frac{1}{\sqrt{3}} \][/tex]

[tex]\[ \text{Side Length} = \frac{9.6}{\sqrt{3}} \][/tex]

[tex]\[ \text{Side Length} = \frac{9.6\sqrt{3}}{3} \][/tex]

[tex]\[ \text{Side Length} = 3.2\sqrt{3} \][/tex]

c. **What is the perimeter of the hexagon?**

Since a regular hexagon has six equal sides, its perimeter is simply:

[tex]\[ \text{Perimeter} = 6 \times \text{Side Length} \][/tex]

[tex]\[ \text{Perimeter} = 6 \times 3.2\sqrt{3} \][/tex]

[tex]\[ \text{Perimeter} = 19.2\sqrt{3} \][/tex]

Answer:

mi/hr • min does not equal mi. You must convert 30 minutes to hours.

Step-by-step explanation:

I got it correct on TTM

Answer:

24

Step-by-step explanation:

A 2024-T4 aluminum tube with an outside diameter of 3.6 in. will be used to support a 24-kip load. If the axial stress in the member must be limited to 6.4 ksi, determine the wall thickness required for the tube.

Answer:

-60

Step-by-step explanation:

The objective is to state whether or not the following limit exists

                                [tex]\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2}[/tex].

First, we simplify the expression in the numerator of the fraction.

[tex]3(2x-1)^2 -75 = 3(4x^2 - 4x +1) -75 = 12x^2 - 12x + 3 - 75 = 12x^2 - 12x -72[/tex]

Now, we obtain

                         [tex]12(x^2-x-6) = 12(x+2)(x-3)[/tex]

and the fraction is transformed into

                       [tex]\frac{3(2x-1)^2 - 75}{x+2} = \frac{12(x+2)(x-3)}{x+2} = 12 (x-3)[/tex]

Therefore, the following limit is

       [tex]\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2} = \lim_{x \to -2} 12(x-3) = 12 \lim_{x \to -2} (x-3)[/tex]

You can plug in [tex]-2[/tex] in the equation, hence

                        [tex]12 \lim_{x \to -2} (x-3) = 12 (-2-3) = -60[/tex]

The limit does not exist.

To find the limit of the given function as x approaches -2, we can simply substitute -2 into the function and simplify.

Start by replacing x with -2 in the function:

lim as x → -2 (3(2x - 1)2 - 75) / (x + 2)

Substitute -2 for x:

(3(2(-2) - 1)2 - 75) / (-2 + 2)

Simplify:

(3(-4 - 1)2 - 75) / 0

Continue simplifying:

(3(-5)2 - 75) / 0

(3(25) - 75) / 0

(75 - 75) / 0

0 / 0

Since we end up with 0/0, the limit is undefined, or it does not exist.

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Answer:

we CANNOT DIVIDE 3 with 6.

Step-by-step explanation:

Here,as given in the question:

Starting value = 6

Constant ratio = 1/3

Now, exponential function is obtained by the product of starting value and the constant ratio repeatedly.

⇒ f(x) = (Starting value) x (ratio)... x times

[tex]\implies f(x) = 6 (\frac{1}{3} )^x = 6 (\frac{1}{3} ) (\frac{1}{3} ) (\frac{1}{3} ) .... x[/tex]  

Now, we CANNOT DIVIDE 3 with 6 as it is in the power of x.

Hence, [tex]\implies f(x) = 6 (\frac{1}{3} )^x[/tex] and [tex]f(x) \neq 2^x[/tex]

Answer:

The student made an error by applying the exponent to the product of a and b instead of just b. The final answer should be ​f(x)=6(1/3)^x

Step-by-step explanation:

U can't multiply 6 by 1/3

Answer:

cannot say

Step-by-step explanation:

Given that Kari owns 28 golf balls. Some are green, some are blue and several are orange. One fourth are red.

With the available information we know that 7 balls are red (1/4 of 28)

No mention is made about any of the other colour.

With the available information we cannot say whether 7 gold balls are definitely green.

We are given only some are green, some blue and several other are organe.

