In the year 1985, a house was valued at $113,000. By the year 2005, the value had appreciated exponentially to $155,000. What was the annual growth rate between 1985 and 2005 (Rourid your answer to two decimal places.) Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010? (Round your answer to the nearest dolar.)

Answer:

$1850.

Step-by-step explanation:

We are asked to find amount of interest earned in one year on an amount of $5000.

We will use simple interest formula to solve our given problem.

[tex]I=Prt[/tex], where,

I = Amount of interest earned,

r = Annual interest rate in decimal form.

t = Time in years.

Let us convert our given interest rate in decimal form.

[tex]37\%=\frac{37}{100}=0.37[/tex]

Upon substituting our given value in simple interest formula, we will get:

[tex]I=\$5000\times 0.37\times 1[/tex]

[tex]I=\$5000\times 0.37[/tex]

[tex]I=\$1850[/tex]

Therefore, an amount of $1850 is earned as interest in one year.

Answer:

<CBA

Step-by-step explanation:

The angle name could be either

<ABC or <CBA

The vertex must be in the middle

Answer:

Yes, the answer is CBA

Step-by-step explanation:

The answer for the sets corresponding to the given conditions is as follows:

a) A ∩ B = {4, 6, 8, 10, 12}.

b) A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

c)  B ∩ C = {}.

d) B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8...}.

e) Set A = {3, 5, 7}   and  Set B = {2,4, 6, 8, 10, 12}

Given:

Set A =  {x ∈ N : 3 ≤ x ≤ 13}

Set B =  {x ∈ N : x is even}

Set C  = {x ∈ N : x is odd}.

Solve each option:

(a) Find A ∩ B (the intersection of sets A and B):

Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

The intersection of A and B includes even numbers that are between 3 and 13: A ∩ B = {4, 6, 8, 10, 12}.

(b) Find A ∪ B (the union of sets A and B):

Set A contains natural numbers from 3 to 13 (inclusive): A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

The union of A and B includes all numbers from both sets: A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(c) Find B ∩ C (the intersection of sets B and C):

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.

The intersection of B and C is the empty set, as there are no numbers that are both even and odd.

(d) Find B ∪ C (the union of sets B and C):

Set B contains even natural numbers: B = {2, 4, 6, 8, 10, 12, ...}.

Set C contains odd natural numbers: C = {1, 3, 5, 7, 9, 11, ...}.

The union of B and C includes all natural numbers: B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...}.

(e) For the given example:

Set A = {3, 5, 7}

Set B = {2,4, 6, 8, 10, 12}

This example satisfies the conditions A ∩ B = {3, 5} and A ∪ B = {2, 3, 5, 7, 8}

The intersection and Unioun of all the sets is found from A and B.

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The intersection of sets A and B is {4, 6, 8, 10, 12}. The union of sets A and B is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. The intersection of sets B and C is {}. The union of sets B and C is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(a) To find the intersection of sets A and B, we need to identify the elements that are common to both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. The elements that are common to both sets are: {4, 6, 8, 10, 12}. Therefore, A ∩ B = {4, 6, 8, 10, 12}.

(b) To find the union of sets A and B, we need to combine all the elements from both sets. In set A, we have: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. In set B, we have: {4, 6, 8, 10, 12}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(c) To find the intersection of sets B and C, we need to identify the elements that are common to both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. The elements that are common to both sets are: {}. Therefore, B ∩ C = {}.

(d) To find the union of sets B and C, we need to combine all the elements from both sets. In set B, we have: {4, 6, 8, 10, 12}. In set C, we have: {3, 5, 7, 9, 11, 13}. Combining these sets gives us: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. Therefore, B ∪ C = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

Answer:

1577.20 ft³

Step-by-step explanation:

Cube of length = 12 ft = a

Hole diameter which is cutout = 4 ft = d

Hole radius which is cutout = 4/2 =2 ft = r

Volume of the cube = a³

⇒Volume of the cube = a×a×a

⇒Volume of the cube = 12×12×12

⇒Volume of the cube = 1728 ft³

The hole cut out will be in the shape of a cylinder

Volume of cylinder = πr²h

⇒Volume of cylinder = π×2²×12

⇒Volume of cylinder = 150.79 ft³

Now volume of the solid figure with hole cut out is

Volume of the cube - Volume of cylinder

=1728 - 150.79

=1577.20 ft³

∴ Volume of solid figure not including hole cutout is 1577.20 ft³

Answer:  The required solution of the given IVP is

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

[tex]y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.[/tex]

Let [tex]y=e^{mx}[/tex] be an auxiliary solution of the given differential equation.

