Recently many companies have been experimenting with telecommuting, allowing employees to work at home on their computers. Among other things, telecommuting is supposed to reduce the number of sick days taken. Suppose that at one firm, it is known that over the past few years employees have taken a mean of 5.4 sick days. This year, the firm introduces telecommuting. Management chooses a simple random sample of 80 employees to follow in detail, and, at the end of the year, these employees average 4.4 sick days with a standard deviation of 2.8 days. Let μ represent the mean number of sick days for all employees of the firm.

(a). Find the P-value for testing H0 :? ? 5.4 versus H1 :? < 5.4.
(b). Do you believe it is plausible that the mean number of sick days is at least 5.4, or are you convinced that it is less than 5.4? Explain your reasoning.

Answer:

Remember, if [tex]\{x_1,x_2,...,x_k\}[/tex] is a basis for a subspace W of [tex]\mathbb{R}^n[/tex] then [tex]\{q_1,q_2,...,q_k\}[/tex] is an orthonormal basis of W, where

[tex]q_i=\frac{1}{||v_i||}[/tex] and [tex]v_i[/tex] is defined as:

[tex]v_1=x_1\\v_2=x_2-\frac{v_1\cdot x_2}{v_1\cdot v_1}v_1\\v_k=x_k-\frac{v_1\cdot x_k}{v_1\cdot v_1}v_1 - \cdots - \frac{v_{k-1}\cdot x_k}{v_{k-1}\cdot v_{k-1}}v_{k-1}[/tex]

Then, to find a orthonormal basis of [tex]\{x_1=(8, -6, 0), x_2=(5, 1, 0), x_3=(0, 0, 2)\}[/tex] we will find first the [tex]v_i[/tex]'s.

[tex]v_1=x_1[/tex]

[tex]v_2=x_2-\frac{v_1\cdot x_2}{v_1\cdot v_1}v_1=(5,1,0)-\frac{(8,-6,0)\cdot (5,1,0)}{(8,-6,0)\cdot (8,-6,0)}(8,-6,0)=\\=(5,1,0)-\frac{17}{50}(8,-6,0)=(\frac{57}{25},\frac{76}{25},0)[/tex]

[tex]v_3=x_3-\frac{v_1\cdot x_3}{v_1\cdot v_1}v_1-\frac{v_2\cdot x_3}{v_2\cdot v_2}v_2=\\=(0,0,2)-\frac{(8-6,0)\cdot (0,0,2)}{(8,-6,0)\cdot (8,-6,0)}(8,-6,0) -\frac{(\frac{57}{25},\frac{76}{25},0)\cdot (0,0,2)}{(\frac{57}{25},\frac{76}{25},0)\cdot (\frac{57}{25},\frac{76}{25},0)}(\frac{57}{25},\frac{76}{25},0)=\\=(0,0,2)-0-0=\\=(0,0,2)[/tex]

Therefore, [tex]\{v_1=(8,-6,0),v_2=(\frac{57}{25},\frac{76}{25},0),v_3=(0,0,2)\}[/tex] is a ortogonal basis for [tex]\mathbb{R}^3[/tex]. But we need a orthonormal basis. Then is enough find the corresponding unit vector of the ortogonal basis found.

[tex]q_1=\frac{1}{||v_1||}v_1=\frac{1}{\sqrt{100}}(8,-6,0)\\q_2=\frac{1}{||v_2||}v_2=\frac{1}{\frac{19}{5}}(\frac{57}{25},\frac{76}{25},0)=\frac{5}{19}(\frac{57}{25},\frac{76}{25},0)\\q_3=\frac{1}{||v_3||}v_3=\frac{1}{2}(0,0,2)[/tex]

Hence

[tex]\{q_1=({\frac{8}{\sqrt{100}},\frac{-6}{\sqrt{100}},0), q_2=(\frac{3}{5},\frac{4}{5},0), q_3=(0,0,1)\}[/tex] is a orthonormal basis for [tex]\mathbb{R}^3[/tex]

Final answer:

To transform the given basis B into an orthonormal basis using the Gram-Schmidt orthonormalization process, we can follow the steps outlined.

Explanation:

To apply the Gram-Schmidt orthonormalization process to the given basis B = {(8, −6, 0), (5, 1, 0), (0, 0, 2)} in Rn, we'll follow these steps:

By applying this process to B, we find that U1 is (8/10, -6/10, 0), U2 is (1/2, 7/10, 0), and U3 is (0, 0, 2/√10).