Nothing mentioned about red in that but given 1/4 are red.

The term some is a relative term and mathematically we cannot define whether this equals 7 or not

Cannot say.

Answer: A differentiable function [tex]f(x)[/tex] has a local minimum at the point [tex]x_0[/tex] if two conditions are met: the value of its first derivative is equal to zero at that point and the value of its second derivative is negative at that point.

Step-by-step explanation: The procedure for finding the local minima of the function [tex]f(x)[/tex] is the following.

Step 1. Find the first derivative of the function [tex]f(x)[/tex], denoted by [tex]f'(x)[/tex] according to the rules of derivation.

Step 2. Find all [tex]x[/tex] such that [tex]f'(x)=0.[/tex] Denote these solutons by [tex]x_1, x_2\ldots[/tex].

Step 3. Find the second derivative of the function [tex]f(x)[/tex], denoted by [tex]f''(x)[/tex]. Evaluate this derivative at each point found in step 2. Only If, say [tex]f''(x_1)>0[/tex] then [tex]x_1[/tex] is the local minimum and the same goes for all other values of [tex]x[/tex] you found in step 2.

For what value(s) of x does f(x) have a local minimum?

Using the example below to explain

f(x) = x2 − 6x + 5.  

Answer:

The point x on the function f(x) is a local minimum if and only if the following conditions are satisfied

1. f'(x) = 0 (at that point df(x)/dx must be equal to zero)

2. f"(x)>0 (the second derivative of the function must be greater than zero, it must be positive)

Using the example below to explain

f(x) = x2 − 6x + 5.  

Since f'(x)= 0 and f"(x) greater than 0 (positive), then we can now confirm that the function f(x) has a local minimum at x = 3

Step-by-step explanation:

The point x on the function f(x) is a local minimum if and only if the following conditions are satisfied

1. f'(x) = 0 (at that point df(x)/dx must be equal to zero)

2. f"(x)>0 (the second derivative of the function must be greater than zero, it must be positive)

For the example above:

f(x) = x2 − 6x + 5

f'(x) = 2x - 6

Condition 1:

f'(x) = 0

So,

f'(x) = 2x - 6 = 0

Solving for x

2x - 6 = 0

2x = 6

x = 3

Therefore, at x = 3, f(x) has a critical point.

We need to determine whether it is a local minimum, local maximum or saddle point.

Condition 2:

f"(x) > 0

f"(x) = f'(f'(x)) = d/dx (2x - 6) = 2

So,

f"(x) = 2 >0

Note: in some cases we would need to substitute x into f"(x) to determine the value.

Since f'(x)= 0 and f"(x) greater than 0 (positive), then we can now confirm that the function f(x) has a local minimum at x = 3

Answer:

The coefficient is 3 and the constant is 4 in

relation to the equation mx + c where m is the coefficient of x and c is the constant.

Answer:

[tex]P(junior)=\frac{9}{35}[/tex]

[tex]P(freshmen)=\frac{5}{35}=\frac{1}{7}[/tex]

Step-by-step explanation:

In a class there are 13 seniors, 9 juniors, 8 sophomores and 5 freshmen

Total number of students= [tex]13+9+8+5=35[/tex]

one student is selected at random from this class, selected student is Junior

there 9 juniors in the class

[tex]P(junior)= \frac{juniors}{total} =\frac{9}{35}[/tex]

there 5 freshmen in the class

[tex]P(freshmen)= \frac{freshmen}{total} =\frac{5}{35}=\frac{1}{7}[/tex]

Answer:

We conclude that there is not enough evidence to support the claim  compute technique performs differently than the traditional method.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 101 hours

Sample mean, [tex]\bar{x}[/tex] =100 hours

Sample size, n = 140

Alpha, α = 0.01

Population standard deviation, σ = 6 hours

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 101\text{ hours}\\H_A: \mu \neq 101\text{ hours}[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{100 - 101}{\frac{6}{\sqrt{140}} } = -1.972[/tex]

Now, [tex]z_{critical} \text{ at 0.01 level of significance } = \pm 2.58[/tex]

Since,  

The calculated z statistic lies in the acceptance region, we fail to reject the null hypothesis and accept it.