Then, we have

[tex]y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.[/tex]

Substituting these values in the given differential equation, we have

[tex]m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.[/tex]

So, the general solution of the given equation is

[tex]y(x)=Ae^x+Be^{-x},[/tex] where A and B are constants.

This gives, after differentiating with respect to x that

[tex]y^\prime(x)=Ae^x-Be^{-x}.[/tex]

The given conditions implies that

[tex]y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

and

[tex]y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Adding equations (i) and (ii), we get

[tex]2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.[/tex]

From equation (i), we get

[tex]\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.[/tex]

Substituting the values of A and B in the general solution, we get

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Thus, the required solution of the given IVP is

[tex]y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.[/tex]

Answer:

The required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].

Step-by-step explanation:

The given differential equation is

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

Substitute u=x-y+1 in the above equation.

[tex]\frac{du}{dx}=1-\frac{dy}{dx}[/tex]

[tex]\frac{dy}{dx}=1-\frac{du}{dx}[/tex]

[tex]1-\frac{du}{dx}=u^2[/tex]

[tex]1-u^2=\frac{du}{dx}[/tex]

Using variable separable method, we get

[tex]dx=\frac{du}{1-u^2}[/tex]

Integrate both the sides.

[tex]\int dx=\int \frac{du}{1-u^2}[/tex]

[tex]x+C=\frac{1}{2}\ln|\frac{1+u}{1-u}|[/tex]      [tex][\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\n|\frac{a+x}{a-x}|+C][/tex]

Substitute u=x-y+1 in the above equation.

[tex]x+C=\frac{1}{2}\ln|\frac{1+x-y+1}{1-(x-y+1)}|[/tex]

[tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex]

Therefore the required answer is [tex]x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|[/tex].

Answer:

The length of a side of the square is 22 inches.

Step-by-step explanation:

Let each side of square be = s

Let each side of triangle be = s - 8

Perimeter of square, p₁ = 4s

Perimeter of triangle = p₂ = 3s

                                          = 3(s-8)

                                          = 3s - 24

Therefore, according to the question

p₁ - p₂ = 46

4s - (3s - 24) = 46

4s - 3s + 24 = 46

s = 46 - 24

s = 22

The length of a side of the square is 22 inches.

Answer:

We have a normal distribution with a mean of 19 minutes and a standard deviation of 3 minutes. To solve the problem we're going to need the help of a calculator:

P(z>20) = 0.3694

Therefore, the percentage of costumbers that will receive the service for half-price is: 36.94%.

Also, we've found that p(z>25.16) = 0.02. Therefore, if they only want to offer half-price discount to only 2% of its costumber, the time limit should be 25.16 minutes.

Let [tex]\vec h[/tex] and [tex]\vec\eta[/tex] be two vectors in [tex]H[/tex].

[tex]H[/tex] is a subspace of [tex]V[/tex] if (1) [tex]\vec h+\vec\eta\in H[/tex] and (2) for any scalar [tex]k[/tex], we have [tex]k\vec h\in H[/tex].

(1) True;

[tex]\mathrm{tr}(\vec h+\vec\eta)=\mathrm{tr}(\vec h)+\mathrm{tr}(\vec eta)=0[/tex]

so [tex]\vec h+\vec\eta\in H[/tex].

(2) Also true, since

[tex]\mathrm{tr}(k\vec h)=0k=k[/tex]

Therefore [tex]H[/tex] is a subspace of [tex]V[/tex].

Answer: Yes, H is a subspace of V

Step-by-step explanation:

We know that V is the space of all the 2x2 matrices with real entries.

H is the set of all 2x2 matrices with real entries that have trace equal to 0.

Obviusly the matrices that are in the space H also belong in the space V (because in H you have some selected matrices and in V you have all of them). The thing we need to prove is if H is an actual subspace.

