There is a probability of 20% that a milk container is underweight throughout of packaging line. Suppose milk containers are shipped to retail outlets in boxes of 10 containers. What is the probability that at least nine milk containers in a box are properly filled?

The amount of salt in the tank changes with rate according to

[tex]Q'(t)=\left(1\dfrac{\rm lb}{\rm gal}\right)\left(4\dfrac{\rm gal}{\rm min}\right)-\left(\dfrac{Q(t)}{300+(4-1)t}\dfrac{\rm lb}{\rm gal}\right)\left(1\dfrac{\rm gal}{\rm min}\right)[/tex]

[tex]\implies Q'+\dfrac Q{300+3t}=4[/tex]

which is a linear ODE in [tex]Q(t)[/tex]. Multiplying both sides by [tex](300+3t)^{1/3}[/tex] gives

[tex](300+3t)^{1/3}Q'+(300+3t)^{-2/3}Q=4(300+3t)^{1/3}[/tex]

so that the left side condenses into the derivative of a product,

[tex]\big((300+3t)^{1/3}Q\big)'=4(300+3t)^{1/3}[/tex]

Integrate both sides and solve for [tex]Q(t)[/tex] to get

[tex](300+3t)^{1/3}Q=(300+3t)^{4/3}+C[/tex]

[tex]\implies Q(t)=300+3t+C(300+3t)^{-1/3}[/tex]

Given that [tex]Q(0)=100[/tex], we find

[tex]100=300+C\cdot300^{-1/3}\implies C=-200\cdot300^{1/3}[/tex]

and we get the particular solution

[tex]Q(t)=300+3t-200\cdot300^{1/3}(300+3t)^{-1/3}[/tex]

[tex]\boxed{Q(t)=300+3t-2\cdot100^{4/3}(100+t)^{-1/3}}[/tex]

Answer:

DNE

Step-by-step explanation:

Given that two particles travel along the space curves

[tex]r_1(t) = (t, t^2, t^3)\\ r_2(t) = (1 + 2t, 1 + 6t, 1 + 14t )[/tex]

To find the points of intersection:

At points of intersection both coordinates should be equal.

i.e. r1 =r2

Equate corresponding coordinates

[tex]t=1+2t\\t^2=1+6t\\t^3=1+14t[/tex]

I equation gives t =-1

Substitute in II equation to get [tex]t^2 = -5[/tex]

i.e. t cannot be real

Hence no point of intersection

DNE

Answer:

You will requiere 0.34375 oz of the solution for 22 ounces of water.

Step-by-step explanation:

We should start by matching the units: in the example we have 2 gallons of water, while the question refers to 22 ounces of water.

So, by the equivalence we are given, we now know that 2 gallons of water are equal to 256 ounces.

From this point the question can be solved using an arithmetic method known as cross-multiply (sometimes refered as the Rule of Three). This consist of wrinting an equation of the form:

[tex]\frac{a}{b} = \frac{c}{x}\\[/tex]

Where b is the amount of solution required for a ounces of water, and x (unknown variable) is the amount of solution for c ounces of water.

So we have the next equation:

[tex]\frac{256}{4} =\frac{22}{x}[/tex]

We apply the correspondent arithmetic rules so we can calculate the x as follows:

[tex]x=\frac{(4)(22)}{256}[/tex]

[tex]x=\frac{88}{256}[/tex]

[tex]x=0.34375[/tex]

This way we can reply that 0.34375 oz of solution are required for 22 oz of water.

Answer:

A

Step-by-step explanation:

Given

u = <6, -1>

u = 6i-j

and

v=<7,-4>

v=7i-4j

The formula for angle is:

Let x be the angle

[tex]cos\ x = \frac{u.v}{||u||.||v||}[/tex]

where ||u|| is the length and u.v is the dot product or scalar product of both vectors

So,

[tex]||u|| = \sqrt{(6)^2+(-1)^2}\\ = \sqrt{36+1}\\ = \sqrt{37}\\ ||v||=\sqrt{(7)^2+(-4)^2}\\ = \sqrt{49+16}\\ = \sqrt{65}\\[/tex]

[tex]u.v = u_1u_2+v_1v_2\\= (6)(7)+(-1)(-4)\\=42+4\\=46[/tex]

[tex]cos\ x=\frac{46}{\sqrt{37}\sqrt{65}} \\= \frac{46}{\sqrt{2405} }\\Can\ also\ be\ written\ as:\\= \frac{46}{\sqrt{2405} } * \frac{\sqrt{2405} }{\sqrt{2405}} \\=\frac{46\sqrt{2405} }{2405}[/tex]

The calculated angle will be in radians. To find the angle in degrees:

[tex]x = \frac{180}{\pi} cos^{-1} (\frac{46\sqrt{2405} }{2405})\\x = 20.282\\x= 20.3\\[/tex]

Hence Option A is correct ..

