Please help me with this question.
Will mark brainliest
Thanks so much

Express the matrix [tex]I[/tex] like [tex](\delta_{ij})_{n\times n}[/tex], where [tex]\delta_{ij}=1[/tex] if [tex]i=j[/tex] and [tex]\delta_{ij}=0[/tex] if [tex]i\neq j[/tex]. ([tex]\delta_{ij}[/tex] is known as kronecker's delta)

In the same form we express the matrix [tex]A=(a_{ij})_{n\times n}[/tex].

The firs index indicate the row and the second the column.

By the multiplication of matrices we have [tex]AI=(c_{ij})_{n\times n}[/tex], where

[tex]c_{ij}=\sum_{k=1}^n a_{ik}\delta _{kj} = a_{ij}[/tex]

because only [tex]\delta_{jj}[/tex] is non-zero in the last sum.

therfore we have [tex]AI=(c_{ij})_{n\times n}=(a_{ij})_{n\times n}=A[/tex].

In the same manner we have

[tex]IA=(d_{ij})_{n\times n}[/tex], where

[tex]d_{ij}=\sum_{k=1}^n \delta _{ik}a_{kj} = a_{ij}[/tex]

And so, [tex]IA=A[/tex]

Matrix multiplication with the identity matrix leaves the original matrix unchanged. This is analogous to multiplying a number by 1, which leaves the number unchanged. Thus, for any square matrix A, AI = IA = A.

The n×n identity matrix, denoted by I, is a special square matrix that has ones on its main diagonal and zeros elsewhere. It's called the 'identity' matrix because multiplication with it leaves a matrix unchanged, similar to how multiplying a number by 1 leaves it unchanged.

Let's prove that for any n×n matrix A, we have AI=IA=A. As A is an n×n matrix, suppose it has elements aij. Because of the definition of matrix multiplication, element at ith row and jth column of the product of two matrices (say A and I) is given by summation of the products of corresponding elements in ith row of first matrix and jth column of the second matrix.

In the case of AI, for any element in the result, we get it by summation of product of corresponding elements in ith row of A and ith column of I. Now, since except for the ith position, rest of the elements in ith column of I are zero, we only get aii in the summation. Hence, AI = A. Similar argument can be made for IA. Therefore, AI=IA=A.

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The confidence interval for population mean is given by :-

[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where [tex]\hat{p}[/tex] is the sample proportion, n is the sample size , [tex]z_{\alpha/2}[/tex] is the critical z-value.

The  values needed to calculate a confidence interval at the 99% confidence level are :

Given : Significance level : [tex]\alpha:1-0.99=0.01[/tex]

Sample size : n=450

Critical value : [tex]z_{\alpha/2}=2.576[/tex]

Sample proportion: [tex]\hat{p}=\dfrac{280}{450}\approx0.62[/tex]

Now, the  99% confidence level will be :

[tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.62\pm(2.576)\sqrt{\dfrac{0.62(1-0.62)}{450}}\\\\\approx0.62\pm0.023\\\\=(0.62-0.023,\ 0.62+0.023)=(0.597,\ 0.643)[/tex]

Hence, the  99% confidence interval is [tex](0.597,\ 0.643)[/tex]

To calculate a 99% confidence interval for the true proportion, find the values needed and apply the formula. In this case, the confidence interval is between 0.5601 and 0.6811.

1. Sample proportion [tex](\( \hat{p} \))[/tex]:

[tex]\[ \hat{p} = \frac{x}{n} = \frac{280}{450} \approx 0.6222 \][/tex]

2. Margin of error  E :

For a 99% confidence level, z = 2.576.

[tex]\[ E = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]\[ E = 2.576 \times \sqrt{\frac{0.6222 \times (1-0.6222)}{450}} \]\[ E = 2.576 \times \sqrt{\frac{0.6222 \times 0.3778}{450}} \]\[ E = 2.576 \times \sqrt{\frac{0.235052}{450}} \]\[ E \approx 2.576 \times \sqrt{0.00052233} \]\[ E \approx 2.576 \times 0.022854 \]\[ E \approx 0.058893 \][/tex]

3. Confidence interval:

[tex]\[ \text{Confidence interval} = \hat{p} \pm E \]\[ \text{Confidence interval} = 0.6222 \pm 0.0589 \][/tex]

Therefore, the confidence interval for the percentage of elementary school children who have a social media account at the 99% confidence level is approximately (0.5633, 0.6811) .

Answer:

a) There are 3,628,800 different strings.

b) There are 28,800 different strings if the letters ad digits must alternate.

c)There are 86,400 different string if all five letters must be adjacent in each string.

Step-by-step explanation:

There are 10 digits.

Our strings have the following format:

C1 - C2 - C3 - C4 - C5 - C6 - C7 - C8 - C9 - C10

repitition of letters is not allowed.

a) How many different strings are there?

C1 can be any of the 10, C2 can be 9, C3 can be 8, ...

