A committee has 6 members who have decided on how many votes each should get, but not on the quota to be used for the system. The system so far is [q : 6, 4, 4, 3, 2, 1 ].
a. What value of q would represent a simple majority?
b. What value of q would represent a two-thirds majority?

By gcd, I think you mean gcf ( Greatest Common Factor).

To find the gcf find all the factors of each number:

Factors of 20: 1, 2, 4,5 ,10 , 20

Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56

The largest common factor is 4.

LCM = Least Common Multiple

This is the smallest number that both numbers divide into evenly

Find the prime factors of each number:

Prime factors of 20: 2 * 2 * 5

Prime factors of 56: 2 * 2 * 2 * 7

To find the LCM, multiply all prime factors the most number of times they occur.

In the prime factor of 56, 2 appears 3 times and 7 appears once.

In the prime factor of 20 5 appears once.

LCM = 2 * 2 *2 * 7 * 5 = 280

Answer:

H(3) = 288 feet

H(4) = 320 feet

H(5) = 320 feet

H(6) = 288 feet

Step-by-step explanation:

A ball is shot out of a cannon at ground level so the ball will follow a parabolic path.

Since height H and time t of the ball have been described by a function H(t) = 144t - 16t²

Then we have to find the values of H(3), H(4), H(5) and H(6).

H(3) = 144×3 - 16(3)²

       = 432 - 144

       = 288 feet

H(4) = 144×4 - 16(4)²

       = 576 - 256

       = 320 feet

H(5) = 144×5 - 16(5)²

       = 720 - 400

       = 320 feet

H(6) = 144×6 - 16(6)²

       = 864 - 576

       = 288 feet

Here we are getting the value like H(3), H(6) and H(4), H(5) are same because in a parabolic path ball first increase in the height above the ground then after the maximum height it decreases.

Therefore, after t = 3 and t = 6 heights of the canon ball are same. Similarly after t = 4 and t = 5 heights of the canon above the ground are same.

The heights of the ball at different times are calculated by substituting the times into the quadratic function H(t) = 144t - 16t². Some heights are equal because the ball reaches the same height on its way up and on its way down due to the parabolic path of the projectile motion.

The height function for the ball being shot out of a cannon is H(t) = 144t - 16t². This is a quadratic function, which models projectile motion. To find the heights at specific times we substitute these times into the function.

Notice that H(4) = H(5) = 320 feet. This is because the path of the ball follows parabolic motion. The ball reaches the same height of 320 feet on its way up (at 4 seconds) and on its way down (at 5 seconds), which is why some output values are equal.

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Answer:

2/21

Step-by-step explanation:

We start out with a full tank.  Once the trucks take from it, it is down to 1/3 of a tank.  Therefore,

[tex]\frac{3}{3} -\frac{1}{3} =\frac{2}{3}[/tex]

So the trucks took 2/3 of the gas.  

If there were 7 trucks and we need to know how much of that 2/3 was taken by each truck, we divide 2/3 by 7:

[tex]\frac{\frac{2}{3} }{7}[/tex]

When dividing fractions, we bring up the lower fraction and flip it and multiply:

[tex]\frac{2}{3}*\frac{1}{7}=\frac{2}{21}[/tex]

The capacity of each truck is 2/21 of the total gasoline tank, which is calculated by dividing the used part of the gasoline tank (2/3) by the number of trucks (7).

Let's denote the total gasoline tank volume as one unit, or 1. Seven trucks share 2/3 of the gasoline tank capacity (since 1/3 is still left); each truck capacity could be gotten by dividing this 2/3 equally among the seven trucks. By dividing 2/3 by 7, we get each truck's capacity as 2/21 of the total gasoline tank capacity. Therefore, the correct answer is d) 2/21.

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Frequency--

It is the number of times an outcome occurs while performing an experiment some " n " number of times.

Relative frequency--

It is the ratio of the frequency of an outcomes to the total number of times an experiment is been performed.

Here there are TOTAL : 120 responses and three outcomes A , B and C.