Answer with explanation:

[tex]\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}[/tex]

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

 [tex]=e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}[/tex]

Multiplying both sides by Integrating Factor  

[tex]x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)[/tex]

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

Answer:

General Solution is [tex]y=x^{3}+cx^{2}[/tex] and the particular solution is  [tex]y=x^{3}-\frac{1}{2}x^{2}[/tex]

Step-by-step explanation:

[tex]x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}[/tex]

This is a linear diffrential equation of type

[tex]\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)[/tex]..................(i)

here [tex]p(x)=\frac{-2}{x}[/tex]

[tex]q(x)=x^{2}[/tex]

The solution of equation i is given by

[tex]y\times e^{\int p(x)dx}=\int  e^{\int p(x)dx}\times q(x)dx[/tex]

we have [tex]e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}[/tex]

Thus the solution becomes

[tex]\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c[/tex][tex]y=x^{3}+cx^{2[/tex]

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have [tex]6=8+4c[/tex]

Thus solving for c we get c = -1/2

Thus particular solution becomes

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

Answer:

The probability is 0.0643

Step-by-step explanation:

* Lets revise some definition to solve the problem

- The standard deviation of the distribution of sample means is called σM

- σM = σ/√n , where σ is the standard deviation and n is the sample size

- z-score = (M - μ)/σM, where M is the mean of the sample , μ is the mean

 of the population  

* Lets solve the problem

- The mean of the washing machine is 9.3 years

∴ μ = 9.3

- The standard deviation is 1.1 years

∴ σ = 1.1

- There are 70 washing machines randomly selected

∴ n = 70

- The mean replacement time less than 9.1 years

∴ M = 9.1

- Lets calculate z-score

∵ σM = σ/√n

∴ σM = 1.1/√70 = 0.1315

∵ z-score = (M - μ)/σM

∴ z-score = (9.1 - 9.3)/0.1315 = - 1.5209

- Use the normal distribution table of z to find P(z < -1.5209)

∴ P(z < -1.5209) = 0.06426

∵ P(M < 9.1) = P(z < -1.5209)

∴ P(M < 9.1) = 0.0643

* The probability is 0.0643

Answer:

Step-by-step explanation:

The pigeonhole principal states that if n items are put into m containers, with n > m then at least one container must contain more than one item.

In other words, to find the case where maximum attempts are required, we eliminate all the cases where our criterion is not met and we will be left with the desired result.

For this case, the box has 10 black socks and 12 blue socks.

10 black socks = 5 left foot ones and  5 right foot ones

12 blue socks = 6 left foot ones and 6 right foot ones

If we draw all left foot or right foot ones then we will not have a pair till 11 draws have been made.

The next socks drawn will be a right foot one of blue or black color and a pair will be made.

Therefore, the maximum number of socks needed to be drawn from the box are 12.

Answer:

Step-by-step explanation:

Let x and y represent the numbers of churches and unions contacted in the month, respectively. Then Jayanta's limit on letter writing hours is ...

  2x +2y ≤ 20

and her limit on follow-up call hours is ...

  x + 3y ≤ 14

Graphing these inequalities (see below) results in a feasible region with vertices at (x, y) = (0, 4 2/3), (8, 2), and (10, 0). Of these, the mixture of groups producing the most money is ...

  8 churches and 2 unions.

The money she can raise from that mixture is ...

  8×$125 +2×$175 = $1350 in a month

Answer:

Sqdancefan's answer is correct.

Step-by-step explanation:

I  misread the question.

Answer:

See below.

Step-by-step explanation:

The set of  all positive integers N is countable so we need to show that there is a 1 to 1 correspondence between the elements in N  with the set of all odd positive integers. This is the case as shown below:

1 2 3 4 5 6 ...

|  |  |   |   |  | ....

1 3 5 7 9 11....

Answer:

He did better on biology test.