So in total there are:

[tex]T = 10*9*8*7*6*5*4*3*2*1 = 3,628,800[/tex]

There are 3,628,800 different strings.

b) How many different strings are there if the letters, i.e. A, E, I, O, U and the digits, i.e. 1,3, 5, 7, 9 must alternate?

There are the following possiblities:

5(l) - 5(d) - 4(l) - (4d) - ...

Or

5(d) - 5(l) - 4(d) - 4(l) - ...

So:

[tex]T = 2*(5*5*4*4*3*3*2*2*1*1) = 28,800[/tex]

There are 28,800 different strings if the letters ad digits must alternate.

c) How many different strings are there if all five letters must be adjacent in each string?

L - L - L - L - L - D - D - D - D - D

D - L - L - L - L - L - D - D - D - D

D - D - L - L - L - L - L - D - D - D

D - D - D - L - L - L - L - L - D - D

D - D - D - D - L - L - L - L - L - D

D - D - D - D - D - L - L - L - L - L

There are [tex]T = 6*(5*5*4*4*3*3*2*2*1*1) = 86,400[/tex]

There are 86,400 different string if all five letters must be adjacent in each string.

Answer:

  (x, y) = (4, -3)

Step-by-step explanation:

Relative to straight east with angles measured CCW, the vector is ...

  5∠-36.9° = 5(cos(-36.9°), sin(-36.9°)) = 5(0.8, -0.6) = (4, -3)

The components of vector A that has a magnitude of 5.00 and is at a 36.9 degrees angle south of east, are 3.96 towards east (x-direction), and -3.00 towards south (y-direction).

The components of a vector can be calculated by using trigonometry. Where the x-component or the horizontal component can be found using the cosine of the angle and the y-component or the vertical component can be found using the sine of the angle. However, there are different coordinate systems and conventions. In this case, positive x-direction is to the east and positive y-direction is to the north, and the angle is measured from the positive x-axis (east) but the vector points towards the negative y-axis (south). So you need to apply a negative to the sine of the angle.

For the vector A⃗:

Therefore, the components of vector A⃗ are 3.96 in x-direction (east) and -3.00 in y-direction (south).

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Answer:

[tex]y=c_0(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(2*3)(5*6)...(3k-1)(3k)} x^{3k}])+c_1(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(3*4)(6*7)...(3k)(3k+1)} x^{3k+1}])[/tex]

Step-by-step explanation:

[tex]y"+4xy=0[/tex]

Assume that the problem has a solution of the form:

[tex]y=f(x)=\sum_0^{\infty}(c_nx^n)[/tex]

then calculate the derivative:

[tex]y'=\sum_1^{\infty}(nc_nx^{n-1})[/tex]

and the second derivative:

[tex]y"=\sum_2^{\infty}(n(n-1)c_nx^{n-2})[/tex]

In these sums the subindex indicates the first non cero term.

Making the substitution in the original equation:

[tex]\sum_2^{\infty}(n(n-1)c_nx^{n-2})+4x\sum_0^{\infty}(c_nx^{n})=0\\[/tex]

Now look at the subindex of the sum: they do not match; you need to make them match:

[tex]\sum_0^{\infty}((n+2)(n+1)c_{n+2}x^{n})+4\sum_0^{\infty}(c_nx^{n+1})=0\\[/tex]

Now they match and they could be joined together on the same sum symbol, however the x term doesn't have the same exponent (after multiplying the x from the original formula with the one inside the sum). Therefore you need to expand the second derivative term and displace the other one (as we did in the previous step) to match them both:

[tex]2c_2+\sum_1^{\infty}((n+2)(n+1)c_{n+2}x^{n})+4\sum_1^{\infty}(c_{n-1}x^{n})=0[/tex]

Now you can join them together:

[tex]2c_2+\sum_1^{\infty}[(n+2)(n+1)c_{n+2}x^{n}+4(c_{n-1}x^{n})]=0[/tex]

[tex]2c_2+\sum_1^{\infty}x^n=0[/tex]

If the solution is valid for all x, then all terms must be independantly 0. therefore:

[tex]a_2=0\\(n+2)(n+1)c_{n+2}+4(c_{n-1})=0\\c_{n+2}=\frac{-4(c_{n-1})}{(n+2)(n+1)}[/tex]

Now you need to use that recurrent function to calculate the coefficients:

[tex]n=1 \rightarrow c_3=\frac{-4c_0}{3*2}=-\frac{2}{3} c_0\\n=2 \rightarrow c_4=\frac{-4c_1}{4*3}=-\frac{1}{3} c_1\\n=3 \rightarrow c_5=\frac{-4c_2}{5*4}=-\frac{1}{5} c_2=0\\n=4 \rightarrow c_6=\frac{-4c_3}{6*5}=-\frac{2}{15} c_3\\n=5 \rightarrow c_7=\frac{-4c_4}{7*6}=-\frac{2}{21} c_4\\n=6 \rightarrow c_8=\frac{-4c_5}{8*7}=0\\n=7 \rightarrow c_9=\frac{-4c_6}{9*8}=-\frac{1}{18} c_6[/tex]

you found that for each n divisible by 3, therefore for each [tex]c_{3n-1}[/tex] the coefficient is 0:

[tex]c_2=c_5=c_8....=c_{3n-1}=0[/tex]

now look at the terms [tex]a_3,a_6,a_9[/tex], they are all recurrent to c_0, therefore you can wirte a rule for them:

[tex]c_{3n} = \frac{(-4)^nc_0}{(2*3)(5*6)...(3n-1)(3n)}[/tex]

Finally, look at the terms [tex]a_4,a_7[/tex], they recurr to c_1, therefore:

[tex]c_{3n+1} = \frac{(-4)^nc_1}{(3*4)(6*7)...(3n)(3n+1)}[/tex]

So you have two constants that are independant, c_0 and c_1.