Answer:

The sides of the container should be 3 cm and height should be 6 cm to minimize the cost

Step-by-step explanation:

Data provided in the question:

costs  for the material used on the side = $3/cm²

costs for the material used for the top lid and the base = $6/cm²

Volume of the container = 54 cm³

Now,

let the side of the base be 'x' and the height of the box be 'y'

Thus,

x × x × y = 54 cm³

or

x²y = 54

or

y = [tex]\frac{54}{x^2}[/tex]  ............(1)

Now,

The total cost of material, C

C = $3 × ( 4 side area of the box ) + $6 × (Area of the top and bottom)

or

C = ( $3 × 4xy ) + ( $6 × (x²)  )

substituting the value of y in the above equation, we get

C = [tex]3x\times4\times\frac{54}{x^2}+2\times6x^2[/tex]

or

C = [tex]\frac{648}{x}+12x^2[/tex]

Differentiating with respect to x and putting it equals to zero to find the point of maxima of minima

thus,

C' = [tex]-\frac{648}{x^2}+2\times12x[/tex] = 0

or

[tex]2\times12x=\frac{648}{x^2}[/tex]

or

24x³ = 648

or

x = 3 cm

also,

C'' = [tex]+\frac{2\times648}{x^3}+2\times12[/tex]

or

C''(3) = [tex]+\frac{2\times648}{3^3}+2\times12[/tex] > 0

Hence,

x = 3 cm is point of minima

Therefore,

y = [tex]\frac{54}{x^2}[/tex]               [from 1]

or

y = [tex]\frac{54}{3^2}[/tex]

or

y = 6 cm

Hence,

The sides of the container should be 3 cm and height should be 6 cm to minimize the cost

Answer:

6 times we need to transmit the message over this unreliable channel so that with probability 63/64.

Step-by-step explanation:

Consider the provided information.

Let x is the number of times massage received.

It is given that the probability of successfully is 1/2.

Thus p = 1/2 and q = 1/2

We want the number of times do we need to transmit the message over this unreliable channel so that with probability 63/64 the message is received at least once.

According to the binomial distribution:

[tex]P(X=x)=\frac{n!}{r!(n-r)!}p^rq^{n-r}[/tex]

We want message is received at least once. This can be written as:

[tex]P(X\geq 1)=1-P(x=0)[/tex]

The probability of at least once is given as 63/64 we need to find the number of times we need to send the massage.

[tex]\frac{63}{64}=1-\frac{n!}{0!(n-0)!}\frac{1}{2}^0\frac{1}{2}^{n-0}[/tex]

[tex]\frac{63}{64}=1-\frac{n!}{n!}\frac{1}{2}^{n}[/tex]

[tex]\frac{63}{64}=1-\frac{1}{2}^{n}[/tex]

[tex]\frac{1}{2}^{n}=1-\frac{63}{64}[/tex]

[tex]\frac{1}{2}^{n}=\frac{1}{64}[/tex]

By comparing the value number we find that the value of n should be 6.

Hence, 6 times we need to transmit the message over this unreliable channel so that with probability 63/64.

Answer:

0.08x + 0.09y = 1700,     where x is the amount that has 8% interest

                                               y is the amount that has 9% interest

                              -------equation (1)

x+y = 19000

x = 19000 - y ,         -------equation   (2)

substitute (2) into (1):

0.08(19000 - y) + 0.09y = 1700

1520-0.08y+0.09y = 1700

so, y = 18000

x = 19000-y=1000

(1000,   18000) is the answer    

Just to let you know I’m in 7th grade

The problem can be solved by creating a system of linear equations that represent the conditions given in the problem. Solve this system of equations to determine the amounts invested in each of the two accounts.

This is a problem of algebra, specifically a type of word problem called a system of linear equations. You have a total of $19,000 invested across two accounts. Let's denote the amount invested at 5% as x and the amount invested at 9% as y. So, we can write two equations:

The next step is to solve this system of equations. Multiply the first equation by 0.05 (so it matches 5% in the second equation), then subtract the resulting equation from the second equation to eliminate x. By doing this, you find the value for y (money in the 9% account). Once you have the value for y, you can substitute it into the first equation to find the value for x (money in the 5% account).

https://brainly.com/question/33609849

#SPJ11

The distance from earth to sun is 91,673,351.94 miles  

Given that the earth, moon, and sun form a right triangle shown in figure

The figure is attached below

Point A represents moon

Point B represents sun

Point C represents earth

The right angle is the angle at the moon. This is represented by point A in right angle triangle

Given that angle at earth is 89.85 degree

Angle BCA = 89.85 degree

Distance between earth and moon = AC = 240,000 miles

Since, Right angled triangle is formed we can use trigonometric identities

[tex]\cos \theta=\frac{B a s e}{\text {Hypotenuse}}[/tex]

The angle θ means angle created on earth

i.e. angle BCA = θ = 89.85 degree

[tex]\cos \left(89.85^{\circ}\right)=\frac{A C}{B C}[/tex]

Let BC = d, which is the distance between Earth and the sun

[tex]\mathrm{D}=\frac{240,000}{\cos (89.85)}=91,673,351.94 \mathrm{miles}[/tex]

So the distance from earth to sun is 91,673,351.94 miles  

To estimate the distance from the Earth to the Sun when the Moon is exactly half full and the Earth, Moon, and Sun form a right triangle, we can use the concept of similar triangles. By setting up a proportion and using the tangent function, we can calculate that the estimated distance is approximately 243,480,000 miles.