The frequency table is given as follows:

Outcome         A           B            C

Frequency       64         23           33

and the Relative frequency table is given as follows:

Outcome                          A              B               C

Relative frequency         64/120     23/120      33/120

i.e. the Relative frequency table is given by:

Outcome                         A              B               C

Relative frequency         0.53         0.19          0.28    

The frequency distribution for the given sample is: A: 64, B: 23, C: 33. The relative frequency distribution is: A: 0.53, B: 0.19, C: 0.28.

To find the frequency distribution, we simply count the number of occurrences of each response. For the given sample of 120 responses, we have:

A: 64 responses

B: 23 responses

C: 33 responses

To find the relative frequency distribution, we divide the frequency of each response by the total number of responses (120). The relative frequencies, rounded to two decimal places, are:

A: 0.53

B: 0.19

C: 0.28

Answer:

t= 3.322 years

Step-by-step explanation:

investment made= $5000 (Principal)

amount obtained after a specific time= $6400

rate %= 7.5% per year compounded quarterly which means

r= 7.5/(100*4)= 0.01875

time = 4t ( compounded quarterly)

we know that Amount obtained is given by

[tex]A= P(1+r)^{4t}[/tex]

[tex]6400= 5000(1+0.01875)^{4t}[/tex]

[tex](1.01875)^{4t}=1.28[/tex]

taking log on both sides and solving we get

t= 3.322 years

hence my answer t= 3.322 years

Final answer:

To find the probability that a computer memory chip will have a lifetime of less than 8 years, we can use the properties of the Gamma distribution.

Explanation:

To find the probability that a computer memory chip will have a lifetime of less than 8 years, we can use the properties of the Gamma distribution. The average lifetime of the chip is given as 4 years, which corresponds to the mean. The variance is given as 16/3 years squared, which is equal to the mean squared.

Using these values, we can determine the shape and rate parameters of the Gamma distribution. The shape (α) is equal to the mean squared divided by the variance, which in this case is 16/3. The rate (β) is equal to the mean divided by the variance, which in this case is 4/(16/3).

To find the probability that the chip will have a lifetime of less than 8 years, we can calculate the cumulative distribution function (CDF) of the Gamma distribution with the shape and rate parameters we obtained.

Answer:

So the vectors are linearly independent.

Step-by-step explanation:

So if they are linearly independent then the following scalars in will have the condition a=b=c=0:

a(2,3,1)+b(2,-5,-3)+c(-3,8,-5)=(0,0,0).

We have three equations:

2a+2b-3c=0

3a-5b+8c=0

1a-3b-5c=0

Multiply last equation by -2:

2a+2b-3c=0

3a-5b+8c=0

-2a+6b+10c=0

Add equation 1 and 3:

0a+8b+7c=0

3a-5b+8c=0

-2a+6b+10c=0

Divide equation 3 by 2:

0a+8b+7c=0

3a-5b+8c=0

-a+3b+2c=0

Multiply equation 3 by 3:

0a+8b+7c=0

3a-5b+8c=0

-3a+9b+6c=0

Add equation 2 and 3:

0a+8b+7c=0

3a-5b+8c=0

0a+4b+13c=0

Multiply equation 3 by -2:

0a+8b+7c=0

3a-5b+8c=0

0a-8b-26c=0

Add equation 1 and 3:

0a+0b-19c=0

3a-5b+8c=0

0a-8b-26c=0

The first equation tells us -19c=0 which implies c=0.

If c=0 we have from the second and third equation:

3a-5b=0

0a-8b=0

0a-8b=0

0-8b=0

-8b=0 implies b=0

We have b=0 and c=0.

So what is a?

3a-5b=0 where b=0

3a-5(0)=0

3a-0=0

3a=0 implies a=0

So we have a=b=c=0.

So the vectors are linearly independent.

Final answer:

To find out if the vectors (2, 3, l), (2, -5, -3), and (-3, 8, -5) are linearly dependent or independent, set up a linear system with the vectors and look for non-trivial solutions.