Therefore the solution must be writen in terms of these two arbitrary constants:

[tex]y=c_0(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(2*3)(5*6)...(3k-1)(3k)} x^{3k}])+c_1(1+\sum_{k=1}^\infty[\frac{(-4)^k}{(3*4)(6*7)...(3k)(3k+1)} x^{3k+1}])[/tex]

Answer:

1/5525

Step-by-step explanation:

We now that a standard deck has 52 different cards. Also we know that a standard deck has four different suits, i.e., Spades, Hearts, Diamonds and Clubs.  We can find the following cards for each suit: Ace, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen and King.

Now, the probability of getting any of these cards off the top of a standard deck of well-shuffled cards is 1/52. As we have 4 different sixes, we have that the probability of getting a six is 4/52. When we get a six, in the deck only remains 3 sixes and 51 cards, so, the probability of getting another six later is 3/51. When we get the second six, in the deck only remains 2 sixes and 50 cards, so, the probability of getting the third six is 2/50. As we have independet events, we should have that the probability of getting 3 sixes off the top of a standard deck of well-shuffled cards is

(4/52)(3/51)(2/50)=

24/132600=

12/66300=

6/33150=

3/16575=

1/5525

The probability of being dealt three sixes off the top of a standard deck of well-shuffled cards is approximately 1/5513 or 0.00018 when rounded to five significant digits.

The subject of this question is probability in mathematics. In a standard deck of 52 playing cards, there are four sixes: one each of hearts, diamonds, clubs, and spades. When looking at the probability of being dealt three sixes off the top of a well-shuffled deck, looking at drawing one card at a time in succession grants us the solution.

For the first card, the probability of drawing a six is 4/52. If you draw a six, there are now three sixes left in a 51-card deck. So, the probability of drawing a six on the second draw is 3/51. Using the same logic, the probability of drawing a six on the third draw is 2/50.

The probability of these three events happening in succession is the product of their individual probabilities, which is calculated as follows: (4/52) * (3/51) * (2/50) = 24/132600 = 0.00018 when rounded to 5 significant digits, or simplified, this is approximately 1/5513.

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Answer:

y = [tex]C_{1}e^{2x} + C_{2}e^{-2x}[/tex]

Step-by-step explanation:

We are given the differential equation: y'' - 4y = 0

We have to find the general solution.

The auxiliary equation for the above differential equation can be written as:

m² - 4 = 0

We solve for m.

⇒m² = 4

⇒m = ±2

⇒[tex]m_{1}[/tex] = +2 and [tex]m_{2}[/tex] = -2

Thus, we have two distinct roots or we have two distinct values of m.

Thus, the general solution will be of the form:

y = [tex]C_{1}e^{m_{1}x} + C_{2}e^{m_{2}x}[/tex]

y = [tex]C_{1}e^{2x} + C_{2}e^{-2x}[/tex]

Answer:

30 million.

Step-by-step explanation:

Let x be the number of accountants who e-filed income tax returns in 2003.

We have been given that in 2004, 34.2 million accountants e-filed income tax returns. That was 114% of the number who e-filed in 2003.

We can represent our given information in an equation as:

[tex]\frac{114}{100}\cdot x=34.2[/tex]

[tex]1.14x=34.2[/tex]

[tex]\frac{1.14x}{1.14}=\frac{34.2}{1.14}[/tex]

[tex]x=30[/tex]

Since the number is given in millions, therefore, 30 million accountants e-filed income tax returns in 2003.

Answer: 0.9575465

Step-by-step explanation:

Let the random variable X is normally distributed with mean [tex]\mu=50[/tex] and standard deviation[tex]\sigma=7[/tex] .

