The lines shown below are perpendicular. If the green line has a slope of a
what is the slope of the red line?
O A 22
OB.

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▹ Answer

0.25 = 1/4 because 25/100 = 1/4

▹ Step-by-Step Explanation

0.25 to a fraction → 25/100

25/100 = 1/4

Therefore, this statement is true. (0.25 = 1/4 because 25/100 = 1/4)

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

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Answer:

The answer is 0.4

Step-by-step explanation:

2/5 is equal to 0.4

Answer:

0.4

Step-by-step explanation:

start with 2/5.

multipy the 5 by 20 to get 100, and the 2 by 20 to get 40.

so, now your fraction should look like this:

40/100

then, shift the decimal in 40 over 2 spots to get 0.40, or 0.4

hope this helps :)

Answer:

Quantity that will maximize profit=1000

Step-by-step explanation:

Assume quantity=x

Revenue=price*quantity

=(4200-6x)x

=4200x-6x^2

Marginal revenue(MR) =4200-12x

Cost(x)= 16000 + 600x − 1.8x2 + 0.004x3

Marginal cost(MC) =600-3.6x+0.012x^2

Marginal cost=Marginal revenue

600-3.6x+0.012x^2=4200-12x

600-3.6x+0.012x^2-4200+12x=0

0.012x^2-8.4x-3600=0

Solve the quadratic equation using

x= -b +or- √b^2-4ac/2a

a=0.012

b=-8.4

c=-3600

x=-(-8.4) +or- √(-8.4)^2- (4)(0.012)(-3600) / (2)(0.012)

= 8.4 +or- √70.56-(-172.8) / 0.024

= 8.4 +or- √70.56+172.8 / 0.024

= 8.4 +or- √243.36 / 0.024

= 8.4 +or- 15.6/0.024

= 8.4/0.024 +15.6/0.024

= 350+650

x=1000

OR

= 8.4/0.024 -15.6/0.024

= 350 - 650

= -300

x=1000 or -300

Quantity that maximises profits can not be negative

So, quantity that maximises profits=1000

Answer:

[tex]\frac{n+300}{2n}[/tex] is the correct answer.

Step-by-step explanation:

Given that total number of people = n

Let the number of people who responded to the survey = x

Let the number of people who did not respond to the survey = y

[tex]x+y=n ...... (1)[/tex]

As per question statement:

The number of people who did not respond to the survey was 300 less than the number of people  who did respond.

i.e. [tex]x-y =300[/tex] ...... (2).

We need to solve the equations (1) and (2).

Adding (1) and (2):

[tex]2x=n+300\\\Rightarrow x = \dfrac{n+300}{2}[/tex]

The fraction of people who received the survey did respond:

[tex]\dfrac{\text{Number of people who responded}}{\text{Total number of people who received the survey}}\\[/tex]

So, the answer is:

[tex]\dfrac{x}{n}[/tex]

Putting the values of x, we get the answer as:

[tex]\dfrac{n+300}{2n}[/tex]

Answer:

y=6x+31

Step-by-step explanation:

Since we are given a point and a slope, we can use the slope-intercept formula.

[tex]y-y_{1} =m(x-x_{1})[/tex]

where (x1,y1) is a point on the line and m is the slope.

The point given is (-6,-5) and the slope is 6.

x1= -6

y1= -5

m=6

[tex]y--5=6(x--6)[/tex]

A negative number subtracted from another number, or two negative signs, becomes a positive.

[tex]y+5=6(x+6)[/tex]

We want to find the equation of the line, which is y=mx+b (m is the slope and b is the y-intercept). Therefore, we must get y by itself on one side of the equation.

First, distribute the 6. Multiply each term inside the parentheses by 6.

[tex]y+5=(6*x)+(6*6)[/tex]

[tex]y+5=6x+36[/tex]

Subtract 5 from both sides, because it is being added on to y.