To estimate the distance from the Earth to the Sun, we can use the concept of similar triangles. When the Moon is exactly half full, the Earth, Moon, and Sun form a right triangle, with the right angle at the Moon. We know that the angle formed by the Sun, Earth, and Moon is 89.85 degrees.

We can use this information to set up a proportion between the distance from the Earth to the Moon and the distance from the Earth to the Sun. Let x be the distance from the Earth to the Sun. We can use the tangent function to find x:

Tan(89.85) = (240,000 miles) / x

Solving for x, we get:

x = (240,000 miles) / Tan(89.85)

Using a calculator, we find that x is approximately 243,480,000 miles.

Answer:

a) [tex]df_{num}=k-1=5-1=4[/tex]

b) [tex]df_{den}=N-k=77-5=72[/tex]

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]

And we have this property

[tex]SST=SS_{between}+SS_{within}[/tex]

The degrees for the numerator are [tex]df_{num}=k-1=5-1=4[/tex], where k represent the number of groups on this case k =5.

The degrees for the denominator are [tex]df_{den}=N-k=77-5=72[/tex], where N represent the total number of people N=17+15+16+18+11=77.

And the total degrees of freedom are given by: [tex]df_{tot}=df_{num}+df_{den}=N-1=4+72=76[/tex]

Final answer:

The degrees of freedom for the F-statistic in a one-way ANOVA with five populations and sample sizes of 17, 15, 16, 18, and 11 are 4 and 72.

Explanation:

The degrees of freedom for the F-statistic in a one-way ANOVA are determined by the sample sizes of the populations being compared. In this case, the sample sizes are 17, 15, 16, 18, and 11.

The degrees of freedom for the numerator (df(num)) is equal to the number of groups being compared minus 1, which is 5-1=4.

The degrees of freedom for the denominator (df(denom)) is equal to the total number of observations minus the number of groups, which is 77-5=72. Therefore, the degrees of freedom for the F-statistic are 4 and 72.

Answer:

(A) 0.4196

(B) 0.2398

(C) 0.0020

Step-by-step explanation:

Given,

Total songs = 15,

Liked songs = 6,

So, not liked songs = 15 - 6 = 9

If any 5 songs are played,

Then the total number of ways =  [tex]^{15}C_5[/tex]

(A) Number of ways of choosing 2 liked songs = [tex]^6C_2\times ^9C_3[/tex]

Since,

[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]

Thus, the probability of choosing 3 females and 2 males = [tex]\frac{ ^6C_2\times ^9C_3}{^{15}C_5}[/tex]

[tex]=\frac{\frac{6!}{2!4!}\times \frac{9!}{3!6!}}{\frac{15!}{10!5!}}[/tex]

= 0.4196

Similarly,

(B)

The probability of choosing 3 liked songs = [tex]\frac{ ^6C_3\times ^9C_2}{^{15}C_5}[/tex]

[tex]=\frac{\frac{6!}{3!3!}\times \frac{9!}{2!7!}}{\frac{15!}{10!5!}}[/tex]

= 0.2398

(C)

The probability of choosing 5 liked songs = [tex]\frac{ ^6C_5\times ^9C_0}{^{15}C_5}[/tex]

[tex]=\frac{\frac{6!}{5!1!}}{\frac{15!}{5!10!}}[/tex]

≈ 0.0020

Answer: the equation is

4x^2 + 4x - 12

Step-by-step explanation:

A quadratic equation is an equation in which the highest power of the unknown is 2.

The general form of a quadratic equation is expressed as

ax^2 + bx + c

Where

a is the leading coefficient

c is a constant

Assuming we want to write the quadratic equation in x, from the information given, the roots which are given are -2 and 1 and the leading coefficient is 4.

Therefore, the linear factors of the quadratic equation will be (x+2) and (x-1)

the equation becomes

(x+2)(x-1)

= x^2 - x +2x - 3

= x^2 + x - 3

Given a leading coefficient of 4, we will multiply the quadratic expression by 4. It becomes

4(x^2 + x - 3)

= 4x^2 + 4x - 12

Final answer:

The quadratic equation with roots -2 and 1 and a leading coefficient of 4 is 4x^2 - 12x + 8 = 0.

Explanation:

To write a quadratic equation with roots -2 and 1 and a leading coefficient of 4, we need to use the fact that if α and β are roots of a quadratic equation, then the equation can be expressed in the form:

a(x - α)(x - β) = 0

Here, α is -2 and β is 1, and the leading coefficient, a, is 4. Therefore, we can write:

4(x + 2)(x - 1) = 0

Multiplying it out, we have:

4(x2 - 1x - 2x + 2) = 0

Combining like terms, we get:

4x2 - 12x + 8 = 0, which is the desired quadratic equation.

The standard deviation of returns is 2.82% by calculating expected value, variance and standard deviation.