Using the formula , [tex]z=\dfrac{x-\mu}{\sigma}[/tex] , we have the z-value for x= 34

[tex]z=\dfrac{34-50}{7}\approx-2.29[/tex]

For x= 63

[tex]z=\dfrac{63-50}{7}\approx1.86[/tex]

P-value : P(34<x<63)=P(-2.29<z<1.86)

[tex]=P(z<1.86)-P(z<-2.29)\\\\=0.9685572-(1-P(z<2.29))\\\\1=0.9685572-(1-0.9889893)\\\\=0.9575465[/tex]

Hence, the required probability = 0.9575465

Answer:

y+0.5x-2=0

Step-by-step explanation:

Given,

X-intercept of line = 4

So, the line intersect the x axis at (4,0)

Y-intercept of the line = 2

So, the line intersect the y axis at (0,2)

So, the slope of the line can be given by

[tex]m\ =\ \dfrac{y_2-y_1}{x_2-x_1}[/tex]

    [tex]=\ \dfrac{2-0}{0-4}[/tex]

      = -0.5

Hence, the equation of line can be given by

y=mx+c

where, m=slope of the line

             c= y-intercept of line

So, after putting the value of m and c in above equation, the equation of line will be

y = -0.5x + 2

=> y+0.5x-2=0

So, the equation of line will be y+0.5x-2=0.

Answer with Step-by-step explanation:

We are given some statements

We have to find that which statements is bi-conditional

We know that

Bi-Conditional statement:It is combination of conditional statement and its converse written as  if and only if .Bi- conditional statement is true if and only both the conditions are true.

1.I am sleeping if and only if I am snoring.

If I am sleeping then I am snoring .Then,it may be true or may not be  true.

If I am snoring then I am sleeping .It is true.

Its one side true result.So, it is not bi conditional true statement.

2.Mary will eat pudding today if and only if is custard.

If Mary will eat pudding today then it is custard.It may or may not be true because pudding can be any soft sweet desserts.It is not necessary that it is custard only.

If it is custard then Mary will east pudding today.It is true, because it is soft sweet dessert.

Hence, it is one side true. Therefore, It is not bi- conditional true.

3.It is raining if and only if it is cloudy.

If it is raining then it is cloudy .It is true.

But if it is cloudy then it is raining. It  may be true or may not be true.

Hence, it is one side true result.Therefore, it is not bi conditional true.

Therefore, any given statement is not both side true.

So, the answer is none of the above because bi conditional statement is true if and only if both the conditions are true.

Answer:

The number of reflexive relations on S  is 64.

The number of reflexive and symmetric relations on S  is 8.

Step-by-step explanation:

Consider the provided set S = {a, b, c}.

The number of elements in the provided set is 3.

Part (a) the number of reflexive relations on S

To calculate the number of reflexive relation on S we can use the formula as shown:

Total number of Reflexive Relations on a set: [tex]2^{n(n-1)}[/tex].

Where, n is the number of elements.

In the provided set we have 3 elements, so substitute the value of n in the above formula:

[tex]2^{3(3-1)}[/tex]

[tex]2^{3(2)}[/tex]

[tex]2^{6}[/tex]

[tex]64[/tex]

Hence, the number of reflexive relations on S  is 64.

Part(b) The number of reflexive and symmetric relations on S.

To calculate the number of reflexive and symmetric relation on S we can use the formula as shown:

Total number of Reflexive and symmetric Relations on a set: [tex]2^{\frac{n(n-1)}{2}}[/tex].

Where, n is the number of elements.

In the provided set we have 3 elements, so substitute the value of n in the above formula:

[tex]2^{\frac{3(3-1)}{2}}[/tex]

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

[tex]2^{\frac{6}{2}}[/tex]

[tex]2^{3}[/tex]

[tex]8[/tex]

Hence, the number of reflexive and symmetric relations on S  is 8.

Answer:

The two events are dependent.

the probability that events A and B both occur is:

[tex]\frac{1}{2424}\times \frac{1}{2423}=1.7026052585\times 10^{-7}\ or\ 0.00000017[/tex]

Step-by-step explanation:

Consider the provided information.

Event A: When a page is randomly selected and ripped from a 2424​-page document and​ destroyed, it is page 2020.

Event B: When a different page is randomly selected and ripped from the​ document, it is page 1616.

Part(A)

The occurring of one event affects the probability of the other event.

Because if we ripped one page then the probability of ripping second page will going to change as the size of sample space will decrease.

For example: The document has 2424 page, that means size of sample space is 2424. If we ripped another page, the sample size will be 2423. That means event B is depending on event A.

Thus, the two events are dependent.

Part(B)

The probability that events A and B both occur.

If we select a page randomly from 2424-page document, the probability will be: 1/2424

Now we have 2423 pages left in the document as one page is destroyed.

The probability of selecting another page is: 1/2423.

Thus, the probability that events A and B both occur is:

[tex]\frac{1}{2424}\times \frac{1}{2423}=1.7026052585\times 10^{-7}\ or\ 0.00000017[/tex]

Answer:

[tex]\frac{25}{8}[/tex]

Step-by-step explanation:

A fraction is a part of a whole .

A proper fraction is a fraction whose numerator is less than denominator .

For example [tex]\frac{2}{3}\,,\,\frac{3}{4}[/tex]

An improper fraction is a fraction whose numerator is greater than denominator .

For example [tex]\frac{5}{4}\,,\,\frac{8}{7}[/tex]

A mixed fraction is made up of whole number and a proper fraction .