Explanation:

To determine whether the vectors (2, 3, l), (2, -5, -3), and (-3, 8, -5) are linearly dependent or linearly independent, we set up the equation a(2, 3, l) + b(2, -5, -3) + c(-3, 8, -5) = (0, 0, 0), where a, b, and c are scalars.

If only the trivial solution exists, where a = b = c = 0, then the vectors are linearly independent. If a non-trivial solution exists, then the vectors are linearly dependent.

Let's solve the system of linear equations generated from the above equation:

Using the methods for solving systems of linear equations, such as Gaussian elimination, we can determine whether a unique solution exists.

If the determinant of the coefficients matrix is non-zero, the system has a unique solution, indicating linear independence. Otherwise, a non-unique solution indicates linear dependence, and we can express

Answer:

Angle made by secant line equals[tex]56.7251^{o}[/tex]

Step-by-step explanation:

Solpe of a line joining points [tex](x_{1},y_{1}),(x_{2},y_{2})[/tex] is given by

[tex]tan(\theta)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

where [tex]y_{i}=f(x_{i})[/tex]

Applying values we get

[tex]tan(\theta)=\frac{7.1-3.9}{3.6-1.5}\\\\\theta =tan^{-1}\frac{32}{21}\\\\\theta=56.7251^{o}[/tex]

Final answer:

To calculate the population in the year 2010 based on exponential growth from 1990, use a growth rate factor and the known population figures from the given years.

Explanation:

Population Growth Calculation:

Determine the growth rate factor from 1990 to 1996: 25 million / 21.7 million = 1.152

Apply the growth rate to find the population in 2010: 21.7 million * (1.152)¹⁴ (14 years from 1996 to 2010) = 48.9 million

Answer:

The critical numbers/values are x = 0, 4/7, 2

Step-by-step explanation:

This is a doozy; no wonder you have it up here for help!

The critical numbers of a function are found where the derivative of the function is equal to 0.  To find these numbers, you have to factor the deriative or simply solve it for 0.  This one is especially difficult since it involves rational exponents that have to be factored.  But this is fun, so let's get to it.

First off, I am assuming that the function is

[tex]f(x)=x^{\frac{4}{5}}*(x-2)^2[/tex] which involves using the product rule to find the derivative.

That derivative is

[tex]f'(x)=x^{\frac{4}{5}}*2(x-2)+\frac{4}{5}x^{-\frac{1}{5}}(x-2)^2[/tex] which simplifies down to

[tex]f'(x)=x^{\frac{4}{5}}(2x^{\frac{5}{5}}-4)+\frac{4}{5}x^{-\frac{1}{5}}(x^{\frac{10}{5}}-4x^{\frac{5}{5}}+4)[/tex] and

[tex]f'(x)=2x^{\frac{9}{5}}-4x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}[/tex]

Let's get everything over the common denominator of 5 so we can easily add and subtract like terms:

[tex]f'(x)=\frac{10}{5}x^{\frac{9}{5}}-\frac{20}{5}x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}[/tex]

Combining like terms gives us

[tex]f'(x)=\frac{14}{5}x^{\frac{9}{5}}-\frac{36}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}[/tex]

This, however, factors so it is easier to solve for x.  First we will set this equal to 0, then we will factor out

[tex]\frac{2}{5}x^{-\frac{1}{5}}[/tex]:

[tex]0=\frac{2}{5}x^{-\frac{1}{5}}(7x^2-18x+8)[/tex]

By the Zero Product Property, one of those terms has to equal 0 for the whole product to equal 0.  So

[tex]\frac{2}{5}x^{-\frac{1}{5}}=0[/tex] when x = 0

And

[tex]7x^2-18x+8=0[/tex] when x = 2 and x = 4/7

Those are the critical numbers/values for that function.  This indicates where there is a max value or a min value.

To find the critical numbers of the function F(x) = x^(4/5)(x - 2)^2, take the derivative, set it equal to zero, and check for undefined values.

To find the critical numbers of the function F(x) = x4/5(x - 2)2, we need to find the values of x where the derivative of the function is equal to zero or undefined.