Given that:

State of Economy Probability Rate of Return if State Occurs

Recession 0.22                         -0.9

Normal         0.47                                 0.105

Boom         0.31                                 0.215

To calculate the standard deviation of returns, by  using the formula:

Standard Deviation = [tex]\sqrt{}[/tex][Σ (Probability * (Rate of Return - Expected Return[tex])^2[/tex])]

Step 1: calculate the Expected Return: Multiply each rate of return by its corresponding probability and sum the results:

E = Σ(Rate of Return* Probability )

Plugging the given data gives:

E = (0.22 * -0.9) + (0.47 * 0.105) + (0.31 * 0.215)

On multiplication gives

E = 0.0225 - 0.04935 + 0.06665

On solving gives:

E = 0.0398.

Step 2: Calculate the Variance: For each rate of return, subtract the expected return, square the result, multiply by its corresponding probability, and sum the results:

Variance ( V) = Σ (Probability * (Rate of Return - Expected Return[tex])^2[/tex]

Plugging the given data gives:

[tex]V = [(0.22 * (-0.9 - 0.0398)^2) + (0.47 * (0.105 - 0.0398)^2) + (0.31 * (0.215 - 0.0398)^2)][/tex]

On squaring and solving gives:

[tex]V = [0.0225 + 0.02741 + 0.0412][/tex]

On adding gives:

[tex]V = 0.09111.[/tex]

Step3: Find the square root of the variance to get the Standard Deviation:

[tex]\sigma = \sqrt V[/tex]

Plugging the given data gives:

[tex]\sigma = \sqrt{0.09111 }[/tex]

On taking square root gives:

[tex]\sigma \approx 0.282[/tex] or 2.82%.

Learn more about standard deviation here:

https://brainly.com/question/13498201

SPJ4

Question:

Based on the following information, what is the standard deviation of returns?

State of Economy Probability Rate of Return if State Occurs

Recession 0.22                         -0.9

Normal         0.47                                 0.105

Boom         0.31                                 0.215

To calculate the standard deviation of returns, follow these steps: 1) Calculate the expected return for each state by multiplying the probability of the state occurring by the rate of return. 2) Calculate the squared difference between each state's return and the expected return. 3) Multiply each squared difference by the probability of the state occurring. 4) Sum up all the values obtained in step 3. 5) Take the square root of the sum obtained in step 4.

To calculate the standard deviation of returns, we need to follow these steps:

Using the given information:

Next, we calculate the squared difference between each state's return and the expected return:

Then, we multiply each squared difference by the probability of the state occurring:

Finally, we sum up all the values obtained:

And take the square root of the sum:

Therefore, the standard deviation of returns is approximately 0.0969.

Answer:

a)[tex]2 + 3 + 5 + 6 = 16[/tex]

b)[tex]8a + b-9[/tex]

c)[tex]8a - 28b + 15c[/tex]

Step-by-step explanation:

We are given the following in the question:

1. Equivalent expression

[tex]2 + 3 + 5 + 6[/tex]

Since all the terms are like terms, we can write,

[tex]2 + 3 + 5 + 6 = 16[/tex]

2. Sum of given expression

[tex](8a - 2b - 4) \text{ and } (3b - 5)\\ (8a - 2b - 4) + (3b -5)\\= 8a -2b -4 + 3b -5\\\text{Collecting the like terms}\\8a + (-2b+3b) + (-4-5)\\= 8a + b-9[/tex]

3. Standard form of the expression

[tex]4(2a) + 7(-4b) + (3 \times c \times 5)\\=(8a) + (-28b) + (15c)\\=8a - 28b + 15c[/tex]

which is the required standard form of the given expression.

Answer:

0.622

Step-by-step explanation:

According to the question,

62.2% of the respondents from Hawaii said that they had exercised more than 30 minutes a day for 3 days out of the week.

So, value of the sample proportion of people from Hawaii who exercised for at least 30 minutes a day for 3 days in a week is given by,

[tex]\frac {62.2}{100}[/tex]

= 0.622  

[tex]f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}[/tex]

[tex]f[/tex] has critical points where the derivative is 0:

[tex]2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1[/tex]

The second derivative is

[tex]f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}[/tex]

and [tex]f''(-1)=\frac2e>0[/tex], which indicates a local minimum at [tex]x=-1[/tex] with a value of [tex]f(-1)=\frac1e[/tex].

At the endpoints of [-2, 2], we have [tex]f(-2)=1[/tex] and [tex]f(2)=e^8[/tex], so that [tex]f[/tex] has an absolute minimum of [tex]\frac1e[/tex] and an absolute maximum of [tex]e^8[/tex] on [-2, 2].

So we have

[tex]\dfrac1e\le f(x)\le e^8[/tex]

[tex]\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx[/tex]

[tex]\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}[/tex]

Answer:

(0,1) and (13,2)

Step-by-step explanation:

We are given that parametric equations

[tex]x=2t^3+3t^2-12t[/tex]

[tex]y=2t^3+3t^2+1[/tex]

a.We have to find the points where tangent line is horizontal.