Given : [tex]4\frac{1}{6}\,,1\frac{1}{3}[/tex]

Solution :

[tex]4\frac{1}{6}=\frac{4\times 6+1}{6}=\frac{25}{6}\\1\frac{1}{3}=\frac{3\times 1+1}{3}=\frac{4}{3}[/tex]

We need to divide [tex]4\frac{1}{6}[/tex] by [tex]1\frac{1}{3}[/tex] .

[tex]4\frac{1}{6}\div 1\frac{1}{3}=\frac{25}{6}\div\frac{4}{3} \\\Rightarrow 4\frac{1}{6}\div 1\frac{1}{3}= \frac{25}{6}\times \frac{3}{4}=\frac{25}{8}[/tex]

Answer:

Step-by-step explanation:

Given that the cost function is linear.

Fixed charges = 4 dollars

Variable charges perhalf hour = 65 cents = 0.65 dollars

Suppose a man leaves for x hours say

then we have he would be charged 4 dollars besides 0.65 for 2x half hours.

Hence

[tex]C(x) = 5+4(2x)\\C(x) = 8x+5[/tex] where x is the number of hours.

The variable charge is 1.3 dollars per hour and the fixed cost is 4 then the linear cost equation is C(x) = 1.3x + 4.

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.

Given

A parking garage charges 4 dollars plus 65 cents per half-hour.

Let C(x) be the linear cost.

The fixed charge is 4 dollars

The variable charge per half-hour is 0.65 dollars.

Suppose the men leave for x half-hours for one hour the x will be 2x.

Then the equation will be

[tex]\rm C(x) = 0.65*2x + 4\\\\C(x) = 1.3x + 4[/tex]

Thus, the equation is [tex]\rm C(x) = 1.3x + 4[/tex].

More about the linear system link is given below.

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Answer: 0.2643

Step-by-step explanation:

Given : The proportion of  adults are unemployed : p=0.077

The sample size = 300

By suing normal approximation to the binomial , we have

[tex]\mu=np=300\times0.077=23.1[/tex]

[tex]\sigma=\sqrt{np(1-p)}=\sqrt{300\times0.077(1-0.077)}\\\\=4.61749932323\approx4.62[/tex]

Now, using formula [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponding to 26 will be :-

[tex]z=\dfrac{26-23.1}{4.62}\approx0.63[/tex]

Using standard distribution table for z , we have

P-value=[tex]P(z\geq0.63)=1-P(z<0.63)[/tex]

[tex]=1-0.7356527=0.2643473\approx0.2643[/tex]

Hence, the probability that at least 26 in the sample are unemployed  =0.2643

Answer:

The output for NAND gate is false if all the inputs are true. it is true if either A or B is False.

Step-by-step explanation:

Step1

NAND operator:

NAND is a logic or Boolean expression and is a combination of NOT and AND Gate.

NAND gate is built using various junction diodes and transistors.

Step2

The output of NAND can be 0 or 1. Here, LOW (0) and HIGH (1). 0 or 1 is commonly named as FALSE or TRUE respectively.

It is basically the complement of AND gate. If all its inputs are True or 1, then the output is False or 0 else if either A or B is 0 or 1 then it will give the output as TRUE.

Step3

Here, A and B are 2 inputs used to make the truth table for NAND.

In 1st row, when A and B are both 0, the output for NAND is 1.

Likewise, in 2st row and 3rd row, when either A or B is 0 or 1, the output for NAND is 1 that is TRUE.

But in last row, when both inputs A and B are true, then the output for NAND gate is FALSE or 0.

The diagram and truth table for NAND is shown below.

Final answer:

A truth table for the NAND logical operator lists all possible binary input combinations of variables A and B, applies the AND operation, and then negates the output to present the NAND result. The resulting truth table will show that a NAND gate outputs true for all cases except when both inputs are true.

Explanation:

Constructing a truth table for the logical operator NAND is a way to visualize how this operator works with different input values.

The NAND operator is essentially the negation of the AND operator, meaning that it outputs the opposite of what an AND gate would. Here is how you construct a truth table for NAND:

The resulting truth table for the NAND operation is as follows:

A B A AND B A NAND B

false false false true

false true false true

true false false true

true true true false

Answer:

a) You can buy 282.88 British pounds.

b) you should use 17.5 cups of flour.

Step-by-step explanation:

a) You arrive in London with $400. How many British pounds can you purchase with your US dollars if the conversion is $1.414 per British pound?

We can solve this problem by using a rule of three

1.414 dollars --- 1 British pound

400 dollars  ---  x British pounds

We solve for x,

x = 400/1.414 = 282.88

We can buy 282.88 British pounds.

b) Your goal is to prepare a recipe for a gathering of 21 people. If the original recipe serves 6 people and requires 5 cups of flour, how much flour should you use?

Again, we can use a rule of three.

6 people --- 5 cups of flour

21 people --- x cups of flour

We solve for x:

x= (21)(5)/6 = 105/6=17.5

We should use 17.5 cups of flour.