Step 1: Find the derivative of F(x) using the product rule and simplify.

Step 2: Set the derivative equal to zero and solve for x.

Step 3: Check if the derivative is undefined at any values of x.

The critical numbers of the function are the values of x where the derivative is equal to zero or undefined.

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Answer:

8000/4*100 = $200'000

Step-by-step explanation:

Answer: -0.9 ; inelastic

Explanation:  

Given:

The average price of wheat per metric ton in 2012 = $305.75

Demand (in millions of metric tons) in 2012 = 672

The average price of wheat per metric ton in 2013 =  $291.56

Demand (in millions of metric tons) in 2013 = 700

We will compute the elasticity using the following formula:

ε = [tex]\frac{\frac{(Q_{2} - Q_{1})}{\frac{(Q_{2} +Q_{1})}{2}}}{\frac{(P_{2} - P_{1})}{\frac{(P_{2} +P_{1})}{2}}}[/tex]

ε = [tex]\frac{\Delta Q}{\Delta P}[/tex]

Now , we'll first compute  [tex]\Delta Q[/tex]

i.e.  [tex]\frac{\Delta Q}{\Delta P}[/tex] = [tex]\frac{(700 - 672)}{\frac{(700 +672)}{2}}[/tex]

[tex]\Delta Q[/tex] = 0.04081

Similarly for  [tex]\Delta P[/tex]

i.e. [tex]\Delta P[/tex] = [tex]\frac{(291.56 - 305.75)}{\frac{(261.56 +305.75)}{2}}[/tex]

[tex]\Delta P[/tex] = -0.0475

ε = [tex]\frac{0.04081}{-0.0475}[/tex]

ε = -0.859 [tex]\simeq[/tex] -0.9

[tex]\because[/tex] we know that ;

If, ε > 1 ⇒ Elastic

ε < 1 ⇒ Inelastic

ε = 1 ⇒ unit elastic

[tex]\because[/tex] Here , ε = -0.859 [tex]\simeq[/tex] -0.9

Therefore ε is inelastic.

Answer: 0.3758

Step-by-step explanation:

Given : The  probability that a milk container is underweight throughout of packaging line: [tex]p = 0.20[/tex]

The number of containers : n= 10

The formula binomial distribution formula :-

[tex]^nC_rp^{n-r}(1-p)^r[/tex]

The probability that at least nine milk containers in a box are properly filled is given by :-

[tex]P(X\geq9)=P(9)+P(10)\\\\=^{10}C_9(0.2)^{10-9}(1-0.20)^9+^{10}C_{10}(0.2)^{10-10}(1-0.2)^{10}\\=10(0.2)(0.8)^9+(1)(0.8)^{10}\\=0.3758096384\approx0.3758[/tex]

Answer:

The answer is

[tex]W=\frac{S-2LH}{2H+2L}[/tex]

Step-by-step explanation:

The formula for the area of a solid rectangle is

[tex]S = 2HW+2LW+2LH[/tex]

Solve it for W

[tex]2HW+2LW=S-2LH\\\\W(2H+2L)=S-2LH\\\\W=\frac{S-2LH}{2H+2L}[/tex]

a. Substitute the given solutions and their derivatives into the ODE.

[tex]y_1=x\implies {y_1}'=1\implies{y_1}''=0[/tex]

[tex]x^2y''-xy'+y=-x+x=0[/tex]

[tex]y_2=x\ln x\implies{y_1}'=\ln x+1\implies{y_1}''=\dfrac1x[/tex]

[tex]x^2y''-xy'+y=x-x(\ln x+1)+x\ln x=0[/tex]

Both solutions satisfy the ODE.

b. The Wronskian determinant is

[tex]\begin{vmatrix}x&x\ln x\\1&\ln x+1\end{vmatrix}=x(\ln x+1)-x\ln x=x\neq0[/tex]

so the solutions are indeed independent.

c. The ODE has general solution [tex]y(t)=C_1x+C_2x\ln x[/tex]. Then with the given initial conditions, the constants satisfy

[tex]y(1)=7\implies 7=C_1[/tex]

[tex]y'(1)=2\implies2=C_1+C_2\implies C_2=-5[/tex]

So the ODE has the particular solution,

[tex]\boxed{y(t)=7x-5x\ln x}[/tex]

Final answer:

The functions y1 = x and y2 = x ln x are verified as solutions to the differential equation x^2y'' - xy' + y = 0. They are confirmed to be linearly independent through a non-zero Wronskian. Lastly, the particular solution is found to be y = 7x - 5x ln x using given initial conditions.