Step-by-step explanation:

For comparing the scores we need to find z-scores for both subjects,

We know that,

z-score or standard score for x is,

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

Where, [tex]\mu[/tex] is mean,

[tex]\sigma[/tex] is standard deviation,

In history test,

[tex]x=77[/tex]

[tex]\mu=70[/tex]

[tex]\sigma = 6[/tex]

Thus, the z-score would be,

[tex]z_1=\frac{77-70}{6}[/tex]

[tex]\approx 1.167[/tex]

In biology test,

[tex]x=79[/tex]

[tex]\mu = 70[/tex]

[tex]\sigma = 4[/tex],

Thus, the z-score would be,

[tex]z_2=\frac{79-70}{4}[/tex]

[tex]= 2.25[/tex]

∵ [tex]z_2>z_1[/tex]

Hence, he did better on biology test.

Answer with  explanation:

Given L: V\rightarrow W is a linear transformation of a vector space  V into a vector space W.

Let Dim V= DimW=n

a.If L is one-one

Then nullity=0 .It means dimension of null space is zero.

By rank- nullity theorem we have

Rank+nullity= Dim V=n

Rank+0=n

Rank=n

Hence, the linear transformation is onto. Because dimension of range is equal to dimension of codomain.

b.If linear transformation is onto.

It means  dimension of range space is equal to dimension of codomain

Rank=n

By rank nullity theorem  we have

Rank + nullity=dimV

n+nullity=n

Nullity=n-n=0

Dimension of null space is zero.Hence, the linear transformation is one-one.

To prove Corollary 6.2, we show that a one-to-one linear transformation between vector spaces of equal dimensions is onto, and conversely, an onto transformation is one-to-one. This conclusion is based on the properties of linear transformations and the significance of preserving vector space dimensions.

To prove Corollary 6.2 regarding a linear transformation L: V → W, where both vector spaces V and W have the same dimension, we must show two parts:

If L is one-to-one, then it is onto. Assuming L is one-to-one, for every vector v in V, there is a unique image in W, guaranteeing that all vectors in W can be reached since dim V = dim W. Thus, L covers all of W, making it onto.

If L is onto, then it is one-to-one. When assuming L is onto, every vector in W is the image of some vector in V. Because dim V = dim W, there can't be more vectors in V than in W, preventing multiple vectors in V from mapping to the same vector in W, thus L is one-to-one.

These parts rely heavily on the concept that linear transformations maintain vector operations, such as the addition and scalar multiplication, and the relationship between dimensions of the domain and codomain for linear transformations.

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Answer:

2500

Step-by-step explanation:

They tell you that only 40% of the total complete the quality test, and you want to know how many represent 100%.

If this 40% is 1000 valid responses use cross multiplication to know how many people are the total.

40%-----1000 valid responses

100%----x=

[tex]\frac{100*1000}{40} =X\\2500=X[/tex]

So, you can tell they need at least 2500 persons to have 1000 valid responses.

Answer:

Number of people = 3473

Step-by-step explanation:

Probability is typically the rate at which an event or occurrence is likely to happen. Whenever we're not so sure about an event outcome, we can then deliberate about the probabilities or possibility of certain outcomes that is how likely they are to happen.

Probability of success, =0.8×0.4×0.9 =0.288

Expected, E=1000

Expected = np

1000=n×0.288

n=​0.288​​1000​​≈34

Answer:

$148.02

Step-by-step explanation:

In the question

Principal = $100, rate(R) = 4% compounded annually, time(T)= 10 years

we know that the formula for compound interest

A=[tex]P\times_(1+\frac{R}{100} )^{T}[/tex]   where A is amount

now putting values in the above formula we get

A=[tex]100\times_(1+\frac{4}{100} )^{10}[/tex]

therefore A= $148.024428

rounding off to the nearest penny we get amount as $148.02 and compound interest will be $48.02

The correct option is to reject the null hypothesis. The mean life spans for whites and non-whites born in 1900 in the certain county are not the same.

To conduct the hypothesis test, we will use a two-sample z-test for the difference in two means. The null hypothesis (H0) states that there is no difference in the mean life spans between whites and non-whites, while the alternative hypothesis (Ha) states that there is a difference.

The formula for the z-test statistic is:

[tex]\[ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \][/tex]

where:

- [tex]\(\bar{x}_1\)[/tex] and [tex]\(\bar{x}_2\)[/tex] are the sample means for whites and non-whites, respectively.

