To review the solution to a similar problem, consult Interactive Solution 1.43. The magnitude of a force vector is 86.4 newtons (N). The x component of this vector is directed along the +x axis and has a magnitude of 72.3 N. The y component points along the +y axis. (a) Find the angle between and the +x axis. (b) Find the component of along the +y axis.

I'll assume the usual definition of set difference, [tex]X-A=\{x\in X,x\not\in A\}[/tex].

Let [tex]x\in X-(A\cup B)[/tex]. Then [tex]x\in X[/tex] and [tex]x\not\in(A\cup B)[/tex]. If [tex]x\not\in(A\cup B)[/tex], then [tex]x\not\in A[/tex] and [tex]x\not\in B[/tex]. This means [tex]x\in X,x\not\in A[/tex] and [tex]x\in X,x\not\in B[/tex], so it follows that [tex]x\in(X-A)\cap(X-B)[/tex]. Hence [tex]X-(A\cup B)\subset(X-A)\cap(X-B)[/tex].

Now let [tex]x\in(X-A)\cap(X-B)[/tex]. Then [tex]x\in X-A[/tex] and [tex]x\in X-B[/tex]. By definition of set difference, [tex]x\in X,x\not\in A[/tex] and [tex]x\in X,x\not\in B[/tex]. Since [tex]x\not A,x\not\in B[/tex], we have [tex]x\not\in(A\cup B)[/tex], and so [tex]x\in X-(A\cup B)[/tex]. Hence [tex](X-A)\cap(X-B)\subset X-(A\cup B)[/tex].

The two sets are subsets of one another, so they must be equal.

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices [tex]i\in I[/tex].

Proof of one direction for example:

Let [tex]x\in X-\left(\bigcup\limits_{i\in I}A_i\right)[/tex]. Then [tex]x\in X[/tex] and [tex]x\not\in\bigcup\limits_{i\in I}A_i[/tex], which in turn means [tex]x\not\in A_i[/tex] for all [tex]i\in I[/tex]. This means [tex]x\in X,x\not\in A_{i_1}[/tex], and [tex]x\in X,x\not\in A_{i_2}[/tex], and so on, where [tex]\{i_1,i_2,\ldots\}\subset I[/tex], for all [tex]i\in I[/tex]. This means [tex]x\in X-A_{i_1}[/tex], and [tex]x\in X-A_{i_2}[/tex], and so on, so [tex]x\in\bigcap\limits_{i\in I}(X-A_i)[/tex]. Hence [tex]X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)[/tex].

Rewriting the expression using m=2p we have:

Answer:

[tex]m^{2} +5 -1[/tex] is an odd integer but the converse is not true.

Step-by-step explanation:

Even numbers are written as 2n where n is any integer, while odd numbers are written as 2n-1 where n is any integer.

a) To prove that [tex]m^{2} +5m-1[/tex] is an odd integer, we have to prove that it can be written as 2n-1.

By hypothesis, m is an even integer so we will write it as 2p.

Rewriting the original expression using [tex]m=2p[/tex] we have:

[tex]m^{2} +5m-1 = (2p)^{2} +5(2p)-1[/tex]

Solving the expression and factorizing it we get

[tex]4p^{2} +10p -1 = 2(2p^{2}+5p) -1\\ \\[/tex]

And this last expression is an expression of the form 2n-1, and therefore [tex]m^{2} +5m-1[/tex] is an odd integer.

b) The converse would be: if [tex]m^{2} +5m-1[/tex] is an odd integer, then m is an even integer.

We'll give a counterexample, let's make [tex]m=3[/tex], then

[tex]m^{2} +5m-1[/tex]

[tex]3^{2}+5(3)-1 = 23[/tex] is an odd integer but m is odd.

Therefore, the converse is not true.

Answer:

The confidence interval is 6.6<μ<6.8.

Step-by-step explanation:

We have:

Number of observations = 601

Mean = 6.7

Standard deviation σ = 1.5

The z-score for a 95% confidence interval is 1.96.

The limits of the confidence interval can be calculated as

[tex]X \pm z*\frac{\sigma}{\sqrt{n}}\\\\LL=X-z*\frac{\sigma}{\sqrt{n}}=6.7-1.96*\frac{1.5}{\sqrt{601} } =6.7-0.1199=6.6\\\\UL=X+z*\frac{\sigma}{\sqrt{n}}=6.7+1.96*\frac{1.5}{\sqrt{601} } =6.7+0.1199=6.8[/tex]

The confidence interval is 6.6<μ<6.8.