Suppose we have two matrices that belong to H, A and B.

We must see that:

1) if A and B ∈ H, then (A + B)∈H

2) for a scalar number k, k*A ∈ H

lets write this as:

[tex]A = \left[\begin{array}{ccc}a1&a2\\a3&a4\\\end{array}\right] B = \left[\begin{array}{ccc}b1&b2\\b3&b4\\\end{array}\right][/tex]

where a1 + a4 = 0 = b1 + b4

then:

[tex]A + B = \left[\begin{array}{ccc}a1 + b1&a2 + b2\\a3 + b3&a4 + b4\\\end{array}\right][/tex]

the trace is:

a1 + b1 - (a4 + b4) = (a1 - a4) + (b1 - b4) = 0

then the trace is nule, and (A + B) ∈ H

and:

[tex]kA = \left[\begin{array}{ccc}k*a1&k*a2\\k*a3&k*a4\end{array}\right][/tex]

the trace is:

k*a1 - k*a4 = k(a1 - a4) = 0

so kA ∈ H

then H is a subspace of V

Let [tex]A(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex]. We're told that [tex]A(0)=0[/tex].

For [tex]0\le t\le15[/tex], the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for [tex]t>15[/tex]. We can capture this in terms of the unit step function [tex]u(t)[/tex] and Dirac delta function [tex]\delta(t)[/tex] as

[tex]\text{rate in}=u(t)-u(t-15)+20\delta(t-15)[/tex]

(in lb/min)

The salt from the mixed solution flows out at a rate of

[tex]\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}[/tex]

Then the amount of salt in the tank at time [tex]t[/tex] changes according to

[tex]\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}[/tex]

Let [tex]\hat A(s)[/tex] denote the Laplace transform of [tex]A(t)[/tex], [tex]\hat A(s)=\mathcal L_s\{A(t)\}[/tex]. Take the transform of both sides to get

[tex]s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)[/tex]

Solve for [tex]\hat A(s)[/tex], then take the inverse of both sides.

[tex]\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}[/tex]

[tex]\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}[/tex]

Answer with Step-by-step explanation:

We are given S be any set which is countable and nonempty.

We have to prove that their exist a surjection g:N[tex]\rightarrow S[/tex]

Surjection: It is also called onto function .When cardinality of domain set is greater than or equal to cardinality of range set then the function is onto

Cardinality of natural numbers set =[tex]\chi_0[/tex]( Aleph naught)

There are two cases

1.S is finite nonempty set

2.S is countably infinite set

1.When S is finite set and nonempty set

Then cardinality of set S is any constant number which is less than the cardinality of set of natura number

Therefore, their exist a surjection from N to S.

2.When S is countably infinite set and cardinality with aleph naught

Then cardinality of set S is equal to cardinality of set of natural .Therefore, their exist a surjection from N to S.

Hence, proved

Explanation:

"Any normal distribution" may have arbitrary mean and standard deviation. The "standard normal distribution" has a mean of zero and a standard deviation of 1.

I'm going to guess that you meant to include parentheses somewhere, so that the ODE is supposed to be

[tex]y'=\dfrac{y^2x}{y^3+x^3}+\dfrac yx[/tex]

Then substitute [tex]y(x)=xv(x)[/tex] so that [tex]y'(x)=xv'(x)+v(x)[/tex]. Then

[tex]xv'+v=\dfrac{x^3v^2}{x^3v^3+x^3}+v[/tex]

[tex]xv'=\dfrac{v^2}{v^3+1}[/tex]

which is separable as

[tex]\dfrac{v^3+1}{v^2}\,\mathrm dv=\dfrac{\mathrm dx}x[/tex]

Integrate both sides: on the left,

[tex]\displaystyle\int\frac{v^3+1}{v^2}\,\mathrm dv=\int\left(v+\frac1{v^2}\right)\,\mathrm dv=\dfrac12v^2-\dfrac1v[/tex]

The other side is trivial. We end up with

[tex]\dfrac12v^2-\dfrac1v=\ln|x|+C[/tex]

Solve in terms of [tex]y(x)[/tex]:

[tex]\boxed{\dfrac{y^2}{2x^2}-\dfrac xy=\ln|x|+C}[/tex]