Answer:

Considering the equation f(x) = 64-7x+ 4x^2 this are the answers

a) Equation 1:

[tex]y=8x-7[/tex]

b) Equation 2:

[tex]y=\frac{64}{x}-7+4x[/tex]

c) Intersection of Equation 1 and Equation 2:

The lines intersects in the points:

P(x,y)=(4,25) and P(x,y)=(-4,-39)

Step-by-step explanation:

a) Drivate f(x) to find Equation 1:

y=f'(x)

y=0-7+8x

y=8x-7

b) Equation 2 is f(x)/x

y=f(x)/x

[tex]y=\frac{64-7x+4x^2}{x}[/tex]

[tex]y=\frac{64}{x}-7+4x[/tex]

c) The intersection between the two equations is the only point that they have in common, this means that the points (x,y) satisfies both equations

To find it lets write the equations side by side

[tex]\left \{ {{y=8x-7} \atop {y=\frac{64}{x}-7+4x}} \right.[/tex]

Given the fact that the y points are the same for both equations, you can replace the equation 1 into equation 2, this means, instead of write y, write equation 2:

[tex]8x-7=\frac{64}{x}-7+4x[/tex]

now you can solve this for x

[tex]8x-4x-7+7=\frac{64}{x}\\4x=\frac{64}{x}\\4x^2=64[/tex]

[tex]x^2=\frac{64}{4}\\x=\sqrt{16}\\ x=±4[/tex]

With the values of x, you can find the values of y by putting it into equation 1:  

y=8*(+4)-7 and y=8*(-4)-7

y=25 and y=-39

Finally, the points where these two equations intersect are P(x,y)=(4,25) and P(x,y)=(-4,-39).

Answer:

144

Step-by-step explanation:

when you square a negative number, it would become positive.

4*-3 = -12^2 = 144

Answer:

a) There is a 6.69% probability that a randomly selected female student abstains from alcohol.

b) If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.

Step-by-step explanation:

This is a probability problem:

We have these following probabilities:

-20.7% of a woman attending an all-women college abstaining from alcohol.

-6% of a woman attending a coeducational college abstaining from alcohol.

-4.7% of a woman attending an all-women college

- 100%-4.7% = 95.3% of a woman attending a coeducational college.

(a) What is the probability that a randomly selected female student abstains from alcohol?

[tex]P = P_{1} + P_{2}[/tex]

[tex]P_{1}[/tex] is the probability of a woman attending an all-women college being chosen and abstaining from alcohol. There is a 0.047 probability of a woman attending an all-women college being chosen and a 0.207 probability that she abstain from alcohol. So:

[tex]P_{1} = 0.047*0.207 = 0.009729[/tex]

[tex]P_{2}[/tex] is the probability of a woman attending a coeducational college being chosen and abstaining from alcohol. There is a 0.953 probability of a woman attending a coeducational college being chosen and a 0.06 probability that she abstain from alcohol. So:

[tex]P_{2} = 0.953*0.06 = 0.05718[/tex]

So, the probability of a randomly selected female student abstaining from alcohol is:

[tex]P = P_{1} + P_{2} = 0.009729 + 0.05718 = 0.0669[/tex]

There is a 6.69% probability that a randomly selected female student abstains from alcohol.

(b) If a randomly selected female student abstains from alcohol, what is the probability she attends a coedücational colege?

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

Here:

What is the probability of a woman attending a coeducational college, knowing that she abstains from alcohol.

It can be calculated by the following formula:

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

We have the following probabilities:

[tex]P(B)[/tex] is the probability of a woman from a coeducational college being chosen. So [tex]P(B) = 0.953[/tex]

[tex]P(A/B)[/tex] is the probability of a woman abstaining from alcohol, given that she attends a coeducational college. So [tex]P(A/B) = 0.06[/tex]

[tex]P(A)[/tex] is the probability of a woman abstaining from alcohol. From a), [tex]P(A) = 0.0669[/tex]

So, the probability that a randomly selected female student attends a coeducational college, given that she abstains from alcohol is:

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.953)*(0.06)}{(0.0669)} = 0.8287[/tex]

If a randomly selected female student abstains from alcohol, there is a 82.87% probability that she attends a coeducational college.

Answer:

The set {d,o,g}: {c,a,t} is equivalent but not equal.

The set {run} : {r,u,n} is neither equivalent nor equal.

The set {t,o,p} :{p,o,t} is equal.

Step-by-step explanation:

Consider the provided sets.

Two sets are equal if they have the exact same elements.

Two sets are equivalent if they have the same number of elements.

Now consider the provided sets:

{d,o,g}: {c,a,t}

The number of elements in set {d,o,g} is 3.

The number of elements in set {c,a,t}  is 3.

Thus, the sets are equivalent, but they have different elements, so the sets are not equal.

{run} : {r,u,n}

The number of elements in set {run} is 1.

The number of elements in set {r,u,n} is 3.

Thus, the sets are not equivalent, also they have different elements, so the sets are not equal.

{t,o,p} :{p,o,t}

The number of elements in set {t,o,p} is 3.

The number of elements in set {p,o,t}  is 3.

Also they have same elements i.e (o, p and t),

Hence, the sets are equal.