Differentiate x and y w.r.t.time

[tex]\frac{dx}{dt}=6t^2+6t-12[/tex]

[tex]\frac{dy}{dt}=6t^2+6t[/tex]

[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

Substitute the values

[tex]\frac{dy}{dx}=\frac{6t^2+6t}{6t^2+6t-12}[/tex]

We know that when tangent line is horizontal then

[tex]\frac{dy}{dx}=0[/tex]

[tex]\frac{6t^2+6t}{6t^2+6t-12}=0[/tex]

[tex]6t^2+6t=0[/tex]

[tex]t^2+t=0[/tex]

[tex]t(t+1)=0[/tex]

[tex]t=0, t+1=0[/tex]

[tex]t+1=0\implies t=-1[/tex]

Substitute the t=0 then we get

[tex]x=2(0)^3+3(0)^2-12(0)=0[/tex]

[tex]y=2(0)^3+3(0)^2+1=1[/tex]

Substitute t=-1

[tex]x=2(-1)^3+3(-1)^2-12(-1)=-2+3+12=13[/tex]

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

The points  (0,1) and (13,2) where tangent line is horizontal .

To find the points where the tangent line is horizontal, we need to find the values of t that make the derivative of y with respect to x equal to 0. The points where the tangent line is horizontal are (0,1) and (20,0).

To find the points where the tangent line is horizontal, we need to find the values of t that make the derivative of y with respect to x equal to 0. Let's start by finding the derivatives of x and y with respect to t:

[d/dt(x)]=(6t^2)+(6t)-12

[d/dt(y)]=(6t^2)+(6t)

To find the values of t that make [d/dt(y)] equal to 0, we set (6t^2)+(6t)=0. By factoring out a 6t, we get t(t+1)=0. So t=0 or t=-1. We can substitute these values of t back into the parametric equations to find the corresponding points (x,y):

Therefore, the points (x,y) where the tangent line is horizontal are (0,1) and (20,0).

https://brainly.com/question/34884506

#SPJ3

The tetrahedron passes through the intercepts (5, 0, 0), (0, 2, 0), and (0, 0, 10). Parameterize the surface (call it [tex]\Sigma[/tex]) by

[tex]\vec r(u,v)=(1-v)\langle5,0,0\rangle+v\left((1-u)\langle0,2,0\rangle+u\langle0,0,10\rangle\right)[/tex]

[tex]\vec r(u,v)=\langle5(1-v),2(1-u)v,10uv\rangle[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]\Sigma[/tex] to be

[tex]\vec r_v\times\vec r_u=\langle20v,50v,10v\rangle[/tex]

Then the flux of [tex]\vec F(x,y,z)=\langle x,y,z\rangle[/tex] across [tex]\Sigma[/tex] is

[tex]\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\langle5(1-v),2(1-u)v,10uv\rangle\cdot\langle20v,50v,10v\rangle\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\int_0^1\int_0^1100v\,\mathrm du\,\mathrm dv=\boxed{50}[/tex]

Answer:

[tex]W(h)=0.0006h^3-20[/tex].

A Person who is 5 feet, 8 inches tall weighs about 168.7 lbs.

Step-by-step explanation:

We know the rate of change of weight W​ (in pounds) with respect to height h​ (in inches)

[tex]\frac{dW}{dh}=0.0018h^2[/tex]

This is a separable equation. A separable equation is a first-order differential equation in which the expression for [tex]\frac{dy}{dx}[/tex] can be factored as a function of x times a function of y. In other words, it can be written in the form

[tex]\frac{dy}{dx}=g(x)f(y)[/tex]

[tex]\frac{dW}{dh}=0.0018h^2\\\\dW=(0.0018h^2)dh\\\\\int dW=\int (0.0018h^2)dh\\\\W=0.0006h^3+C[/tex]

To find the value of C, we use W(80​) = 287.2 lbs

[tex]287.2=0.0006(80)^3+C\\0.0006\left(80\right)^3+C=287.2\\307.2+C=287.2\\307.2+C-307.2=287.2-307.2\\C=-20[/tex]

Thus,

[tex]W(h)=0.0006h^3-20[/tex]

Convert the 5 feet into inches

[tex]5 \:ft \:\frac{12 \:in}{1\:ft} = 60 \:in[/tex]

Add 60 in and 8 in, to find the total height of the person

h = 68 in

Substitute h = 68 in into [tex]W(h)=0.0006h^3-20[/tex] to find the weight:[tex]W(68)=0.0006(68)^3-20=168.7[/tex]

A Person who is 5 feet, 8 inches tall weighs about 168.7 lbs.

To find the weight function W(h) and the weight of a person who is 5 feet, 8 inches tall, we need to integrate the given rate of change formula and substitute the values into the equation.

To find the weight function W(h), we need to integrate the given rate of change formula dW/dh = 0.0018[tex]h^2[/tex].