[tex]f(t)=\begin{cases}9&\text{for }0\le t<2\\(t-5)^2&\text{for }2\le t<5\\2te^{6t}&\text{for }t\ge5\end{cases}[/tex]

and presumably 0 for [tex]t<0[/tex]. We can express [tex]f(t)[/tex] in terms of the unit step function,

[tex]u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}[/tex]

[tex]f(t)=9(u(t)-u(t-2))+(t-5)^2(u(t-2)-u(t-5))+2te^{6t}u(t-5)[/tex]

Quick explanation: [tex]9u(t)=9[/tex] for [tex]t\ge0[/tex], and [tex]9u(t-2)=9[/tex] for [tex]t\ge2[/tex]. So subtracting these will cancel the value of 9 for all [tex]t\ge2[/tex] and leave us with the value of 9 over the interval we want, [tex]0\le t<2[/tex]. The same reasoning applies for the other 3 terms.

Recall the time displacement theorem:

[tex]\mathcal L_s\{f(t-c)u(t-c)\}=e^{-sc}\mathcal L_s\{f(t)\}[/tex]

By this property, we have

[tex]\mathcal L_s\{9u(t)\}=\mathcal L_s\{9\}=\dfrac9s[/tex]

[tex]\mathcal L_s\{9u(t-2)\}=e^{-2s}\mathcal L_s\{9\}=\dfrac{9e^{-2s}}s[/tex]

[tex]\mathcal L_s\{(t-5)^2u(t-2)\}=\mathcal L_s\{((t-2)-3)^2u(t-2)\}[/tex]

[tex]=e^{-2s}\mathcal L_s\{(t-3)^2\}=\left(\dfrac2{s^3}-\dfrac6{s^2}+\dfrac9s\right)e^{-2s}[/tex]

[tex]\mathcal L_s\{(t-5)^2u(t-5)\}=e^{-5s}\mathcal L_s\{t^2\}=\dfrac{2e^{-5s}}{s^3}[/tex]

[tex]\mathcal L_s\{2te^{6t}u(t-5)\}=\mathcal L_s\{2e^{30}(t-5)e^{6(t-5)}+10e^{30}e^{6(t-5)}\}[/tex]

[tex]=2e^{30-5s}\mathcal L_s\{te^{6t}+5e^{6t}\}=2e^{30-5s}\left(\dfrac1{(s-6)^2}+\dfrac5{s-6}\right)[/tex]

Putting everything together, we end up with

[tex]\boxed{\mathcal L_s\{f(t)\}=\dfrac{(2-6s)e^{-2s}-2e^{-5s}}{s^3}+\dfrac9s-\dfrac{2e^{30-5s}(29-5s)}{(s-6)^2}}[/tex]

Answer:

(8x-1) (x-1)

Step-by-step explanation:

8x² - 9x + 1

(8x -   ) (x - )

We know it is minus because we have -9x

We have +1 so both have to be -

To fill in the blanks we put 1

The only combination is 1*1 =1

(8x-1) (x-1)

Lets check

8x^2 -x -8x +1

(8x^2 -9x+1

Answer: 0.0136

Step-by-step explanation:

Given : Mean : [tex]\mu=25[/tex]

Standard deviation : [tex]\sigma=3.2[/tex]

Sample size : [tex]n=50[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x = 24

[tex]z=\dfrac{24-25}{\dfrac{3.2}{\sqrt{50}}}=-2.21[/tex]

The p-value = [tex]P(z\leq-2.21)= 0.0135526\approx0.0136[/tex]

Hence, the probability that the sample mean age for 50 randomly selected women to marry is at most 24 years = 0.0136

Answer:  Each muffin tin contains 3 oz of batter.

Step-by-step explanation:  Given that a baker pours  108 oz of batter into 36 muffin tins such that each tin has same amount of batter.

We are to calculate the quantity of batter in each tin.

We will be using the UNITARY method to solve the given problem.

Quantity of batter in 36 muffin tins = 108 oz.

Therefore, the quantity of batter in 1 muffin tin is given by

[tex]Q_t=\dfrac{108}{36}=3~\textup{oz}.[/tex]

Thus, each muffin tin contains 3 oz of batter.