- [tex]\(\mu_1\)[/tex] and [tex]\(\mu_2\)[/tex] are the population means for whites and non-whites, respectively.

- [tex]\(\sigma_1\)[/tex] and [tex]\(\sigma_2\)[/tex] are the population standard deviations for whites and non-whites, respectively.

- [tex]\(n_1\)[/tex] and [tex]\(n_2\)[/tex] are the sample sizes for whites and non-whites, respectively.

Given:

- [tex]\(\bar{x}_1 = 45.3\) years, \(s_1 = 12.7\) years, \(n_1 = 124\)[/tex] (for whites)

- [tex]\(\bar{x}_2 = 34.1\)[/tex] years, [tex]\(s_2 = 15.6\) years, \(n_2 = 82\)[/tex] (for non-whites)

- [tex]\(\mu_1 = 47.6\)[/tex] years (for whites)

- [tex]\(\mu_2 = 33.0\)[/tex] years (for non-whites)

Since we do not have the population standard deviations, we will use the sample standard deviations as an estimate. This is appropriate given the sample sizes are large enough (generally [tex]\(n > 30\)[/tex] is considered sufficient).

First, we calculate the standard error (SE) of the difference in means:

[tex]\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{12.7^2}{124} + \frac{15.6^2}{82}} \][/tex]

[tex]\[ SE = \sqrt{\frac{161.29}{124} + \frac{243.36}{82}} \][/tex]

[tex]\[ SE = \sqrt{1.299 + 2.968} \][/tex]

[tex]\[ SE = \sqrt{4.267} \][/tex]

[tex]\[ SE \approx 2.066 \][/tex]

Now, we calculate the z-statistic:

[tex]\[ z = \frac{(45.3 - 34.1) - (47.6 - 33.0)}{2.066} \][/tex]

[tex]\[ z = \frac{11.2 - 14.6}{2.066} \][/tex]

[tex]\[ z = \frac{-3.4}{2.066} \][/tex]

[tex]\[ z \approx -1.646 \][/tex]

Next, we find the p-value for this z-statistic. Since we are conducting a two-tailed test, we will look at the probability of a z-score being less than -1.646 or greater than 1.646. Using a standard normal distribution table or a calculator, we find that the p-value is approximately 0.100.

Finally, we compare the p-value to our significance level (commonly denoted as [tex]\(\alpha\))[/tex]. If the p-value is less than [tex]\(\alpha\)[/tex], we reject the null hypothesis. If we choose [tex]\(\alpha = 0.05\)[/tex], then since [tex]\(0.100 > 0.05\)[/tex], we fail to reject the null hypothesis.

However, if we choose a significance level of [tex]\(\alpha = 0.10\)[/tex], then since [tex]\(0.100 \leq 0.10\)[/tex], we would reject the null hypothesis, indicating that there is a statistically significant difference in the mean life spans between whites and non-whites in the county.

Given the p-value is on the boundary of common significance levels, the conclusion may vary depending on the chosen level of significance. However, the correct option based on the provided conversation is to reject the null hypothesis, suggesting that the mean life spans are not the same for whites and non-whites in the county.

Answer:

4) 50,000

5) 900,000

Step-by-step explanation:

In Front end rounding what we do is we focus on first two digit of the number if the second digit number is greater than or equal to 5 we add the first digit by 1.

4) 50,987

here we can clearly see that the first two digit is 50 and second digit is not equal to 5 then rounding will be equal to 50,000.

5) 851,004

here we can clearly see that the first two digit is 85 and second digit is equal to 5 then we will round it  8+1=9

hence rounded number will be 900,000

Final answer:

Front-end rounding of 50,987 is 50,000 and for 851,004 it's 900,000, based on the most significant digits, using zeros as placeholders after rounding.

Explanation:

Front-end rounding involves rounding numbers based on the most significant digits. Let's apply this method to the numbers provided.

When rounding, if the digit immediately after the place you are rounding to is 5 or greater, you round up. Otherwise, you round down. Placeholder zeros are used to maintain the value's place in the number system.

Answer: 0.4302

Step-by-step explanation:

Given : Mean : [tex]\mu=\text{38.7 miles }[/tex]

Standard deviation : [tex]\sigma=\text{6.7 miles }[/tex]

Sample size : [tex]n=60[/tex]

Also, these distances are normally distributed.