Integrating both sides, we get:

∫ dW = ∫ 0.0018[tex]h^2[/tex] dh

Integrating, we have:

W(h) = 0.0018 * (1/3) * [tex]h^3[/tex] + C

To find the value of C, we use the given information W(80) = 287.2 pounds.

Substituting the values, we have:

287.2 = 0.0018 * (1/3) * [tex]80^3[/tex] + C

Simplifying the equation, we solve for C and find:

C = 287.2 - 0.0018 * (1/3) * [tex]80^3[/tex]

Now, we can write the weight function W(h) as:

W(h) = 0.0018 * (1/3) * [tex]h^3[/tex] + (287.2 - 0.0018 * (1/3) * [tex]80^3[/tex])

To find the weight of a person who is 5 feet, 8 inches tall, we need to convert the height to inches.

5 feet is equal to 60 inches, and 8 inches is 8 inches. So, the total height is 60 + 8 = 68 inches.

Substituting the value into the weight function, we have:

W(68) = 0.0018 * (1/3) * [tex]68^3[/tex] + (287.2 - 0.0018 * (1/3) * [tex]80^3[/tex])

Simplifying the equation, we find the weight of a person who is 5 feet, 8 inches tall.

https://brainly.com/question/38986050

#SPJ3

Final answer:

It would take roughly 22.22 hours to fill an 80,000 L swimming pool with a garden hose at a rate of 60 L/min. However, if diverting a moderate size river flowing at 5000 m³/s, it would fill the pool in approximately 0.016 seconds.

Explanation:

To estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min, we simply divide the total volume of the pool by the flow rate of the hose. For this calculation:

Time (min) = Total Volume (L) / Flow Rate (L/min)

Time = 80,000 L / 60 L/min = 1333.33 min

To convert minutes to hours, we divide by 60:

Time = 1333.33 min / 60 = 22.22 hours

For part (b), if you could divert a moderate size river, flowing at 5000 m³/s, into the pool, we need to convert this flow rate into liters per second (L/s) for consistency:

Flow Rate (L/s) = 5000 m³/s * 1000 L/m³

Flow Rate = 5000000 L/s

Now we can calculate the time it would take to fill the pool with this flow rate:

Time = 80,000 L / 5000000 L/s

Time = 0.016 seconds

Without the exact F-statistic and p-value, it is not possible to definitively choose between rejecting or failing to reject the null hypothesis. The decision would depend on whether the calculated p-value is less than the level of significance (0.10).

To determine whether there is a significant difference in the variance of grading procedures between two professors, we need to conduct an F-test for variance. The null hypothesis (H0) would be that there is no significant difference in the variances of grading, while the alternative hypothesis (H1) would be that there is a significant difference in the variances.

Using the data provided:

The F-statistic is calculated by dividing the variance of the set with the larger variance (squared standard deviation) by the variance of the set with the smaller variance, which means:

F = (22.4)^2 / (12.0)^2

However, the exact F value is not provided in the data, nor is the p-value. To make the decision, we would compare the p-value to the level of significance (
0.10). If the p-value is less than 0.10, we would reject the null hypothesis, indicating that there is a significant difference between the variances. If the p-value is greater than 0.10, we would fail to reject the null hypothesis, indicating there is no significant difference in variances.

Based on the options provided in the question, without the exact p-value or F-statistic, we cannot provide the correct answer. Thus, the answer would depend on the calculated p-value from the F-statistic. The options, therefore, are between A (rejecting the H0) or B (failing to reject the H0), and an exact answer requires additional information from an F-test.

Answer:

[tex]P(\bar X>530)=1-0.808=0.192[/tex]

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the scores, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu=505,\sigma=170)[/tex]  

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

On this case  [tex]\bar X \sim N(505,\frac{170}{\sqrt{35}})[/tex]

2) Calculate the probability

We want this probability:

[tex]P(\bar X>530)=1-P(\bar X<530)[/tex]

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]P(\bar X >530)=1-P(Z<\frac{530-505}{\frac{170}{\sqrt{35}}})=1-P(Z<0.87)[/tex]

[tex]P(\bar X>530)=1-0.808=0.192[/tex]

Using the central limit theorem, the probability of a sample mean of 530 or more is calculated considering the distribution of the sample mean, which is approximately normal due to the sample size.

To calculate the probability that a sample of 35 exams will have a mean score of 530 or more, when the population mean is 505 with a standard deviation of 170, we use the concept of the sampling distribution of the sample mean. Since the sample size is 35, which is less than 10% of all psychology majors (assuming they number in the thousands), and the population distribution is normal or the sample size is large enough (which it is in this case), we can apply the Central Limit Theorem.

This theorem states that the sampling distribution of the sample mean will be approximately normally distributed with a mean equal to the population mean (μ = 505) and a standard deviation equal to σ/√n, where σ is the population standard deviation and n is the sample size (which is √35 in this case).

Answer: The 95% confidence interval for the mean of x is (94.08, 101.92) .