Answer:

The cost of the old ball was $100.

Step-by-step explanation:

The cost of the new ball = $300

The new ball has three times the price of his old ball.

So, let the price of the old ball be = x

As per situation, we get the equation:

[tex]3x=300[/tex]

Dividing both sides by 3:

[tex]\frac{3x}{3}= \frac{300}{3}[/tex]

=> x = 100

Hence, the cost of the old ball was $100.

Answer:

102.8571 :)

Step-by-step explanation:

Answer:102.857 or rounded to 103

Step-by-step explanation:

You divide 720 by 7 which = 102.857

If it asks for a rounded number it would be 103

Answer:

[tex](\sqrt{3} + \frac{1}{\sqrt[4]{6}})^{15}[/tex]

Binomial expansion formula,

[tex](a+b)^n=\sum_{r=0}^{n} ^nC_r (a)^{n-r} (b)^r[/tex]

Where,

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

[tex]\implies (\sqrt{3} + \frac{1}{2})^{15}=\sum_{r=0}^{15} ^{15}C_r (\sqrt{3})^{15-r} (\frac{1}{\sqrt[4]{6}})^r[/tex]

[tex]=(\sqrt{3})^{15}+15(\sqrt{3})^{14}(\frac{1}{\sqrt[4]{6}})^1+105(\sqrt{3})^{13}(\frac{1}{\sqrt[4]{6}})^2+455(\sqrt{3})^{12}(\frac{1}{\sqrt[4]{6}})^3+1365(\sqrt{3})^{11}(\frac{1}{\sqrt[4]{6}})^4+3003(\sqrt{3})^{10}(\frac{1}{\sqrt[4]{6}})^5+5005(\sqrt{3})^{9}(\frac{1}{\sqrt[4]{6}})^6+6435(\sqrt{3})^{8}(\frac{1}{\sqrt[4]{6}})^7+6435(\sqrt{3})^{7}(\frac{1}{\sqrt[4]{6}})^8+5005(\sqrt{3})^{6}(\frac{1}{\sqrt[4]{6}})^9+3003(\sqrt{3})^{5}(\frac{1}{\sqrt[4]{6}})^{10}+1365(\sqrt{3})^{4}(\frac{1}{\sqrt[4]{6}})^{11}+455(\sqrt{3})^{3}(\frac{1}{\sqrt[4]{6}})^{12}+105(\sqrt{3})^{2}(\frac{1}{\sqrt[4]{6}})^{13}+15(\sqrt{3})^{1}(\frac{1}{\sqrt[4]{6}})^{14}+(\frac{1}{\sqrt[4]{6}})^{15}[/tex]

∵ both [tex]\sqrt{3}[/tex] and [tex]\frac{1}{\sqrt[4]{6}}[/tex] are irrational numbers,

And, if the power of √3 is even, it converted to a rational number,

If its power is odd it remained as irrational number,

But, the product of a rational number and irrational number is irrational,

Thus, all terms in the above expansion are irrational. ( which can not expressed in the form of p/q, where, p and q are integers s.t. q ≠ 0 )

Answer with explanation:

For any object having the velocity vector as

[tex]\overrightarrow{v}=v_x\widehat{i}+v_y\widehat{j}+v_z\widehat{k}[/tex]

the magnitude of velocity is given by

[tex]|v|=\sqrt{v_x^2+v_y^2+v_z^2}[/tex]

For car 1 the velocity vector is

[tex]\overrightarrow{v}_1=20\widehat{i}+25\widehat{j}[/tex]

Therefore

[tex]|v_1|=\sqrt{20^2+25^2}\\\\\therefore v_1=32.0156units[/tex]

Similarly for car 2 we have

[tex]\overrightarrow{v}_2=30\widehat{i}[/tex]

Therefore

[tex]|v_2|=\sqrt{30^2}\\\\\therefore v_2=30.0units[/tex]

Comparing both the values we find that car 1 has the greater speed.

The correct answer is that the car with velocity vector 30i is traveling faster, and its speed is 30 units.

To determine which car is traveling faster, we need to calculate the magnitude (or speed) of each velocity vector. The magnitude of a velocity vector in two dimensions is given by the square root of the sum of the squares of its components.