By dividing 108 oz of batter by 36 muffin tins, you find that each tin contains 3 oz of batter. This simple division problem helps distribute the batter evenly. Each tin thus gets exactly 3 oz.

To find out how much batter is in each muffin tin, you need to divide the total amount of batter by the number of muffin tins.

Here are the steps:

So, there are 3 oz of batter in each muffin tin.

Answer:

d

Step-by-step explanation:

13 cant be divided equally nor cubed  because its not an even number u can try to give all thirteen of then

The simpler version of the initial problem is arranging three people in a line. There are three choices for the first spot, two for the second, and one for the third, which results in a total of six possible arrangements. This involves the principle of permutation in combinatorics.

The subject of the given problem can be defined as permutations. If we're looking for a simpler version of it, we should choose a problem which still involves line-up or arrangement of a smaller number of people. Hence, the best option is: 'In how many ways can three people be lined up for a picture?'

To solve this simpler problem, we consider the number of available spots for each person in the line. For the first spot, there are 3 people that could be selected. After the first person is chosen, there are only 2 people left for the second spot. Lastly, there is only 1 person left for the third spot. So, the total number of ways we can line up 3 people for a picture is 3*2*1 = 6 ways.

This is a basic principle called permutation in combinatorics which is a fundamental concept in mathematics that deals with counting, both as a means and an end in obtaining results.

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Answer:

The given expression is divisible by 3 for all natural values of x.

Step-by-step explanation:

The given expression is

[tex]2^{2x+1}+1[/tex]

For x=1,

[tex]2^{2(1)+1}+1=2^{3}+18+1=9[/tex]

9 is divisible by 3. So, the given statement is true for x=1.

Assumed that the given statement is true for n=k.

[tex]2^{2k+1}+1[/tex]

This expression is divisible by 3. So,

[tex]2^{2k+1}+1=3n[/tex]              .... (1)

For x=k+1

[tex]2^{2(k+1)+1}+1[/tex]

[tex]2^{2k+2+1}+1[/tex]

[tex]2^{(2k+1)+2}+1[/tex]

[tex]2^{2k+1}2^2+1[/tex]

Using equation (1), we get

[tex](3n-1)2^2+1[/tex]

[tex](3n)2^2-2^2+1[/tex]

[tex](3n)2^2-4+1[/tex]

[tex](3n)4-3[/tex]

[tex]3(4n-1)[/tex]

This expression is also divisible by 3.

Therefore the given expression is divisible by 3 for all natural values of x.

The probability that the average age of the 100 residents selected is less than 68.5 years is approximately 0.1949 or 19.49%.

The subject of this problem refers to statistics, specifically the concept of the sampling distribution of sample means. It is related to the central limit theorem, which states that if you take sufficiently large random samples from a population, the distribution of the sample means will approximate a normal distribution, regardless of the shape of the population's distribution. We know that the population average (mean) is 69 and the standard deviation is 5.8.

We are given a sample size of 100, and hence we calculate its standard deviation as 5.8/√100 = 0.58. With this value, we can use the z-score formula (Z = (X - μ) / σ), where X is the sample mean, μ is the population mean, and σ is the standard deviation of the sample mean, to find the z-score for a sample mean of 68.5 years: Z = (68.5 - 69) / 0.58 ≈ -0.86.

Finally, this Z score is used to find the probability that the sample mean age is less than 68.5 years, by referring to a standard normal distribution table, also known as the Z-table. It should be taken into account that this table provides the probability that a value is less than the given Z score, which is exactly what we need in this case. Consulting the Z table with Z=-0.86, we find that the probability is approximately 0.1949 or 19.49%.

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Answer:

190578024 ways.

Step-by-step explanation:

We are asked to find the number of ways in which a committee of 5 be chosen from 120 employees to interview prospective applicants.

We will use combinations to solve our given problem.

[tex]_{r}^{n}\textrm{C}=\frac{n!}{(n-r)!r!}[/tex], where,

n = Total number of items,

r = Number of items being chosen at a time.