Explanation:

To verify that y1 = x and y2 = x ln x are solutions to the differential equation x2y'' - xy' + y = 0, we need to substitute each function into the equation and show that the left-hand side reduces to zero.

For y1 = x, its derivatives are y1' = 1 and y1'' = 0. Substituting these into the equation gives x2(0) - x(1) + x = 0, which simplifies to 0, confirming that y1 is a solution.

For y2 = x ln x, its first derivative is y2' = ln x + 1, and the second derivative is y2'' = 1/x. Substituting these into the equation gives x2(1/x) - x(ln x + 1) + x ln x = 0, which also simplifies to 0, confirming that y2 is a solution.

To demonstrate that y1 and y2 are linearly independent, we must calculate the Wronskian, W(y1,y2), and show that it is non-zero. The Wronskian is:
W(y1,y2) = y1y2' - y1'y2 = x(ln x + 1) - (x ln x) = x.
Since the Wronskian is not zero for all x
e 0, y1 and y2 are linearly independent.

For the particular solution of the differential equation with initial conditions y(1) = 7, y'(1) = 2, we express y as a linear combination of y1 and y2:
y = c1y1 + c2y2 = c1x + c2x ln x.
Applying the initial conditions, we get two equations:
1) y(1) = c1(1) + c2(1 ln 1) = 7
2) y'(1) = c1 + c2(ln 1 + 1) = 2
Simplifying these equations gives us c1 = 7 and c2 = -5, therefore the particular solution is y = 7x - 5x ln x.

Answer:

4013.45

Step-by-step explanation:

Given,

Purchased amount of the bond, P = 3500,

Annual rate of simple interest, r = 4.89% = 0.0489,

Time ( in years ), t = 3,

Since, the total amount of a bond that earns simple interest is,

[tex]A=P(1+r\times t)[/tex]

By substituting values,

The amount of the bond would be,

[tex]=3500(1+0.0489\times 3)[/tex]

[tex]=3500(1.1467)[/tex]

[tex]=4013.45[/tex]

Answer:

For maximum volume of the box, squares with 4.79 inches should be cut off.

Step-by-step explanation:

A candy box is made from a piece of a cardboard that measures 43 × 23 inches.

Let squares of equal size will be cut out of each corner with the measure of x inches.  

Therefore, measures of each side of the candy box will become

Length = (43 - 2x)

Width = (23 - 2x)

Height = x

Now we have to calculate the value of x for which volume of the box should be maximum.

Volume (V) = Length×Width×Height

V = (43 -2x)(23 - 2x)(x)

  = [(43)×(23) - 46x - 86x + 4x²]x

  = [989 - 132x + 4x²]x

  = 4x³- 132x² + 989x

Now we find the derivative of V and equate it to 0

[tex]\frac{dV}{dx}=12x^{2}-264x+989[/tex] = 0

Now we get values of x by quadratic formula

[tex]x=\frac{264\pm \sqrt{264^{2}-4\times 12\times 989}}{2\times 12}[/tex]

x = 17.212, 4.79

Now we test it by second derivative test for the maximum volume.

[tex]\frac{d"V}{dx}= 24x - 264[/tex]

For x = 17.212

[tex]\frac{d"V}{dx}= 24(17.212)-264=413.088-264=149.088[/tex]

This value is > 0 so volume will be minimum.

For x = 4.79

[tex]\frac{d"V}{dx}=24(4.79)-264=-149.04[/tex]

-149.04 < 0, so volume of the box will be maximum.

Therefore, for x = 4.79 inches volume of the box will be maximum.