For the first car, the velocity vector is[tex]\( \mathbf{v}_1 = 20\mathbf{i} + 25\mathbf{j} \). The magnitude of this vector is calculated as follows:\[ ||\mathbf{v}_1|| = \sqrt{(20)^2 + (25)^2} = \sqrt{400 + 625} = \sqrt{1025} \][/tex]

For the second car, the velocity vector is [tex]\( \mathbf{v}_2 = 30\mathbf{i} + 0\mathbf{j} \).[/tex]

The magnitude of this vector is calculated as:[tex]\[ ||\mathbf{v}_2|| = \sqrt{(30)^2 + (0)^2} = \sqrt{900 + 0} = \sqrt{900} = 30 \]Now, comparing the magnitudes of the two vectors:\[ ||\mathbf{v}_1|| = \sqrt{1025} \approx 32.0156 \]\[ ||\mathbf{v}_2|| = 30 \]It is clear that \( ||\mathbf{v}_1|| \) is approximately 32.0156 units, while \( ||\mathbf{v}_2|| \)[/tex] is exactly 30 units. Since 32.0156 is greater than 30, the car with velocity vector[tex]\( \mathbf{v}_2 = 30\mathbf{i} \)[/tex]  is traveling faster.Therefore, the faster car is traveling at a speed of 30 units.

Answer: Each person will take [tex]\dfrac{1}{6}[/tex] of the pie.

Step-by-step explanation:

Given : One pie is shared equally by 6 people.

The total number of persons = 6

Now, the fraction of the whole pie each person will take :-

[tex]\dfrac{\text{Number of pie}}{\text{total persons}}\\\\=\dfrac{1}{6}[/tex]

Therefore, the fraction of the whole pie each person will take= [tex]\dfrac{1}{6}[/tex]

Answer:

nth term of this sequence is [tex](197+(n+6)\times 3^{27})[/tex]

and 100th term is [tex](197+106\times 3^{27})[/tex].

Step-by-step explanation:

The given sequence is [tex](197+7\times 3^{27}),(197+8\times 3^{27}),(197+9\times 3^{27})[/tex]

Now we will find the difference between each successive term.

Second term - First term = [tex](197+8\times 3^{27})-(197+7\times 3^{27})[/tex]

                                         = [tex](8\times 3^{27}-7\times 3^{27})[/tex]

                                         = [tex]3^{27}(8-7)[/tex]

                                         = [tex]3^{27}[/tex]

Similarly third term - second term = [tex]3^{27}[/tex]

So there is a common difference of [tex]3^{27}[/tex].

It is an arithmetic sequence for which the explicit formula will be

[tex]T_{n}[/tex]=a + (n - 1)d

where [tex]T_{n}[/tex] = nth term of the arithmetic sequence

where a = first term of the arithmetic sequence

n = number of term

d = common difference in each successive term

Now we plug in the values to get the 100th term of the sequence.

[tex]T_{n}=(197+7\times 3^{27})+(n-1)\times 3^{27}[/tex]

               = [tex](197+(n+6)\times 3^{27})[/tex]

[tex]T_{100}=(197+7\times 3^{27})+(100-1)\times 3^{27}[/tex]

                   = [tex]197+7\times 3^{27}+99\times 3^{27}[/tex]

                   = [tex]197+106\times 3^{27}[/tex]

Therefore, nth term of this sequence is [tex](197+(n+6)\times 3^{27})[/tex]

and 100th term is [tex](197+106\times 3^{27})[/tex].

Answer:

Step-by-step explanation:

Given that Z the set of integers is the universal set and

A is given in set builder form.

[tex]A = {x | x^2 ≤ 5}[/tex]

To convert this into roster form, we can find solutions for x

When [tex]x^2\leq 5\\|x|\leq \sqrt{5} =2.236[/tex]

i.e. all integers lying between -2.236 and 2.236

The only integers satisfying this conditions are

-2,-1,0,1,2

Hence A in roster form is

A=[tex]{-2,-1,0,1,2}[/tex]

Final answer:

The set A = {x | x^2 ≤ 5}, which includes all integers whose squares are less than or equal to 5, is expressed using the roster method as A = { -2, -1, 0, 1, 2 }.

Explanation:

The set A includes all integers x such that x squared is less than or equal to 5. To list the set using the roster method, we identify all integers which, when squared, give a result that does not exceed 5.

The integers satisfying x2 ≤ 5 are -2, -1, 0, 1, and 2 because:

Therefore, using the roster method, the set A is written as A = { -2, -1, 0, 1, 2 }.