Then , the formula to calculate the z-score is given by :-

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

For x=32.5

[tex]\\\\ z=\dfrac{32.5-38.7}{6.7}=-0.925373134\approx-0.93[/tex]

For x=40.5

[tex]\\\\ z=\dfrac{40.5-38.7}{6.7}=0.268656\approx0.27[/tex]

The p-value = [tex]P(-0.93<z<0.27)[/tex]

[tex]=P(0.27)-P(-0.93)=0.6064198- 0.1761855=0.4302343\approx0.4302[/tex]

Hence, the required probability :-0.4302

[tex]f(x,y,z)=2z^2x+y^3[/tex]

[tex]f[/tex] has gradient

[tex]\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k[/tex]

which at the point (-1, 4, 3) has a value of

[tex]\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k[/tex]

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say [tex]\vec u=15\,\vec\imath+25\,\vec\jmath[/tex], in which case we have

[tex]\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}[/tex]

Then the derivative of [tex]f[/tex] at (-1, 4, 3) in the direction of [tex]\vec u[/tex] is

[tex]D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}[/tex]

To find the directional derivative, we need to find the gradient of the function and dot product it with the given direction vector.

To find the directional derivative, we need to find the gradient of the function and dot product it with the given direction vector. The gradient of the function f(x, y, z) = 2z^2x + y^3 is (∂f/∂x, ∂f/∂y, ∂f/∂z) = (4zx, 3y^2, 4xz). The directional derivative in the direction of the vector (15-√i + 25-√j) is given by the dot product of the gradient and the direction vector: (4(-1)(15-√) + 3(4^2)(25-√) + 4(3)(-1)(15-√))/√((15-√)^2 + (25-√)^2). Simplifying this expression gives the directional derivative.

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Answer:

(x² -7)/(2x + 1)

Step-by-step explanation:

f(x) = 2x+1 and g(x) = x² -7

thus: (g/f)(x) = g(x)/f(x) = x² -7/2x + 1

The answer is:

[tex](g \circ f)(x)=4x^{2}-4x-1[/tex]

To solve the problem, we need to remember that composing functions means evaluate a function into another different function.

Also, we need to remember how to solve the following notable product:

[tex](a-b)^{2}=a^{2}-2ab+b^{2}[/tex]

We have that:

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

Now, we are given the equations:

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

So, composing we have:

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

[tex](g \circ f)(x)=(2x-1)^{2}-2[/tex]

Now, we have to solve the notable product:

[tex](g \circ f)(x)=((2x)^{2}-2(2x*1)+1^{2})-2[/tex]

[tex](g \circ f)(x)=4x^{2}-4x+1-2[/tex]

Hence, we have that:

[tex](g \circ f)(x)=4x^{2}-4x-1[/tex]

Have a nice day!

Answer:[tex]\frac{^{10}C_5}{^{12}C_5}[/tex]

Step-by-step explanation:

Given a total of 12 diesel engines

out of which 2 are defective

Company have to choose 5 good engines to ship.

Therefore no of ways in good engines are selected is [tex]^{10}C_5[/tex]

no of ways in which any 5 engines are selected from a total of 12 engines is

[tex]^{12}C_5[/tex]

Therefore the required probability =[tex]\frac{^{10}C_5}{^{12}C_5}[/tex]

=[tex]\frac{7}{22}[/tex]=0.318

Answer:

There are three transformation i.e. vertically stretch, downward and towards left.            

Step-by-step explanation:

Given : Parent function [tex]f(x)=|x|[/tex] and transformed function [tex]f(x)=2|x+3|-5[/tex]

To find : Name the type of function and describe each of the three transformations?

Solution :

Parent function [tex]f(x)=|x|[/tex] get transformed into [tex]f(x)=2|x+3|-5[/tex]

The transformations are as follow :

1) The function is multiplied by 2 which means there is a transformation vertically by a stretch factor '2'.

2) The function is subtracted by 5 which means there is a transformation downward by a factor '5'.

3) The function inside x value get added by 3 which means there is a transformation left by a factor '3'.

Therefore, There are three transformation i.e. vertically stretch, downward and towards left.

Answer:

By 2086

Step-by-step explanation:

The provided equation is:

[tex]A=A0*e^{k*t}[/tex] , where:

A=total of population after t years

A0=initial population

k= rate of growth

t= time in years

Given information:

The final population will be 15 million, then A=15.

We start in 2000 with a 5.82 million population, then A0=5.82.