Upon substituting our given values in above formula, we will get:

[tex]_{5}^{120}\textrm{C}=\frac{120!}{(120-5)!5!}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120!}{115!*5!}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116*115!}{115!*5*4*3*2*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{5*4*3*2*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{120*119*118*117*116}{120*1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{119*118*117*116}{1}[/tex]

[tex]_{5}^{120}\textrm{C}=\frac{190578024}{1}[/tex]

Therefore, the committee of five can be chosen from 120 employees in 190578024 ways.

Final answer:

2 to the power of three halves is equivalent to the square root of 2 cubed, which is approximately 2.83.

Explanation:

In mathematics, when we raise a number to a fraction exponent, we are essentially taking the root of that number. In this case, 2 to the power of three halves is equivalent to the square root of 2 cubed.

2 to the power of three halves = [tex]\sqrt{(2^3)}[/tex] = [tex]\sqrt{8}[/tex] = 2.83

Answer:

  about 34.9%

Step-by-step explanation:

The probability of not making a marking error is 0.9. The probability of doing that 10 times independently is 0.9^10 ≈ 0.34868 ≈ 34.9%.

Non zero vector x perpendicular to u and v : x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]

Given, v= [-2,-8,-7,2] u= [6,7,-2,8]

Let the vector be x = [[tex]x_1 , x_2 , x_3, x_4[/tex]]

Now x is non xero vector perpendicular to vector 'v' and 'u' .

So,

x . v = 0

[tex]-2x_1 - 8x_2 - 7x_3 + 2x_4 = 0[/tex] .........1

x . u = 0

[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex] .........2

Solve 1 and 2 to eliminate [tex]x_4[/tex] .

Multiply 1 with 4 to make the coefficients of [tex]x_4[/tex] same .

[tex]-8x_1 - 32x_2 - 28x_3 + 8x_4 = 0[/tex]

[tex]6x_1 + 7x_2 -2x_3 + 8x_4 = 0[/tex]

Subtract two equations,

[tex]-14x_1 -39x_2 -26x_3 = 0[/tex]

[tex]-14x_1 = 39x_2 + 26x_3[/tex]

[tex]x_1 = \frac{-39}{14} x_2 - \frac{26}{14} x_3[/tex]

From equation 1,

[tex]x_4 = x_1 + 4x_2 + \frac{7}{2} x_3[/tex]

[tex]x_4 = \frac{-39}{14} x_2 - \frac{26}{14} x_3+ 4x_2 + \frac{7}{2} x_3\\\\x_4 = \frac{17}{14}x_2 + \frac{23}{14} x_3[/tex]

Thus x = [tex][ \frac{-39}{14} x_2 - \frac{26}{14} x_3 , x_2 , x_3, \frac{17}{14}x_2 + \frac{23}{14} x_3 ][/tex]

[tex]x_2 = [-39/14 , 1 , 0 , 17/14] + x_3[-26/14, 0 , 1 , 23/14 ][/tex]

[tex]x_1 , x_3[/tex] are arbitrary .

For every value of [tex]x_2 , x_3[/tex] vector x is obtained.

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To find a vector x that is perpendicular to vectors v and u, we can use the cross product.

To find a vector x that is perpendicular to vectors v and u, we can use the cross product. The cross product of two vectors is a vector that is perpendicular to both of them. To find the cross-product, we can use the formula:

x = (v2u3 - v3u2, v3u1 - v1u3, v1u2 - v2u1)

Plugging in the values, we get:

x = (-8(-2) - (-7)(7), (-7)(6) - (-2)(-2), (-2)(8) - (-8)(6)) = (1, 52, -32)

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Answer:

  105/252 = 0.41666...

Step-by-step explanation:

There are (7C4)(3C1) = (35)(3) = 105 ways to choose exactly 4 boys. There are 10C5 = 252 ways to choose 5 youths, so the probability that a randomly chosen team will consist of exactly 4 boys is ...

  105/252

_____

nCk = n!/(k!(n-k!))

Answer:

There is a 41.67% probability that exactly 4 boys are picked in this team of 5.

Step-by-step explanation:

The order is not important, so we use the combinations formula.

[tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

Number of desired outcomes.

Four boys and one girl: So

[tex]C_{7,4}*C_{3,1} = \frac{7!}{4!(7-4)!}*\frac{3!}{1!(3-1)!} = 35*3 = 105[/tex]

Number of total outcomes:

Combination of five from a set of 10.