Answer:

B. [tex]1.83\text{ m}^2[/tex]

Step-by-step explanation:

We are asked to find the body surface area of a person whose height is 150 cm and who weighs 80 kg.

[tex]\text{Body surface area}( m^2)=\sqrt{\frac{\text{Height (cm)}\times \text{Weight (kg)}}{3600}}[/tex]

Substitute the values:

[tex]\text{Body surface area}( m^2)=\sqrt{\frac{150\text{ cm}\times 80\text{(kg)}}{3600}}[/tex]

[tex]\text{Body surface area}( m^2)=\sqrt{\frac{12,000}{3600}}[/tex]

[tex]\text{Body surface area}( m^2)=\sqrt{3.3333333}[/tex]

[tex]\text{Body surface area}( m^2)=1.825741[/tex]

[tex]\text{Body surface area}( m^2)=1.83[/tex]

Therefore, the body surface area of the person would be 1.83 square meters.

Answer:

x = 4

y = 1

z= -3

Step-by-step explanation:

Given equations are

-2x + 3y - 4z = 7

5x - y + 2z = 13

3x + 2y - z = 17

We can write the above equations in matrix augmented form as

[tex]\left[\begin{array}{ccc}-2&3&-4:7\\5&-1&2:13\\3&2&-1:17\end{array}\right][/tex]

[tex]R_1=>\dfrac{R_1}{-2}[/tex]

[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\5&-1&2:13\\3&2&-1:17\end{array}\right][/tex]

[tex]R_2=>R_2-5R_1\ and\ R_3=>\ R_3-3R_1[/tex]

[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\0&-1+\dfrac{15}{2}&-8:13+\dfrac{35}{2}\\0&0&-7:17+\dfrac{21}{2}\end{array}\right][/tex]

[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\\\0&\dfrac{13}{2}&-8:\dfrac{61}{2}\\\\0&\dfrac{13}{2}&-7:\dfrac{55}{2}+\dfrac{21}{2}\end{array}\right][/tex]

[tex]R_2=>\ \dfrac{2}{13}R_2[/tex]

[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\\\0&1&\dfrac{-16}{13}:\dfrac{61}{13}\\\\0&\dfrac{13}{2}&-7:\dfrac{55}{2}\end{array}\right][/tex]

[tex]R_3=>R_3-\dfrac{13}{2}R_2[/tex]

[tex]=\ \left[\begin{array}{ccc}1&\dfrac{-3}{2}&2:\dfrac{-7}{2}\\\\0&1&\dfrac{-16}{13}:\dfrac{61}{13}\\\\0&0&1:-3\end{array}\right][/tex]

So, from the above augmented matrix, we can write

[tex]x+\dfrac{-3}{2}y+2z=\dfrac{-7}{2}.......(1)[/tex]

[tex]y+\dfrac{-16}{13}z=\dfrac{61}{13}......(2)[/tex]

z= -3.....(3)

From eq(2) and (3)

[tex]y+\dfrac{-16}{13}(-3)=\dfrac{61}{13}[/tex]

=> y = 1

Now, by putting the value of y and z in equation (1), we will get

[tex]x+\dfrac{-3}{2}(1)+2(-3)=\dfrac{-7}{2}[/tex]

=> x = 4

Hence, the value of

x = 4

y = 1

z= -3

Answer: There are 20 ways and 1680 ways respectively.

Step-by-step explanation:

Since we have given that

Total number of blocks = 6

Number of red blocks = 3

Number of green blocks = 3

So, Number of patterns she can make by placing them all in a line is given by

[tex]\dfrac{6!}{3!\times 3!}\\\\=20[/tex]

If there are 3 white blocks

so, total number of white blocks becomes 9

So, Number of total pattern she can make by placing all nine blocks in a line is given by

[tex]\dfrac{9!}{3!\times 3!\times 3!}\\\\=1680\ ways[/tex]

Hence, there are 20 ways and 1680 ways respectively.

The child can create 20 different patterns if she uses just the 6 blocks (3 red, 3 green), and she can create 14,040 different patterns if she uses all 9 blocks (3 red, 3 green, 3 white). This is calculated using a branch of mathematics known as combinatorics.

In this math problem, we are dealing with a concept known as permutations in combination, which is part of combinatorics branch of Mathematics. When placing the blocks in a line, the order in which you arrange them matters, which makes this a permutation problem.