Step-by-step explanation:

We are given that ,

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

i.e. [tex]\sigma= 12[/tex]

Also, it is given that , Sample mean [tex]\overline{x}=98[/tex] having sample size : n= 36

For 95% confidence ,

Significance level : [tex]\alpha=1-0.95=0.05[/tex]

By using the z-value table , the two-tailed critical value for 95% Confidence interval :

[tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]

We know that the confidence interval for unknown population mean[tex](\mu)[/tex] is given by :-

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]\overline{x}[/tex] = Sample mean

[tex]\sigma[/tex] = Population standard deviation

[tex]z_{\alpha/2}[/tex] = Critical z-value.

Substitute all the given values, then the required confidence interval will be :

[tex]98\pm (1.96)\dfrac{12}{\sqrt{36}}[/tex]

[tex]=98\pm (1.96)\dfrac{12}{6}[/tex]

[tex]=98\pm (1.96)(2)[/tex]

[tex]=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)[/tex]

Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .

Answer:

Step-by-step explanation:

Hello!

You want to test two samples of batteries to see is the mean voltage of these battery types are different.

Sample 1 (type K)

n₁= 37

sample mean x₁[bar]= 8.54

standard deviation S₁= 0.225

Sample 2 (Type Q)

n₂= 58

sample mean x₂[bar]= 8.69

standard deviation S₂= 0.725

1. The test hypothesis are:

H₀: μ₁ = μ₂

H₁: μ₁ ≠ μ₂

2. I'll apply the Central Limit Theorem and approximate the distribution of the sample means to normal so that I can use an approximate Z statistic for this test.

Z: (x₁[bar] - x₂[bar]) - (μ₁ - μ₂) ≈ N(0;1)

        √ (S₁²/n₁) + (S₂²/n₂)

[tex]Z_{H0}[/tex]= (8.54 - 8.69) / [√ (0.225²/37) + (0.725²/58)]

[tex]Z_{H0}[/tex]= -1.468 ≅ -1.47

3. This is a two tailed test, so you'll have two critical values

[tex]Z_{\alpha/2} = Z_{0.025} = - 1.96[/tex]

[tex]Z_{1 - \alpha/2} = Z_{0.975} = 1.96[/tex]

You'll reject the null hypothesis if [tex]Z_{H0}[/tex] ≤ -1.96 or if [tex]Z_{H0}[/tex] ≥ 1.96

You'll not reject the null hypothesis if -1.96 < [tex]Z_{H0}[/tex] < 1.96

4.

Since the value [tex]Z_{H0}[/tex] = -1.47 is in the acceptance region, the decision is to not reject the null hypothesis.

I hope it helps!

Answer:

Step-by-step explanation:

Answer:

z= 5.08

The p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .

Step-by-step explanation:

1) Data given and notation

n=750 represent the random sample taken

X=589 represent the adults that said that it is morally wrong to not report all income on tax returns

[tex]\hat p=\frac{589}{750}=0.785[/tex] estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

[tex]p_o=0.7[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

pv represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:  

Null hypothesis:[tex]p=0.7[/tex]  

Alternative hypothesis:[tex]p \neq 0.7[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.785 -0.7}{\sqrt{\frac{0.7(1-0.7)}{750}}}=5.08[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z>5.08)=0.000000377[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .  

- Make an equation representing the number of vehicles needed.

We have six drivers so

x + y ≤ 6

That's not really an equation; it's an inequality.  We want to use all our drivers so we can use the small vans, so

x + y = 6

- Make an equation representing the total number of seats in vehicles for the orchestra members.

s = 25x + 12y

That's how many seats total; it has to be at least 111 so again an inequality,

25x + 12y ≥ 111

We solve it like a system of equations.  

x + y = 6

y = 6 - x

111 = 25x + 12y = 25x + 12(6-x)

111 = 25x + 72 - 12x

111 - 72 = 13 x

39 = 13 x

x = 3

Look at that,  it worked out exactly.  It didn't have to.

y = 6 - x = 3

Answer: 3 buses, 3 vans

Using a System of equations, the number of buses and vans required are 3 respectively.

Using the system of equations :

Total number of buses :

Total number of passengers :

From (1)

Substitute (3) into (2)

25(6-v) + 12v = 111

150 - 25v + 12v = 111

-13v = 111 - 150

-13v = - 39

v = 39/13

v = 3

From (3)

b = 6 - 3

b = 3

Hence, the number of vans and buses required are 3 and 3 respectively.

Learn more : https://brainly.com/question/16144029

Answer:

Step-by-step explanation:

Demand = 19,500 units per year (D)

Ordering cost = $25 per order (O)

Holding cost = $4 per unit per year (C)

a) [tex]EOQ=\sqrt{ \frac{2\times D\times O}{C}}[/tex]

   [tex]EOQ=\sqrt{ \frac{2\times 19,500\times 25}{4}}[/tex]

   = [tex]\sqrt{\frac{975000}{4}}[/tex]

   = [tex]\sqrt{243,750}[/tex]

   = 493.71044 ≈ 494

b) Annual holding cost = [tex]4\times(\frac{Q}{2})[/tex]

                                      = [tex]4\times(\frac{494}{2})[/tex]

                                      = 4 × 247

                                      = 988

c) Annual ordering cost = [tex]O\times(\frac{D}{Q} )[/tex]

                                       = [tex]25\times(\frac{19,500}{494} )[/tex]

                                       = 25 × 39.47

                                       = 986.75

AOQ = 494

Annual holding cost = 988

Annual ordering cost = 986.75

There are 125,970 ways to select the players who will travel to an away game. There are 79,833,600 ways to select the starting line-up. With a restriction on the center position, there are 12 ways to select the starting line-up.