So

[tex]C_{10,5} = \frac{10!}{5!(10-5)!} = 252[/tex]

What is the probability that exactly 4 boys are picked in this team of 5?

[tex]P = \frac{105}{252} = 0.4167[/tex]

There is a 41.67% probability that exactly 4 boys are picked in this team of 5.

Answer:

Step-by-step explanation:

Given : The readings on thermometers are normally distributed with

Mean : [tex]\mu=\ 0[/tex]

Standard deviation : [tex]\sigma= 1[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = -1.52

[tex]z=\dfrac{-1.52-0}{1}=-1.52[/tex]

For x = -0.81

[tex]z=\dfrac{-0.81-0}{1}=-0.81[/tex]

The p-value = [tex]P(-1.52<z<-0.81)=P(z<-0.81)-P(z<-1.52)[/tex]

[tex]0.2089701-0.0642555=0.1447146\approx0.1447[/tex]

Hence, the probability that a randomly selected thermometer reads between negative 1.52 and negative 0.81 = 0.1447

To find the probability, standardize the values using z-scores and find the area under the normal curve between the z-scores.

To find the probability that a randomly selected thermometer reads between -1.52 and -0.81, we need to find the area under the normal curve between these two values. First, we need to standardize the values by finding the z-scores for these values using the formula z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation. After finding the z-scores, we can then use the normal distribution table or a calculator to find the area between these z-scores.

The z-score for -1.52 is z = (-1.52 - 0) / 1.00 = -1.52 and the z-score for -0.81 is z = (-0.81 - 0) / 1.00 = -0.81. Using a normal distribution table or a calculator, we can find the area to the left of -1.52 and the area to the left of -0.81. The probability that a randomly selected thermometer reads between -1.52 and -0.81 is the difference between these two areas: P(-1.52 < X < -0.81) = P(X < -0.81) - P(X < -1.52).

Using the normal distribution table or a calculator, we can find that P(X < -0.81) is approximately 0.2123 and P(X < -1.52) is approximately 0.0655. Therefore, the probability that a randomly selected thermometer reads between -1.52 and -0.81 is approximately 0.2123 - 0.0655 = 0.1468, or 14.68%. The sketch of the region would be a shaded area under the standard normal curve between -1.52 and -0.81.

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Answer:

The distance is 9.17 feet.

Step-by-step explanation:

The ramp, vertical distance it is lifted, and the ground form a right triangle, whose hypotenuse the ramp, and whose base and perpendicular are the ground and the lifted distance respectively.

Thus we have a triangle whose hypotenuse [tex]H[/tex] is 10 feet, the perpendicular [tex]P[/tex] is 4 feet, and a base [tex]B[/tex] feet.

The Pythagorean theorem gives:

[tex]H^2=P^2+B^2[/tex]

We substitute the values [tex]H=10[/tex], [tex]P =4[/tex] and solve for B:

[tex]B=\sqrt{H^2-P^2} =\sqrt{10^2-4^2} =9.17.[/tex]

Thus the distance is 9.17 feet.

Answer:

the Answer is ≈ 9.17 feet

Step-by-step explanation:

it is correct on edge  2020

Answer:

  28 inches high by 7 inches wide

Step-by-step explanation:

Let x represent the width of the poster with margins. Then the printable width is (x -2). The printable height will be 100/(x-2), so the overall poster height is ...

  height = 100/(x -2) +8 = (8x +84)/(x -2)

The poster's overall area is the product of its width and height, so is ...

  A = x(8x +84)/(x -2)

The derivative of this with respect to x is ...

  A' = ((16x +84)(x -2) -(8x^2 +84x)(1))/(x -2)^2

This is zero when the numerator is zero, so ...

  8x^2 -32x -168 = 0

  x^2 -4x -21 = 0 . . . . . . divide by 8

  (x +3)(x -7) = 0 . . . . . . . factor

The values of x that make these factors be zero are -3 and +7. The height corresponding to a width of 7 is ...

  height = 100/(7 -2) +8 = 28

The amount of paper is minimized when the poster is 7 inches wide by 28 inches tall.