For the first case where she has 6 blocks, 3 red and 3 green, the number of different patterns she can create is calculated by the equation 6! / (3! * 3!). Here, the '!' character means factorial, which is the product of all positive integers up to that number. So, 6! = 6 * 5 * 4 * 3 * 2 * 1 and similarly 3! = 3 * 2 * 1. Plugging these in, the equation becomes 720 / (6*6) = 20 patterns.

For the second case, if she is given 3 white blocks, she then has a total of 9 blocks (3 red, 3 green, 3 white). The number of different patterns she can create is calculated similarly, but this time the equation is 9! / (3! * 3! * 3!). Plugging in the factorials, we have 362,880 / (6 * 6 * 6) = 14,040 patterns.

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Answer:

[tex]\frac{{10 \choose 8}2^8}{{20 \choose 8}}\approx 0.091[/tex]

Step-by-step explanation:

We can think of the 10 pairs of gloves as simply being gloves of different colors. Picking no matching pair is the same as picking no 2 gloves of the same color. To compute the probability of doing so, we can compute the number of ways to select 8 gloves from different colors, and divide that by the total number of ways to select 8 random gloves out of the 20 gloves.

To compute the number of ways in which we can select 8 gloves from different colors, we can think of the choosing procedure as follows:

1st step- We choose from which 8 colors are we going to pick gloves from. So we have to pick 8 out of 10 colors. This can be done in [tex]{10 \choose 8}[/tex] ways.

2nd step - We now have to choose which glove are we going to pick from each of the chosen colors. Either the left one or the right one. For the first chosen color we have 2 choices, for the second chosen color we have 2 choices, for the third chosen color we have 2 choices, and so on. Therefore the number of ways in which we could choose gloves from the chosen colors is [tex]2^8[/tex]

And so the total number of ways in which we could choose 8 gloves from different colors is

[tex]{10 \choose 8 }2^8 [/tex]

Now, the total numer of ways in which we could choose 8 gloves out of the 20 gloves is simply [tex] {20 \choose 8}[/tex]

So the probability of picking no mathing pair is

[tex]\frac{{10 \choose 8}2^8}{{20 \choose 8}}\approx 0.091[/tex]

Answer:

C. 36 in.

Step-by-step explanation:

The wall is 66 inches high by 72 inches wide.

The tiles are

a) 9 inch square = 3 by 3

b) 16 inch square = 4 by 4

c) 36 inch square = 6 by 6

d) 64 inch square = 8 by 8

Factors of 66 = 2 x 3 x 11

Factors of 72 = 2 x 2 x 2 x 3 x 3

Now, we can see that in both 66 and 72 , we have 2 x 3 common that is 6.

And square of 6 is 36.

So, the answer is option C.

Answer: 10.6

Step-by-step explanation:

divide the number she spent (2.65) by the amount the store charges per ounce (.25)

2.65/ .25 = 10.6

Answer:

1) Prime Factorization

2) Technology

3) Existence of prime number in nature

Step-by-step explanation:

Prime numbers are the numbers whose divisors are 1 and the number itself, For example: 2, 3, 7, 11,...

Prime Numbers are a significant part of our life and are widely used in daily life.

1) Prime Factorization

This method help us to break a number into products of prime Number. This approach help us to find the LCM(Lowest Common Multiple) and GCD(Greatest Common Divisor)

2) Technology

Prime factorization forms the basis oh cryptography. Prime numbers play an important role in password protection and security purposes. They give the basis for many cryptographic algorithms.

3) Existence of prime number in nature

Many scientist have claimed that prime numbers exist in our life in unexpected form. For example, the number of petals in a flower, number of hexes in beehive, the pattern in pineapple are all related to prime number.

Answer:

Mid point will be 0.9887

Step-by-step explanation:

We have given a =0.9876 and b = 0.9887

And N = 2

We have to find midpoint

We know that formula for finding mid point that is

Midpoint [tex]=\frac{a+b}{2}[/tex]

So mid point will be

Midpoint [tex]=\frac{0.9876+0.9887}{2}=0.98815[/tex]

So the mid point between a = 0.9876 and b=0.9887 for N =2 will be 0.9887

Answer:

Price of the drill kit should be set as $4180.

Step-by-step explanation:

Daily supply of the most popular cordless drill kit is represented by the equation

S(q) = 100q² + 100q + 20

where S(q) = price of the kits at which q units are supplied

q = number of drill kits supplied

Now we have to calculate the price of the drill kits if company plans to supply 16 kits a day.