To find the size of the square to cut from each corner to attain maximum volume, one needs to create a function for the volume based on the size of the cut, derive it, and solve for x when the derivative equals zero. However, this solution might require advanced calculus.

In this case, we're dealing with a problem in maximum volume. Let's say the size of the square cut is x. The length, width, and height of the box would then be 43-2x, 23-2x, and x, respectively. The volume of the box will then be (43-2x)(23-2x)*x.

To find the maximum volume, we take the derivative of this function (V'(x)) and find for which value of x it equals zero. But as this is a somewhat complex calculus problem, an alternative approach might be to solve it graphically or computationally, seeking for what value of x the volume becomes maximum.

For more detailed calculus solution, consult a mathematics teacher or resources that delve deeper into maximum and minimum problems within calculus.

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Answer:

a) 0; b) 4[tex]x^{2}[/tex]

Step-by-step explanation:

a) To compute f o g (0), first evaluate g(x) for x=0 and then evaluate f for x=g(0).

[tex]f \circ g (0)=f(2 \cdot 0)=f(0)=0^2[/tex]

b) To compute a mathematical expression for f o g do the same but instead of 0 use x,

[tex]f \circ g (x) = f( 2 \cdot x)= (2 \cdot x )^2[/tex]

In the question, f(x) = x², g(x) = 2x. We need to determine the value of function f composed with function g at 0 (f o g(0)), and the general expression for f o g(x). f o g(0) = 0 and (f o g)(x) = 4x².

To solve this problem, we first need to understand that 'f o g' denotes the composition of function f and function g, defined as (f o g)(x) = f(g(x)). In this case, function f(x) = x^2 and function g(x) = 2x.

(a) To evaluate f o g at 0, we substitute x = 0 into g(x), giving us g(0) = 2*0 = 0. Substituting g(0) into f(x), we get f(g(0)) = f(0) = 0. So, f o g(0) = 0.

(b) For a general form of f o g, we substitute g(x) = 2x into f(x), resulting in (f o g)(x) = f(2x) = (2x)^2 = 4x^2.

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Answer:

f(-4)=60 and g(-6)=17

Step-by-step explanation:

Plug in -4 for x-values

Square -4

Multiply 3 by 16 and -3 by -4 then solve

Simplify

Plug in -6 for x

Multiply -3 by -6

Subtract 1

Simplify

For this case we have the following functions:

[tex]f (x) = 3x ^ 2-3x\\g (x) = 3x-1[/tex]

We must find the value of the function [tex]f (x)[/tex] when [tex]x = -4[/tex], then:

[tex]f (-4) = 3 (-4) ^ 2-3 (-4)\\f (-4) = 3 * 16 12\\f (-4) = 48 12\\f (-4) = 60[/tex]

We must find the value of the function g (x) when [tex]x = -6[/tex], then:

[tex]g (-6) = 3 (-6) -1\\g (-6) = - 18-1\\g (-6) = - 19[/tex]

Answer:

[tex]f (-4) = 60\\g (-6) = - 19[/tex]

Answer:

A=331.84 m2

Step-by-step explanation:

Hello

The density of a object is defined by

[tex]d=\frac{m}{v}[/tex]

d is the density

m is the mass of the object

v es the volume of the object

We  have

[tex]thickness= t =2.87 * 10^{-7} m\\\ A=unknown=area\\ Volume=Area*thickness\\m=1 kg \\\\\\d=10500kg/m^{3}[/tex]

[tex]d=\frac{m}{v}\\ v=\frac{m}{d}\\ A*t=\frac{m}{d}\\ A=\frac{m}{d*t}\\ \\A=\frac{1 kg}{10500\frac{kg}{m^{3} }*2.87*10^{-7}m}\\A=331.84m^{2}[/tex]

I hope it helps

Answer:

The quotient is (-x³ + 4x² + 4x - 8) and the remainder is 0

Step-by-step explanation:

Look to the attached file

Answer:

The total number of samples that contain at least 3 red balls is 115.