Sets {d,o,g} and {c,a,t} are equivalent, {run} and {{r,u,n}} are neither, and {t,o,p} and {p,o,t} are equal. Equality refers to having identical elements, while equivalence means having the same number of elements.

We are asked to determine if each pair of sets is equal, equivalent, or neither. Let's consider each pair:

So for the given sets:

Answer:

$ 30

Step-by-step explanation:

Let x be the cost of tie ( in dollars ),

∵ The  shirt costs $21 more than the tie,

So, the cost of shirt = x + 21

Thus, the cost of a tie and a shirt = x + x + 21 = 2x + 21,

According to the question,

2x + 21 = 57

2x = 57 - 21

2x = 36

x = 18

Hence, the cost of the shirt = 18 + 12 = $ 30

The problem is solved using Polya's four-step problem-solving process. By setting up two equations based on the problem, we determine that the tie costs $18 and the shirt costs $39 which is confirmed by substitution into the original problem statement.

According to Polya's four-step problem solving strategy, we first need to understand the problem. We know that a shirt and tie cost $57 together, and the shirt costs $21 more than the tie.

Step two of Polya's strategy involves devising a plan. In this case, we can write two expressions to represent the cost of the two items: S (the cost of the shirt) and T (the cost of the tie). The total cost (S+T) equals $57. We also know that S (the cost of the shirt) is T (the cost of the tie) plus $21.

For step three, we carry out the plan. We can substitute S from the second expression into the first expression: (T + $21) + T = $57. Solving for T gives us T = $18. Hence, the tie costs $18.

Finally, for step four, we review our answer. We can plug T back into the second expression to get S = T + $21 = $18 + $21 = $39. So, the shirt costs $39, which sounds reasonable given that the shirt is supposed to be more expensive than the tie.

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Answer: It is 1\2x and the next one is 1\2x+10 then the last one is 1\2x+10=22

Step-by-step explanation:For sure they are right

The algebraic expression of half a number is expressed as:  x/2

Algebraic expressions are defined as the idea of expressing numbers using letters or alphabets without specifying their actual values.

Now, we can represent a number in general by x.

Halving a number means dividing it by two, so the expression that shows the half of a number is:

x/2, since you divide a number by two.

Thus, we conclude that the algebraic expression of half a number is x/2

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Answer:

For all n ≥ 4, 2n < n!

Step-by-step explanation:

Let's use the induction method to prove this statement.

In the induction method, first we prove the statement for n=4

1) If n = 4 ⇒2(4) < 4! ⇒2(4) < 24 ⇒8 < 24.

Therefore the statement holds for n=4

2) Now we assume that the statement is valid for  n = k

⇒2k < k!

3) Now we will prove the statement holds for n = k +1

We will prove that 2(k + 1) < (k +1)!

(k + 1)! = (k+1) (k) (k-1) .... (3) (2) (1)

If the statement is valid for k + 1, then it would mean that

2 (k + 1) < (k+1) (k) (k-1) ... (3) (2) (1)

2 < (k) (k-1).... (3) (2) (1)

which is clearly true since k ≥4

Therefore the statement n ≥4, 2n < n! is true.

Using proof by induction, we establish the base case for n = 4 and then assume 2k < k! to show 2(k+1) < (k+1)! holding true for k ≥ 4, which proves the inequality for all n ≥ 4.

To prove that 2n < n! for all n ≥ 4, we will use proof by induction.

We start by establishing the base case for n = 4, where we compute 2^4 = 16 and 4! = 24 which confirms that 16 < 24.

Now, assume the inequality 2k < k! is true for some k ≥ 4.

For our inductive step, we need to show that if 2k < k! then 2(k+1) < (k+1)!.

Multiplying both sides of our assumption by 2 gives us 2*2k < 2*k!, and since k ≥ 4, we know that 2 < k + 1.

Multiplying our original inductive assumption by (k + 1) gives us (k+1)*2k < (k+1)*k!, which simplifies to 2(k+1) < (k+1)! as required.

The important thing to note here is that a limit is defined when the value of f(x) as it approaches x is the same when approaching from both the left and the right sides (unless a specific side from which to approach has been defined in the question, eg. lim (x -> 10⁻) f(x) or lim (x -> 10⁺) f(x) ).

1. The first line states that the limit of f(x) as x approaches 10 is 0. This means that the value of f(x) as x approaches 10 from both the left and the right sides must be 0.

Looking at the graph, we can see that this is indeed the case, and may write this as:

lim (x -> 10⁻) f(x) = lim (x -> 10⁺) f(x) = 0

Thus, the statement is True.

2. The second line states that the limit of f(x) as x approaches -2 from the right side is 3.

Looking at the graph, whilst f(x) approaches 3 as x approaches -2 from the left side, f(x) approaches 7 as x approaches -2 from the right side. We can write this as:

lim (x -> -2⁺) f(x) = 7

Thus, since lim (x -> -2⁺) f(x) ≠ 3, the statement is False.

3. The third line states that the limit of f(x) as x approaches -8 is the value of f(x) at x = -8 (ie. the value of f(x) for which the graph exists at x = -8).