S(16) = 10(16)² + 100(16) + 20

       = 10×256 + 1600 + 20

       = 2560 + 1600 + 20

       = $4180

Therefore, cost of the drill set should be set as $4180.

To determine the price for 16 drill kits, substitute q = 16 into the supply equation S(q) = 10q^2 + 100q + 20, resulting in a price of $4180.

To find the price at which the home improvement company should set the drill kit if they plan to supply 16 a day, we need to plug the quantity (q) into the given supply equation S(q) = 10q2 + 100q + 20.

Substituting q = 16, we get:
S(16) = 10(16)2 + 100(16) + 20
= 10(256) + 1600 + 20
= 2560 + 1600 + 20
= 4180.

So, the company should set the price of the cordless drill kit at $4180 if they plan to supply 16 units a day.

Answer:

The correct option is 5) {Paul, Ringo, Pete, John, George, Stuart}.

Step-by-step explanation:

Consider the provided sets:

B = {John, Paul, George, Ringo, Pete, Stuart}

Two sets are equal if all the elements of Sets are same.

Set B has the elements: John, Paul, George, Ringo, Pete and Stuart

Now consider the provided options of sets.

From the provided options of set only option 4 has all the elements of set B but the order is different.

Thus, the correct option is 5) {Paul, Ringo, Pete, John, George, Stuart}.

Answer:

12 ; 12 dollars

Step-by-step explanation:

Data provided in the question:

Revenue function, R = 12x

R is in dollars

Now,

The slope can be found out by differentiating the above revenue function w.r.t 'x'

thus,

[tex]\frac{\textup{dR}}{\textup{dx}}[/tex]= [tex]\frac{\textup{d(12x)}}{\textup{dx}}[/tex]

or

slope = 12

Now, for the second case of selling one more unit i.e x = 1, the revenue can be obtained by substituting x = 1 in revenue function

therefore,

R = 12 × 1 = 12 dollars

Answer: $ 67.03

Step-by-step explanation:

Given : The average expenditure in a sample survey of 50 male consumers was $135.67, and the average expenditure in a sample survey of 38 female consumers was $68.64.

i.e. [tex]\overline{x}_1=\$135.67\ \ \&\ \ \overline{x}_2=\$68.64[/tex]

The best point estimate of the difference between the two population means is given by :-

[tex]\overline{x}_1-\overline{x}_2\\\\=135.67-68.64=67.03[/tex]

Hence, the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females : $ 67.03

The point estimate of the difference between the average expenditure of male and female consumers for Valentine's Day is $67.03.

The subject of your question is related to comparative statistical analysis between two groups, in this case, male and female consumers on Valentine's Day expenditures. Your question focuses on finding the point estimate for the difference between the population mean expenditure of males and females.

The point estimate is calculated by simply subtracting one mean from the other. According to your data, the average expenditure of the male consumers is $135.67 and of female consumers is $68.64. So, the calculation looks like this: $135.67 - $68.64 = $67.03. Therefore, the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females is $67.03.

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Answer:

The kinetic energy of car with mass 1500 kg and with speed of 35 miles/hour is KE=183598 J and when the car increases its speed to 70 miles/hour the kinetic energy changes by a factor of 4.

Step-by-step explanation:

The first step is to convert the speed miles/hour to m/s.

[tex]35\frac{miles}{hour} *\frac{1609.34 \>m}{1 \>miles}*\frac{1 \>hour}{3600 \> s}=15.646 \frac{m}{s}[/tex]

Next, the formula for the kinetic energy is

[tex]KE=\frac{1}{2} mv^{2}[/tex]

So input the values given:

[tex]KE=\frac{1}{2} (1500)(15.646)^{2}\\KE=750 \cdot (15.646)^{2}\\KE=183597.987 = 183598 \frac{kg \cdot \>m^{2}}{s^{2}} \\KE=183598 \>J[/tex]

Notice that the speed of 70 miles/hour is the double of 35 miles/hour so we can say that [tex]v_{2}=2v_{1}[/tex] and use the formula for the kinetic energy

[tex]KE_{2} =\frac{1}{2} m(v_{2}) ^{2}\\if \: v_{2}= 2v_{1}, then \:\\KE_{2} =\frac{1}{2} m(2v_{1}) ^{2}\\KE_{2} =\frac{1}{2} m4(v_{1})^{2}\\KE_{2} =4(\frac{1}{2} m(v_{1})^{2})\\We \:know \:that \:KE_{1} =\frac{1}{2} m(v_{1})^{2} so\\KE_{2} =4(KE_{1})[/tex]

We can see that when the car increases its speed to 70 miles/hour the kinetic energy changes by a factor of 4.