Step-by-step explanation:

Total number of balls = 11

Total number of red balls = 6

Total number of white balls = 5

A sample of 4 balls is to be selected that contain at least 3 red. It means either 3 out of 4 balls are red or 4 out of 4 ball are red.

[tex]\text{Total ways}=\text{Three balls are red}+\text{Four balls are red}[/tex]

[tex]\text{Total ways}=^6C_3\times ^5C_1+^6C_4\times ^5C_0[/tex]

Combination formula:

[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Using this formula we get

[tex]\text{Total ways}=\frac{6!}{3!(6-3)!}\times \frac{5!}{1!(5-1)!}+\frac{6!}{4!(6-4)!}\times \frac{5!}{0!(5-0)!}[/tex]

[tex]\text{Total ways}=20\times 5+15\times 1[/tex]

[tex]\text{Total ways}=115[/tex]

Therefore the total number of samples that contain at least 3 red balls is 115.

Using the combination formula, it is found that 115 samples contain at least 3 red balls.

The balls are chosen without replacement, which is why the combination formula is used.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, the outcomes with at least 3 red balls are:

Hence:

[tex]T = C_{6,3}C_{5,1} + C_{6,4} = \frac{6!}{3!3!}\frac{5!}{1!4!} + \frac{6!}{4!2!} = 20(5) + 15 = 100 + 15 = 115[/tex]

115 samples contain at least 3 red balls.

A similar problem is given at https://brainly.com/question/24437717

Answer: a) 48

b) 43

Step-by-step explanation:

Given : The number of students Professor Date has in his Calculus class = 31

The number of students Professor Date has in his Discrete Mathematics class = 17

(a) If we assume that there are no students who take both classes, then the total number of students Professor Date Has = 31+17=48

(b) If we assume that there are five students who take both classes, then the total number of students Professor Date Has = 31+17-5=43

Answer:

B. Meaningful use

Step-by-step explanation:

Meaningful use is the use of EHRs in a meaningful manner.

Answer:

.008000

Step-by-step explanation:

The first digit is either 0,1,2,3,4

P( right guess) = 1/5

The second digit is either 0,1,2,3,4

P( right guess) = 1/5

The third digit is either 0,1,2,3,4

P( right guess) = 1/5

Since they are independent

P( right,right,right) = 1/5*1/5*1/5 = 1/125 =.008

To six decimal places = .008000

To solve the problem, let's consider each piece of information step by step:
1. The password is 3 digits long.
2. Each digit can be any number from 0 to 4.
Since there are 5 choices for each digit (0, 1, 2, 3, or 4), we calculate the total number of distinct combinations possible for a 3-digit password where each digit has 5 possibilities.
For each place of the three digits, we have 5 choices, which gives us a total combination count using the Multiplication Principle:
- First digit: 5 choices (0-4)
- Second digit: 5 choices (0-4)
- Third digit: 5 choices (0-4)
To find the total number of different password combinations, we multiply the number of choices for each digit:
Total combinations = 5 (choices for the first digit) × 5 (choices for the second digit) × 5 (choices for the third digit) = \( 5^3 = 125 \) possible password combinations.
Each of these combinations is equally likely if the hacker guesses at random. Hence, the probability that the hacker guesses the correct password on the first try is 1 out of the total number of combinations.
Therefore, the probability is:
\( P(\text{correct on first try}) = \frac{1}{125} \)
Let's convert this probability to a decimal and then round it to six decimal places:
\( P(\text{correct on first try}) = \frac{1}{125} = 0.008 \)
When rounded to six decimal places, the probability is:
\( P(\text{correct on first try}) \approx 0.008000 \)
So, the probability that the hacker guesses the password correctly on the first try is approximately 0.008000.