We conclude that there is not enough evidence to support the claim that compute technique performs differently than the traditional method.

To determine if the technique performs differently than the traditional method, a hypothesis test is performed by comparing the test statistic to the critical value(s) from the t-distribution.

To determine if there is sufficient evidence to support the claim that the CAI technique performs differently than the traditional method, we can perform a hypothesis test. First, we state our null hypothesis (H0) and alternative hypothesis (Ha). H0: The mean time for obtaining a flying license using CAI is equal to 101 hours. Ha: The mean time for obtaining a flying license using CAI is different from 101 hours.

Next, we calculate the test statistic using the formula z = (X - µ) / (σ / √n), where X is the sample mean, µ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the values: X = 100, µ = 101, σ = 6, and n = 140, we can calculate the test statistic.

Using a significance level of 0.01, we compare the test statistic to the critical value(s) from the t-distribution. If the test statistic falls outside the critical region, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the technique performs differently than the traditional method.

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Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

[tex]\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}[/tex]

Solve for S(t):

[tex]\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}[/tex]

[tex]e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}[/tex]

The left side is the derivative of a product:

[tex]\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}[/tex]

Integrate both sides:

[tex]e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt[/tex]

[tex]e^{t/800}S(t)=64e^{t/800}+C[/tex]

[tex]S(t)=64+Ce^{-t/800}[/tex]

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

[tex]0=64+C\impleis C=-64[/tex]

and so (b) the amount of sugar in the tank at time t is

[tex]S(t)=64\left(1-e^{-t/800}\right)[/tex]

As [tex]t\to\infty[/tex], the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

The tank initially contains 64 kg of sugar. The amount of sugar after t minutes is given by the equation: Amount of sugar = Initial amount of sugar + (Rate of sugar entering - Rate of solution leaving) × t. As t becomes large, the value of y(t) approaches the concentration of sugar in the solution entering the tank (0.04 kg/L).

(a) How much sugar is in the tank at the beginning?

To find the amount of sugar in the tank at the beginning, we need to calculate the total mass of sugar in the tank.

Mass of sugar = Volume of solution × Concentration of sugar = 1600 L × 0.04 kg/L = 64 kg

Therefore, there is 64 kg of sugar in the tank at the beginning.

(b) Find the amount of sugar after t minutes.

To find the amount of sugar after t minutes, we need to know the rate of sugar entering the tank and the rate of solution leaving the tank.

The rate of sugar entering the tank is given as 0.04 kg/L.

The rate of solution entering and leaving the tank is given as 2 L/min.

Therefore, the amount of sugar after t minutes is given by the equation: Amount of sugar = Initial amount of sugar + (Rate of sugar entering - Rate of solution leaving) × t = 64 kg + (0.04 kg/L - 2 L/min) × t

(c) As t becomes large, what value is y(t) approaching?

As t becomes large, the value of y(t) is approaching a constant value, which is the concentration of sugar in the solution entering the tank.

In this case, the concentration of sugar in the solution entering the tank is 0.04 kg/L.

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Answer:

B. Yes, it is equivalent

Step-by-step explanation:

A = (-3, 2, 3)

B = (-3, 5, 2)

/AB/ = (-3-(-3), 2-5, 3-2)

= (0, -3, 1)

P = (2, -3, 2)

Q = (2, 0, 1)

/PQ/ = (2-2, -3-0, 2-1)

= (0, -3, 1)

So, /AB/ is equivalent to /PQ/

Answer:

D

Step-by-step explanation:

The correlation coefficient r=-0.84 denotes that there is inverse relationship between x and y. It means that as the x values increase the y values decrease whereas as the x values decreases the y-values increases. Also, r=-0.84 denotes the strong relationship between  x and y because it is close to 1. So, r=-0.84 denotes that there is strong linear relationship between x and y with smaller x values tending to be associated with larger y values.