Answer: 6.416%

Step-by-step explanation:

The percentage error formula is given by :-

[tex]\%\text{error}=\dfrac{|\text{Estimate-Actual}|}{\text{Actual}}\times100[/tex]

Given : The estimated weight of a substance = 475 mg

The actual weight of the substance = 445 mg

Then,

[tex]\%\text{error}=\dfrac{| 475-445|}{445}\times100\\\\=\dfrac{30}{445}\times100=6.74157303371\approx6.416\%[/tex]

Hence, the percentage error in the first weighing. = 6.416%

Final answer:

The percentage error in the pharmacist's first weighing is approximately 6.74%, calculated by subtracting the accurate weight from the inaccurate weight, resulting in an absolute error of 30 mg, and then dividing the absolute error by the accurate weight, multiplying by 100.

Explanation:

To calculate the percentage error of the pharmacist's initial weighing, we first need to determine the absolute error by subtracting the accurate weight from the inaccurate weight. In this instance, the initial weight ( A ) recorded was 475 mg, and upon checking with a high accuracy balance, the true weight ( B ) was found to be 445 mg. Therefore, the absolute error ( Δ ) is the difference between these two measurements:  Δ =  A -  B = 475 mg - 445 mg = 30 mg.

After determining the absolute error, we can calculate the percentage error using the following formula:

Percentage Error = ( Δ /  B ) × 100%

Substituting in the respective values, we get:

Percentage Error = (30 mg / 445 mg) × 100% ≈ 6.74%

Answer:

a) There is a 61,41% of none of the samples containing high levels of contamination.

b)There is a 32.52% probability that exactly one sample contains high levels of contamination.

c) There is a 38.59% probability that at least one contains high levels of contamination

Step-by-step explanation:

The probabilities are independent from each other. It means that the probability of selecting a lab specimen being contaminated is always 15%, no matter how many contaminated lab specimen have been chosen.

a) There are 3 independent samples. For each sample, the probability of it not being contaminated is 85%. So, the probability that none of the sample are contaminated is

[tex]P = (0.85)^3 = 0.6141 = 61,41%[/tex]

There is a 61,41% of none of the samples containing high levels of contamination.

b) There are 3 independent samples. For each sample, the probability of it being contaminated is 15% and not contaminated 85%.

So the probability the exactly one sample contains high levels of contamination is:

[tex]P = (0.85)^2(0.15) = 0.1084 = 10,84%[/tex]

There can be 3 orderings of the sample in these conditions.(C-NC-NC, NC-C-NC, NC-NC,C), so the probability that exactly one contains high levels of contamination is

P = 3*0.1084 = 0.3252 = 32.52%.

There is a 32.52% probability that exactly one sample contains high levels of contamination.

c) The sum of the probabilities is always 100%.

In relation to the existence of a contaminated sample, either:

-None of the samples are contaminated.

-At least one of the samples are contaminated.

So, the probability of at least one of the samples being contaminated is 100% - the probability that none of the samples are contaminated, that we have already found in a).

So, it is

100% - 61.41% = 38.59%

There is a 38.59% probability that at least one contains high levels of contamination

[tex]\mbox{First, we compute the derivative of $y$ at $x_0=1$. So, we get}\\$$ y' = 3x^2 - 6x \, , \, y'(1) = -3 $$[/tex].

Therefore, the linear approximation at the point (1,-3) is

[tex]$$ y = -3 - 3(x -1) \ . $$[/tex]

To find the linear approximation of the non-linear system at the point (1, -3), first find the derivative of the function to get the slope of the tangent line at that point. Then, plug the slope and the point into the linearization formula. For the plotting part in Matlab, it should be a separate discussion as this platform does not support programming languages.

The subject of this question is a non-linear system given by the equation y=x^3 - 3x^2 - 1. The student is asked to find the linear approximation at the point (1, -3). The linear approximation of a function at a given point is the tangent line to the function at the given point, and it's also the best linear approximation of the function near that point.