Answer:

The cost difference between lowest and highest bid $ 251.20

Step-by-step explanation:

Given,

The average quote for the traffic control subcontractor = $ 1570,

Also, the lowest bid is 4 % under average,

That is, lowest bid = average quote - 4% average quote

= 1570 - 4% of 1570

[tex]=1570-\frac{4\times 1570}{100}[/tex]

[tex]=1570-\frac{6280}{100}[/tex]

[tex]=1570-62.80[/tex]

[tex]=\$1507.2[/tex]

While, the highest bid is 12% above average,

That is, the highest bid = average quote + 12% average quote

= 1570 + 12% of 1570

[tex]=1570+\frac{12\times 1570}{100}[/tex]

[tex]=1570+\frac{18840}{100}[/tex]

[tex]=1570+188.4[/tex]

[tex]=\$1758.4[/tex]

Hence,  the cost difference between lowest and highest bid = $ 1758.4 - $ 1507.2 = $ 251.20

Answer:

The probability that the racket is wood or defective is 0.57.

Step-by-step explanation:

Let W represents wood racket, G represents the graphite racket and D represents the defective racket,

Given,

n(W) = 100,

n(G) = 100,

⇒ Total rackets = 100 + 100 = 200

n(W∩D) = 15,

n(G∩D) = 14,

⇒ n(D) = n(W∩D) + n(G∩D) = 15 + 14 = 29,

We know that,

n(W∪D) = n(W) + n(D) - n(W∩D)

= 100 + 29 - 15

= 100 + 14

= 114,

Hence, the probability that the racket is wood or defective,

[tex]P(W\cup D) = \frac{114}{200}[/tex]

[tex]=0.57[/tex]

Answer:

There are going to be 31 matches played in the soccer league.

Step-by-step explanation:

The soccer league has 6 teams, so if every team plays against the others twice, there are going to be played 30 matches:

-Team 1: v Team 2 (2), v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 2: v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 3: v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 4: v Team 5 (2), v Team 6 (2)

-Team 5: v Team 6 (2)

-Team 6: -

If there is a final championship game after the 30 regular season matches, there are going to be 31 matches played in the league.

Answer:

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

Step-by-step explanation:

We are given the following quadratic equation by completing the square:

[tex]z^2 - 3z - 5 = 0[/tex]

Rewriting the equation in the form [tex]x^2+2ax+a^2[/tex] to get:

[tex]z^2 - 3z - 5+(-\frac{3}{2} )^2-(-\frac{3}{2} )^2[/tex]

[tex]z^2-3z+(-\frac{3}{2} )^2=(z-\frac{3}{2} )^2[/tex]

Completing the square to get:

[tex] ( z - \frac{ 3 } { 2 } )^ 2 - 5 - ( - \frac { 3 } { 2 } ) ^ 2[/tex]

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

Answer: [tex]z_1=4.19\\\\z_2=-1.19[/tex]

Step-by-step explanation:

Add 5 to both sides of the equation:

[tex]z^2 - 3z - 5 +5= 0+5\\\\z^2 - 3z = 5[/tex]

Divide the coefficient of [tex]z[/tex] by two and square it:

[tex](\frac{b}{2})^2= (\frac{3}{2})^2[/tex]

Add it to both sides of the equation:

[tex]z^{2} -3z+ (\frac{3}{2})^2=5+ (\frac{3}{2})^2[/tex]

Then, simplifying:

[tex](z- \frac{3}{2})^2=\frac{29}{4}[/tex]

Apply square root to both sides and solve for "z":

[tex]\sqrt{(z- \frac{3}{2})^2}=\±\sqrt{\frac{29}{4} }\\\\z=\±\sqrt{\frac{29}{4}}+ \frac{3}{2}\\\\z_1=4.19\\\\z_2=-1.19[/tex]

Answer:

40514.837 kg

Step-by-step explanation:

The weight of the main axle of the 1893 ferris wheel built by George Washington Gale Ferris Jr. in Chicago, USA was 89,320 pounds (lb).

1 kg=2.20462 pounds (lb)

[tex]\Rightarrow 1 lb=\frac{1}{2.20426}[/tex]

[tex]\Rightarrow 1 lb=0.453592 kg[/tex]

[tex]\Rightarrow 89320 lb=89320\times 0.453592[/tex]

[tex]\therefore 89320 lb=40514.837 kg[/tex]