The correlation coefficient of -0.84 indicates a strong inverse relationship between x and y, with smaller x values generally corresponding to larger y values.

Based on the given correlation coefficient of -0.84, the correct answer is option d. This option states that there is a strong linear relationship between variables x and y, with smaller x values tending to be associated with larger values of the y variable.

A correlation coefficient communicates both the strength and direction of a linear relationship between two variables. In this context, a coefficient of -0.84 indicates a strong relationship (values close to -1 or 1 denote strong relationships), and because the value is negative, it reflects an inverse or negative correlation, meaning y tends to decrease as x increases, and vice versa.

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Answer:

Step-by-step explanation:

The values of x are given as

0, 1, 2, 3

The corresponding values of y are given as

0,5 10,15

Let k represent constant of proportionality

Therefore,

When x = 0, y = 0

When x = 1, y = 5

y/x = k

k = 5/1 = 5

When x = 2, y = 10

y/x = k

k = 10/2 = 5

When x = 3, y = 15

y/x = k

k = 15/3 = 5

Therefore, the constant of proportionality is 5

The equation to represent the table is

y = 5x

Answer:

Population:  The computer chips coming off of an assembly line.

Sample: The selected 36 chips coming off an assembly line.

Step-by-step explanation:

In the undergoing study the population is always consists of set of all possible units and sample is a part or subset of population. The population of interest in the study conducted by engineer to check the whether the computer chips coming off of an assembly line are within the specification consists of all the computer chips coming off of an assembly line. The sample in the undergoing study are the selected 36 computer chips coming off of an assembly line.

In this scenario, the population is all the computer chips coming off the assembly line and the sample is the 36 chips the engineer randomly selects and tests.

In this study, the population refers to all the computer chips coming off of the assembly line. It's important to note that the population in statistical terms does not have to represent an actual, concrete group of physical objects or individuals. Instead, it represents the total set of observations that can be made. In this case, all the chips that come off the assembly line - both those chosen for testing and those not chosen - constitute the population.

On the other hand, the sample in this study would be the 36 random chips that the engineer tests. A sample is a subset of the population that is selected for study. The characteristics of this sample are then used to infer information about the overall population.

So, in the context of this study, the objective is to infer from the testing of 36 chips (sample) whether all chips coming off the assembly line (population) are within specifications.

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Answer:

[tex]x=\sqrt{21}[/tex]

[tex]y=\sqrt{21}[/tex]

Step-by-step explanation:

Let x  and y are the two positive real numbers

product of two numbers is 21

[tex]x y=21[/tex]

[tex]y=\frac{21}{x}[/tex]

The sum of the two numbers is f(x) =x+y

Replace y with 21/x

[tex]f(x) =x+\frac{21}{x}[/tex]

we need to find smallest possible sum , so we take derivative using power rule

[tex]f'(x)= 1-\frac{21}{x^2}[/tex]

when sum is minimum then the derivative is equal to 0

[tex]0= 1-\frac{21}{x^2}[/tex]

[tex]0=\frac{x^2-21}{x^2}[/tex]

multiply both sides by x^2

[tex]x^2-21=0[/tex]

[tex]x^2=21[/tex]

Take square root on both sides

[tex]x=\sqrt{21}[/tex]

[tex]y=\frac{21}{x}[/tex]

[tex]y=\frac{21}{\sqrt{21}}[/tex]

[tex]y=\sqrt{21}[/tex]

The smallest possible sum of two positive real numbers whose product equals 21 is attained when both numbers are equal to the square root of 21.

To solve this problem, we can use algebra and calculus. To minimize the sum of the two numbers, call them x and y, subject to the constraint that their product equals 21 (x*y=21), we can use the formula for a minimum of a function. We first change the problem into the equation x + y = sum and then transform it into a function with a single variable: f(y) = y + 21/y. Solving the derivative of y with respect to this function and setting the result equal to zero, we find that the minimum sum is attained when y = sqrt(21), and consequently, x = sqrt(21) as well. Therefore, the two positive real numbers whose product is 21 have the smallest possible sum are both sqrt(21).

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