Before we begin, let's define some terms. Linear approximation is a process of approximating the values of a nonlinear function using a line near a point. To find the linear approximation, we use the formula for the linearization of a function, L(x) = f(a) + f'(a)(x - a), where 'a' is the x-value of the point of tangency, f(a) is the y-value, and f'(a) is the slope of the tangent line at point 'a'. Tangent line is a straight line that just touches a curve at a given point. The tangent line is the best linear approximation to the curve at that point.

First, we need to find the derivative of the function, f'(x), which is 3x^2 - 6x. Then, evaluate f'(1) to find the slope of the tangent line. Plug these values into the linearization formula to get L(x) =  -3 + (3 - 6)(x - 1). Now, you can plot the original function and the linearization on the same graph.

Please note, for the Matlab portion of the question, it should be a separate discussion as this website is designed to walk through problems in a step-by-step manner and doesn't support running such programming languages directly. However, there are many online resources that can provide specific Matlab example codes for plotting functions and their linear approximations.

https://brainly.com/question/1621850

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Answer:

Step-by-step explanation:

Given that the first difference of a sequence is the arithmetic sequence 1​, 3​, 5​, 7​, 9​, ....

a) When I term a =1

[tex]a_2 =1+1 =2\\a_3 = 4+5 =9\\a_4 = 9+7 =16\\a_5 =16+9 =25\\a_6=25+11 =36[/tex]

Thus first 6 terms are

1,2,5,12,21,32.....

b) Here [tex]a_1+a_2=5\\a_2-a_1 =3\\-------------\\2a_2=8\\a_2 =4\\a_1 =1[/tex]

[tex]a_2 =1+3 =4\\a_3 = 4+5 =9\\a_4 = 9+7 =16\\a_5 =16+9 =25\\a_6=25+11 =36[/tex]

So sequence would be

3,4,9,16,25, 36,...

c) When 5th term is 28

we have the sequences as

a1, a1+1,a1+1+3, ...a1+1+3+5+7

When 5th term is 28 we have

[tex]a_1 +16 =28\\a_1 =12\\[/tex]

Hence first 6 terms would be

12, 13, 16, 21, 28, 37,...

Answer:

a) Degree of E = 2

b) Even vertices: B, C, E

Odd vertices : A, D

c) Vertices A, C, and E are adjacent to D

Step-by-step explanation:

a) The degree of a vertex is given by the number of segments that end there, so in the case of vertex E, there are only two segments that connect it, therefore its degree is 2

b) Following the same idea of degree of a vertex, we can find the number of segments that end on each one of the 5 vertices shown and assign to them their degree:

A (3), B (2), C (4), D (3), E (2)

Therefore the odd vertices are: A and D (both of degree 3)

The even vertices are: B, E (both of degree 2, and C (degree 4)

c) the vertices adjacent to vertex D are those connected directly to it via a segment: that is, vertices A, C, and E

Answer:

The system has solution when:

[tex]a\neq 16[/tex]

The system has no solution when:

[tex]a=16[/tex]

Step-by-step explanation:

First rewrite the system in its augmented matrix form

[tex]\left[\begin{array}{cccc}1&2&-3&4\\3&-1&5&2\\4&1&a-14&a+2\end{array}\right][/tex]

Let´s apply row reduction process to  its associated augmented matrix:

[tex]F2-3F1\\F3-4F1[/tex]

[tex]\left[\begin{array}{cccc}1&2&-3&4\\0&-7&14&-10\\0&-7&a-2&a-14\end{array}\right][/tex]

[tex]F3-F2[/tex]

[tex]\left[\begin{array}{cccc}1&2&-3&4\\0&-7&14&-10\\0&0&a-16&a-4\end{array}\right][/tex]

Now we have this:

[tex]x+2y-3z=4\\0-7y+14=-10\\0+0+(a-16)z=a-4[/tex]

We can conclude now:

The system has no solution when:

[tex]a=16[/tex]

And the system has solution when:

[tex]a\neq 16[/tex]

Answer:

(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)

Step-by-step explanation:

We are given the set (a,b,c,d,e).

Total number of subsets of the above set are [tex]2^5[/tex] = 32

Subsets:

φ

(a,b,c,d,e)

(a), (b), (c), (d), (e)

(a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e)

(a,b,c), (a,b,d), (a,b,e), (a,c,d), (a,c,e), (a,d,e), ( b,c,d), (b,c,e), (b,d,e), (c,d,e)

(a,b,c,d), (a,b,c,e), (a,b,d,e), (a,c,d,e), (b,c,d,e)

Subset having a but not b :

(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)