Missing information:

Although k is not given, we can calculate k by using the following statement, from 2000 to 2040 (within 40 years) population is proyected to grow to 9 million, which means a passage from 5.8 to 9 million (3.2 million increament).

Then we can use the same expression to calculate k:

[tex]A=A0*e^{k*t}[/tex]

[tex]9=5.8*e^{40*k}[/tex]

[tex]ln(9/5.8)/40=k[/tex]

[tex]0.010984166494596147=k[/tex]

[tex]0.011=k[/tex]

Now that we have k=0.011, we can find the time (t) by which population will be 15 million:

[tex]A=A0*e^{k*t}[/tex]

[tex]15=5.8*e^{0.011t}[/tex]

[tex]ln(15/5.8)/0.011=t[/tex]

[tex]86.38111668634878=t[/tex]

[tex]86.38=t[/tex]

Because the starting year is 2000, and we need 86.38 years for increasing the population from 5.8 to 15 million, then by 2086 the population will be 15 million.

Answer:

200 Creamy Vanilla

300 Continental Mocha

Step-by-step explanation:

3 cups cream for each:

500 Quarts total

Creamy Vanilla requires 400 eggs

Continental Mocha requres 300 eggs for a total of 700 eggs

The answer is:

[tex]\[ \boxed{V = 200, M = 300} \][/tex]

To solve this problem, we need to set up a system of equations based on the given information and then solve for the number of quarts of each flavour of ice cream that should be made.

Let [tex]\( V \)[/tex] represent the number of quarts of Creamy Vanilla ice cream and [tex]\( M \)[/tex] represent the number of quarts of Continental Mocha ice cream.

From the information given, we can derive the following equations:

For the eggs:

Each quart of Creamy Vanilla requires 2 eggs, so [tex]\( 2V \)[/tex] eggs are used for Creamy Vanilla.

Each quart of Continental Mocha requires 1 egg, so [tex]\( M \)[/tex] eggs are used for Continental Mocha.

Since there are 700 eggs in total, we have the equation:

[tex]\[ 2V + M = 700 \][/tex]

For the cream:

Each quart of Creamy Vanilla requires 3 cups of cream, so [tex]\( 3V \)[/tex] cups of cream are used for Creamy Vanilla.

Each quart of Continental Mocha also requires 3 cups of cream, so [tex]\( 3M \)[/tex] cups of cream are used for Continental Mocha.

Since there are 1500 cups of cream in total, we have the equation:

[tex]\[ 3V + 3M = 1500 \][/tex]

Now we have a system of two equations with two variables:

[tex]\[ \begin{cases} 2V + M = 700 \\ 3V + 3M = 1500 \end{cases} \][/tex]

To solve this system, we can simplify the second equation by dividing every term by 3:

[tex]\[ V + M = 500 \][/tex]

Now we have a simpler system:

[tex]\[ \begin{cases} 2V + M = 700 \\ V + M = 500 \end{cases} \][/tex]

We can subtract the second equation from the first to eliminate [tex]\( M \)[/tex] and solve for [tex]\( V \)[/tex] :

[tex]\[ (2V + M) - (V + M) = 700 - 500 \][/tex]

[tex]\[ V = 200 \][/tex]

Now that we have the value of [tex]\( V \)[/tex], we can substitute it back into the simplified second equation to find [tex]\( M \)[/tex]:

[tex]\[ 200 + M = 500 \][/tex]

[tex]\[ M = 500 - 200 \][/tex]

[tex]\[ M = 300 \][/tex]

Therefore, the factory should make 200 quarts of Creamy Vanilla and 300 quarts of Continental Mocha to use up all the eggs and cream.

The final answer is:

[tex]\[ \boxed{V = 200, M = 300} \][/tex]

Answer:

1) 0.3456

2) 0.6544.

Step-by-step explanation:

Let X represents the event of recognizing the brand,

Given,

The probability of recognizing the brand, p = 40​% = 0.40,

Thus, the probability of not recognizing the brand, q = 1 - 0.40 = 0.60,

Since, the binomial distribution formula,

[tex]P(x) = ^nC_r (p)^r(q)^{n-r}[/tex]

Where,

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

1) Thus, the probability that exactly 2 of the 5 consumers recognize the brand name is,

[tex]P(X=2)=^5C_2 (0.40)^2 (0.60)^{5-2}[/tex]

[tex]=10 (0.40)^2 (0.60)^3[/tex]

[tex]=0.3456[/tex]

2) Also, the probability that the number who recognize the brand name is not 2 = 1 - P(X=2) = 1 - 0.3456 = 0.6544.