Looking at the graph, we can see that the graph of f(x) approaches -6 as x approaches -8 from both the left and right sides. We can write this as:

lim (x -> -8⁻) f(x) = lim (x -> -8⁺) f(x) = -6

Now, this is where a knowledge of what open and closed circles represent on a graph is crucial. A closed circle means that a point exists on a graph, whereas an open circle means that there is a break in the graph at that point.

If we look at the graph, we can see that the closed circle at x = -8 is actually the value f(x) = -3. We can write this as:

f(-8) = -3

Moving from the line before this, where we defined the limit as -6, we can thus write:

lim (x -> -8) f(x) ≠ -3

Therefor, lim (x -> -8) f(x) ≠ f(-8)

Thus, the statement is False.

4. The fourth line states that the limit of f(x) as x approaches 6 is 5. Thus, for this to be true, f(x) must approach 5 as x approaches 6 from both the left and right sides.

Looking at the graph, we can see that as x approaches 6 from the left side, f(x) approaches 2; whilst as x approaches 6 from the right side, f(x) approaches 5. We can write this as:

lim (x -> 6⁻) f(x) = 2

lim (x -> 6⁺) f(x) = 5

These limits are not the same:

lim (x -> 6⁻) ≠ lim (x -> 6⁺), therefor a limit does not exist at x = 6.

Thus, this statement is False.

Hope this helps, but if anything is unclear, please feel free to leave a comment below :)

Answer:

  The digit 6 in February is worth 1/100 of its value in January.

Step-by-step explanation:

The value of a given digit in a number can be found by setting the other digits to zero.

In February, the digit 6 has a value of 00600 = 600.

In January, the digit 6 has a value of 060000 = 60,000.

The ratio of the value in February to the value in January is ...

  600/60000 = 1/100

The digit 6 in February is one hundredth of the digit 6 in January.

Answer:

Step-by-step explanation:

We can solve this using a Venn Diagram,

First we write down the 91 tourists that didn't visit any of the theme parks.

Then we focus on the 38 tourists that visited all three theme parks and write that down in the intersection of the 3 circles.

80 tourists visited both LEGOLAND and Universal Studios but we already know that 38 visited the three of them, so 80-38= 42 tourists visited only those two parks.

97 tourists visited both Magic Kingdom and Universal Studios, but again, we know that 38 visited the three of them, so 97-38 = 59 tourists visited only those two parks.

94 tourists visited both Magic Kingdom and LEGOLAND, but since 38 visited the three of them, we subtract 94-38 = 56 tourists visited those two parks.

292 tourists visited LEGOLAND so we'll focus on that circle 292 - (56+38+42) = 156 tourists visited only LEGOLAND

280 tourists visited Magic Kingdom so we focus now on the Magic Kingdom circle and we have 290 - (56+38+59) = 127 only visited Magic Kingdom.

Now we will sum up all the quantities we have and subtract them from the total amount of tourists (1116) to find out how many people visit Universal Studios

1116 - (91+156+42+38+56+59+127) = 547

Answer: [tex]\dfrac{3}{4}[/tex]

Step-by-step explanation:

Given : Line k lies in the xy-plane.

The x-intercept of line K is -4. i.e. line k is intersecting the x-axis at (-4,0).

We know that the mid point of a line passing through any two points (a,b) and (c,d) is [tex](\dfrac{a+c}{2},\dfrac{b+d}{2})[/tex].

Then, the  midpoint of the line segment whose endpoints are (2, 9) and (2, 0) will be :-

[tex](\dfrac{2+2}{2},\dfrac{9+0}{2})=(2,4.5)[/tex]

The slope of line passing through (p,q) and (r,s) will be :-

[tex]m=\dfrac{s-q}{r-p}[/tex]

Then, the slope of line passing through (-4,0) and (2,4.5) will be :-

[tex]m=\dfrac{4.5-0}{2-(-4)}=\dfrac{4.5}{2+4}=\dfrac{4.5}{6}\\\\=\dfrac{45}{60}=\dfrac{3}{4}[/tex]

Hence, the slope of line k [tex]=\dfrac{3}{4}[/tex]

Given : An urn contains 2 red marbles and 3 blue marbles.

Total marbles = 2+3=5

a) The general ways in which the person could get a red marble and a blue marble are :

1)  He draws red marble first and then second marble as blue.

2)  He draws blue marble first marble and then second marble as red.

b)  The number of ways to get one red and one blue marble is given by :-

[tex]^2C_1\times^3C_1=2\times3=6[/tex]                          (i)

c) Number of ways to get 2 marbles from 5 is given by :-

[tex]^5C_2=\dfrac{5!}{2!(5-2)!}=\dfrac{5\times4\times3!}{3!\times2}=10[/tex]     (ii)

Now, The probability the person gets a red and a blue marble will be :-

[tex]P(R\ \&B)=\dfrac{6}{10}=0.6[/tex]       [Divide (i) by (ii)]

Hence, the  probability the person gets a red and a blue marble= 0.6