_____

Comment on the problem and solution

You will notice that the poster is 4 times as high as it is wide. It is no accident that this ratio is the ratio of the vertical margin to the horizontal margin. That is, the fraction of the poster devoted to margin is the same in each direction. This is the generic solution to this sort of problem.

Knowing that the margins have a ratio of 4:1 tells you the printable area will have a ratio of 4:1, hence is equivalent to 4 squares, each with an area of 100/4 = 25 square inches. That means the printable area is √25 = 5 inches wide by 4×5 = 20 inches high, so the overall poster area is 28 inches high by 7 inches wide. This arithmetic can be all mental and does not involve derivatives.

To minimize the amount of paper used, the overall dimensions of the rectangular poster should be 102 inches in width and 108 inches in height.

To minimize the amount of paper used, we need to find the dimensions of the rectangle that will enclose 100 in2 of printing. Since there is a 4-inch margin at the top and bottom, the height of the rectangle will be the printing area plus the margins, which is 100 + 4 + 4 = 108 inches. Similarly, there is a 1-inch margin on each side, so the width of the rectangle will be the printing area plus the margins, which is 100 + 1 + 1 = 102 inches. Therefore, the overall dimensions of the rectangle that will minimize the amount of paper used are 102 inches in width and 108 inches in height.

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Answer:  The required solution of the given differential equation is

[tex]x+y\log y=Cy.[/tex]

Step-by-step explanation:  We are given to solve the following ordinary differential equation :

[tex]ydx+(y-x)dy=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We will be using the following formulas for integration and differentiation :

[tex](i)~d\left(\dfrac{x}{y}\right)=\dfrac{ydx-xdy}{y^2},\\\\\\(ii)~\int\dfrac{1}{y}dy=\log y.[/tex]

From equation (i), we have

[tex]ydx+(y-x)dy=0\\\\\Rightarrow ydx+ydy-xdy=0\\\\\\\Rightarrow \dfrac{ydx+ydy-xdy}{y^2}=\dfrac{0}{y^2}~~~~~~~~~~~~~~~~~~~~[\textup{dividing both sides by }y^2]\\\\\\\Rightarrow \dfrac{ydx-xdy}{y^2}+\dfrac{1}{y}dy=0\\\\\\\Rightarrow d\left(\dfrac{x}{y}\right)+d(\log y)=0.[/tex]

Integrating the above equation on both sides, we get

[tex]\int d\left(\dfrac{x}{y}\right)+\int d(\log y)=C~~~~~~~[\textup{where C is the constant of integration}]\\\\\\\Rightarrow \dfrac{x}{y}+\log y=C\\\\\Rightarrow x+y\log y=Cy.[/tex].

Thus, the required solution of the given differential equation is

[tex]x+y\log y=Cy.[/tex].

The probability that has been chosen Bag 1 is 0.2087.

Given that, bag 1 contains 10 blue marbles, while bag 2 contains 15 blue marbles.

Here we have;

Bag 1 contains 10 blue marbles

Bag 2 contains 15 blue marbles

Chosen a bag at random and throw 5 red marbles in it.

[tex]Required Probability = P(\frac{Bag 1}{2 red and 3 blue marbles})[/tex]

= [tex]\frac{P(bag 1)\cap(2 Red \ and \ 3 blue)}{P(2 \ red \ and \ 3 \ blue \ marbles)}[/tex]

= [tex]\frac{\frac{1}{2}\times ^6C_2\times^{10}C_3}{\frac{1}{2}\times^6C_2\times^{10}C_3+\frac{1}{2}\times^6C_2\times^{15}C_3}[/tex]

= 0.2087

Therefore, the probability that has been chosen Bag 1 is 0.2087.

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To find the probability of choosing Bag 1 given there are 2 red marbles among the 5 chosen marbles, we can use Bayes' theorem to calculate the probability.

To solve this problem, we can use Bayes' theorem to find the probability that Bag 1 was chosen given there are exactly 2 red marbles among the 5 chosen marbles. Let's denote Bag 1 as event A and Bag 2 as event B.

The answer to the question is the probability of choosing Bag 1 given there are exactly 2 red marbles among the 5 chosen marbles.