Answer:

Step-by-step explanation:

Put hypotheses as:

[tex]H_0: mu =40 mg\\H_a: mu \neq 40 mg.[/tex]

(Two tailed test at 4%)

Since population std deviation is  given we can do Z test.

[tex]Sample size n =30\\Sample mean = 42.3 mg\\Mean diff = 2.3 mg\\Std error of sample = \frac{7.1}{\sqrt{30} } =1.296[/tex]

Test statistic t = mean diff/se = [tex]\frac{2.3}{1.296} =1.775[/tex]

p value=0.075

Since p >0.04 our alpha, we accept null hypothesis.

At alphaequals0.04​, we cannot reject the​ company's claim

Step-by-step explanation:

Put hypotheses as:

\begin{gathered}H_0: mu =40 mg\\H_a: mu \neq 40 mg.\end{gathered}

H

0

:mu=40mg

H

a

:mu



=40mg.

(Two tailed test at 4%)

Since population std deviation is given we can do Z test.

\begin{gathered}Sample size n =30\\Sample mean = 42.3 mg\\Mean diff = 2.3 mg\\Std error of sample = \frac{7.1}{\sqrt{30} } =1.296\end{gathered}

Samplesizen=30

Samplemean=42.3mg

Meandiff=2.3mg

Stderrorofsample=

30

7.1

=1.296

Test statistic t = mean diff/se = \frac{2.3}{1.296} =1.775

1.296

2.3

=1.775

p value=0.075

Since p >0.04 our alpha, we accept null hypothesis.

At alphaequals0.04, we cannot reject the company's claim

Answer:

C. P69,256.82

Step-by-step explanation:

We know that,

The amount formula in compound interest is,

[tex]A=P(1+\frac{r_1}{n_1})^{n_1t_1} (1+\frac{r_2}{n_2})^{n_2t_2}.......[/tex]

Where, P is the principal amount,

[tex]r_1, r_2....[/tex] are the annual rate for the different periods,

[tex]t_1, t_2,.....[/tex] are the number of year for different periods,

[tex]n_1, n_2, n_3...[/tex] are the number of periods,

Given,

A = P 200,000,

[tex]r_1=10%=0.1[/tex], [tex]n_1=4[/tex], [tex]t_1=5[/tex],[tex]r_2=12%=0.12[/tex], [tex]n_2=1[/tex], [tex]t_2=5[/tex]

Thus, by the above formula the final amount would be,

[tex]200000=P(1+\frac{0.1}{4})^{4\times 5}(1+\frac{0.12}{1})^{1\times 5}[/tex]

[tex]200000=P(1+0.025)^{20}(1+0.12)^5[/tex]

[tex]200000=P(1.025)^{20}(1.12)^5[/tex]

[tex]\implies P=69,256.824\approx 69,256.82[/tex]

Option C is correct.

The correct option is C. P69,256.82. The father should invest approximately P69,256.82 today to ensure his son receives P200,000 in ten years.

To determine the amount the father needs to invest today to give his son P200,000 ten years from now, we will break the problem into two phases due to different interest rates and compounding periods.

Phase 1: First Five Years (10% compounded quarterly)

→ Future Value (FV) needed after 10 years: P200,000

→ Future Value (FV) after first five years at 12% annual interest for the next five years:

Let's use the formula for compound interest to calculate amount required after the first five years.

Here,

→ n is the number of times the interest is compounded per year

→ t is time in years.

→ [tex]FV = PV*(1 + r/n)^{(nt)[/tex]

After the first five years, the amount needs to grow at 12% compounded annually for 5 years to reach P200,000.

We can calculate the present value (PV) at the end of the first five years needed to achieve P200,000 after next 5 years.

→ P200,000 = [tex]PV * (1 + 0.12/1)^{(1*5)[/tex]

→ P200,000 = [tex]PV * (1.12)^5[/tex]

Calculating PV:

→ PV = P200,000 / (1.7623)

        ≈ P113,477.57

Phase 2: First Five Years Investment Calculation

Now, we need to find out the amount the father should invest today to reach P113,477.57 after five years with 10% interest compounded quarterly.