(a) The coach must select 12 players to travel to an away game:

There are 20 members on the basketball team, and the coach needs to select 12 players to travel. This is a combination problem, since the order doesn't matter. The formula for calculating combinations is: nCr = n! / (r! * (n-r)!), where n is the total number of players and r is the number of players to be selected.

Using this formula, we can calculate the number of ways to select the traveling players: 20C12 = 20! / (12! * (20-12)!) = 125,970 ways.

(b) The coach must select her starting line-up:

There are 12 players who will travel, and the coach needs to select 5 players for the starting line-up. This is a permutation problem, since the order does matter. The formula for calculating permutations is: nPr = n! / (n-r)!, where n is the total number of players and r is the number of players to be selected.

Using this formula, we can calculate the number of ways to select the starting line-up: 12P5 = 12! / (12-5)! = 12! / 7! =  79,833,600 ways.

(c) The coach must select her starting line-up, with a restriction on the center position:

There are 12 players who will travel, and only 3 players who can play center. The remaining 4 positions have no restrictions. We can calculate the number of ways to select the starting line-up by first selecting the center player (3 ways), and then selecting the players for the other positions (4P4 = 4 ways).

Therefore, the total number of ways to select the starting line-up with the restriction on the center position is: 3 * 4 = 12 ways.

Answer:

Step-by-step explanation:

The given quadratic equation is

2x^2+3x-8 = 0

To find the roots of the equation. We will apply the general formula for quadratic equations

x = -b ± √b^2 - 4ac]/2a

from the equation,

a = 2

b = 3

c = -8

It becomes

x = [- 3 ± √3^2 - 4(2 × -8)]/2×2

x = - 3 ± √9 - 4(- 16)]/2×2

x = [- 3 ± √9 + 64]/2×2

x = [- 3 ± √73]/4

x = [- 3 ± 8.544]/4

x = (-3 + 8.544) /4 or x = (-3 - 8.544) / 4

x = 5.544/4 or - 11.544/4

x = 1.386 or x = - 2.886

The positive solution is 1.39 rounded up to the nearest hundredth

Answer:

its a. 1.39

Step-by-step explanation:

Answer:

195 mm

Step-by-step explanation:

There are 10 mm in one cm. This is due to metric prefix rules

Answer:

A. AAS

Step-by-step explanation:

The vertical angles are congruent, the adjacent marked angles are congruent, and the marked sides on the far side of both angles are congruent. This geometry matches the conditions for Angle-Angle-Side (AAS) congruence.

The decision rule for rejecting the null hypothesis H₀ is

if t-statistic < -2.093.

First , let's state the null and alternative hypotheses:

Null hypothesis (H₀): The population mean weight of banana chips is 447 grams (the machine works correctly).

Alternative hypothesis (H₁): The population mean weight of banana chips is less than 447 grams (the machine is underfilling).

α = 0.025

Since this is a one-tailed test (we only care if the weight is less than 447 grams), we need to find the t-value for a single-tailed test with α = 0.025 and degrees of freedom (df) = n - 1 = 19 - 1 = 18.

We can use a t-distribution table or a statistical calculator to find this value. In most calculators, you can use the TINV function with df = 18, tails = 1, and p = 0.025. The critical value (t_α) is approximately -2.093.

Reject the null hypothesis (H₀) if the calculated t-statistic (t_stat) is less than the critical value (t_α). In other words, if the sample mean weight is significantly lower than 447 grams, we have evidence to suggest the machine is underfilling.

Reject H₀ if t_stat < -2.093.

This will require  sample data (the weights of the 19 bags). Once we have that, we can calculate the t-statistic using the formula:

t_stat = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size)).

The complete question is : A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 447 gram setting. It is believed that the machine is underfilling the bags. A 19 bag sample had a mean of 443 grams with a standard deviation of 21. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer: We can conclude that the average containers are mislabelled.

Step-by-step explanation:

From the question, the significant level of 0.01 or 1% given is less than 0.05 or 5%. So we say it is significant.

Answer:

H ₀ : p ≤ 0.25

H ₁: p > 0.25

Test statics value  z = 12.91

Decision Rule:

Reject the value of null hypothesis

Conclusion:

If survey is online then there is sufficient evidence to claim . If the sample is  voluntary response sample then conclusion is not valid.

Step-by-step explanation:

step by step explanation is shown in attachment.