Here,

→ r is the quarterly rate,

→ nt is the total number of quarters.

We use the same compound interest formula:

→ [tex]FV = PV * (1 + r/n)^{(nt)[/tex]

→ P113,477.57 = [tex]PV * (1 + 0.10/4)^{(4*5)[/tex]

→ P113,477.57 = [tex]PV * (1.025)^{20[/tex]

→ [tex](1.025)^{20} \approx 1.6386[/tex]

Calculating PV:

→ PV = P113,477.57 / 1.6386

        ≈ P69,256.82

So, the father needs to invest approximately P69,256.82 today.

Answer:  The proof is done below.

Step-by-step explanation:  Given that U and V are subspaces of a vector space W.

We are to prove that the intersection U ∩ V is also a subspace of W.

(a) Since U and V are subspaces of the vector space W, so we must have

0 ∈ U and 0 ∈ V.

Then, 0 ∈ U ∩ V.

That is, zero vector is in the intersection of U and V.

(b) Now, let x, y ∈ U ∩ V.

This implies that x ∈ U, x ∈ V, y ∈ U and y ∈ V.

Since U and V are subspaces of U and V, so we get

x + y ∈ U  and  x + y ∈ V.

This implies that x + y ∈ U ∩ V.

(c) Also, for a ∈ R (a real number), we have

ax ∈ U and ax ∈ V (since U and V are subspaces of W).

So, ax ∈ U∩ V.

Therefore, 0 ∈ U ∩ V and for x, y ∈ U ∩ V, a ∈ R, we have

x + y and ax ∈ U ∩ V.

Thus, U ∩ V is also a subspace of W.

Hence proved.

Answer:

  $21012.50

Step-by-step explanation:

$20,000 today will be inflated to $20,000·(1.025)^2 ≈ $21012.50 in 2 years. We presume this is the nominal terminal value you want.

Answer with explanation:

It is given that, a, b and c belong to the set of integers.

→ gcd(a,c)=d

→GCD=Greatest Common Divisor

→The greatest number which divides both a and c is d.

It means d divides a, and d divides c.

 a=d k, for some integer k.-------(1)

  c= d m, for some integer m.-------(2)

Now, it is given that, a divides bc.

So,→ 'a' will divide "bdm".--------[using 2, as c=d m]

It shows that, a divides bd, that is a| bd.

Hence proved

Answer:

1.  [tex]x=3[/tex]; 2. Domain:  [tex]x>1[/tex]  Range: all real numbers

Step-by-step explanation:

Let's find the solutions.

1. Solve the equation [tex]x=\sqrt{2x+3}[/tex] so:

[tex](x)^2=(\sqrt{2x+3})^2[/tex]

[tex]x^2=2x+3[/tex]

[tex]x^2-2x-3=0[/tex]

[tex]x1=\frac{-b+\sqrt{b^{2}-4ac}}{2a}[/tex]

[tex]x1=\frac{2+\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex]

[tex]x1=3[/tex]

[tex]x2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}[/tex]

[tex]x2=\frac{2-\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex]

[tex]x2=-1[/tex]

Although we have two answers, remember that from the original equation the result of [tex]\sqrt{2x+3} > 0[/tex] is never negative. So -1 do not solve the equation.

In conlcusion, the equation is solved by x=3.

2A. Domains and ranges of f1(x) and f2(x)

[tex]f1(x)=ln(x+1)+ln(x-1)[/tex]

Using logarithmic property [tex]ln(a)+ln(b)=ln(a*b)[/tex] we have:

[tex]f1(x)=ln(x^2-1)[/tex] because:

[tex]ln(x)[/tex] is defined by  [tex]x>0[/tex] then:

[tex]x^2-1>0[/tex]

[tex]x>\sqrt{1}[/tex] so the domain of f1(x) is [tex]x>1[/tex]

Now for the range:

[tex]f1(x)=ln(x+1)+ln(x-1)[/tex]

[tex]y=ln(x^2-1)[/tex]

[tex]e^y=x^2-1[/tex]

[tex]\sqrt{e^y+1}=x^2-1[/tex] notice that [tex]e^y+1[/tex] is always positive, so the range of f1(x) is all real numbers.

Be aware that although point number two of the problem mentioned two equations, f1(x)=f2(x) by logarithmic properties, so their domains and ranges are the same.

2B. Graph of f1(x) is attached. Because f1(x)=f2(x) both functions plot equal.