[tex]y+5-5=6x+36-5[/tex]

[tex]y=6x+36-5[/tex]

[tex]y=6x+31[/tex]

The equation of the line is y=6x+31

Answer:

1²/33

Step-by-step explanation:

3/11×35/9

=35/33

=1²/33

Fractions are written as a ratio of two integers. The simplified form of the expression is 35/33

Fractions are written as a ratio of two integers. Given the expression below;

3/11 of 35/9

Of means multiplication, hence;

3/11 of 35/9 = 3/11 * 35/9

Take the product

3/11 * 35/9 = 105/99

Divide through  by 3

105/99 = 35/33

Hence the simplified form of the expression is 35/33

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Answer:

The area of the remaining board is (x² - 289) sq. ft.

Step-by-step explanation:

Let the sides of the bigger square board be, x feet.

It is provided that a smaller square of side length 17 feet is cut out of the bigger square board.

The area of a square is:

[tex]Area=(side)^{2}[/tex]

Compute the area of the bigger square board as follows:

[tex]A_{b}=(side_{b})^{2}=x^{2}[/tex]

Compute the area of the smaller square board as follows:

[tex]A_{s}=(side_{s})^{2}=(17)^{2}=289[/tex]

Compute the area of the remaining board in square feet as follows:

[tex]\text{Remaining Area}=A_{b}-A_{s}[/tex]

                          [tex]=[x^{2}-289]\ \text{square ft.}[/tex]

Thus, the area of the remaining board is (x² - 289) sq. ft.

Answer:

d. The mean is 43.6°F, and the median is 43.5°F.

Step-by-step explanation:

Hello!

The data corresponds to the low temperatures in Phoenix recorded for April to November.

April: 32ºF

May: 40ºF

June: 50ºF

July: 61ºF

August: 60ºF

September: 47ºF

October: 34ºF

November: 25ºF

Sample size: n= 8 months

The mean or average temperature of the low temperatures in Phoenix can be calculated as:

[tex]\frac{}{X}[/tex]= ∑X/n= (32+40+50+61+60+47+34+25)/8= 43.625ºF (≅ 43.6ºF)

The Median (Me) is the value that separates the data set in two halves, first you have to calculate its position:

PosMe= (n+1)/2= (8+1)/2= 4.5

The value that separates the sample in halves is between the 4th and the 5th observations, so first you have to order the data from least to greatest:

25; 32; 34; 40; 47; 50; 60; 61

The Median is between 40 and 47 ºF, so you have to calculate the average between these two values:

[tex]Me= \frac{(40+47)}{2} = 43.5[/tex] ºF

The correct option is D.

I hope this helps!

Answer:

it is d

Step-by-step explanation:

Answer:

The solution to the given Initial - Value - Problem is [tex]y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - [\frac{-1}{9} + \frac{1}{3}t - \frac{2}{9}e^{-3(t-1)}]u(t-1)[/tex]

Step-by-step explanation:

y' + 3y = f(t).................(1)

f(t) = t      when 0 ≤ t < 1

f(t) = 0     when t ≥ 1

Step 1: Take the Laplace transform of the LHS of equation (1)

That is L(y' + 3y) = sY(s) + 3Y(s) = Y(s)[s + 3]..............(*)

Step 2: Get an expression for f(t)

For f(t) = t      when 0 ≤ t < 1

f₁(t) = t (1 - u(t - 1)) ( there is a time shift of the unit step)

For f(t) = 0     when t ≥ 1

f₂(t) = 0(u(t-1))

f(t) = f₁(t) + f₂(t)

f(t) = t - t u(t-1)................(2)

Step 3: Taking the Laplace transform of equation (2)

[tex]F(s) = \frac{1}{s^2} - e^{-s} ( \frac{1}{s^2} + \frac{1}{s})[/tex]...............(**)

Step 4: Equating * and **

[tex]Y(s) [s + 3]=\frac{1}{s^2} - e^{-s} ( \frac{1}{s^2} + \frac{1}{s}) \\Y(s) = \frac{1}{s^2(s+3)} - e^{-s} ( \frac{1}{s^2(s+3)} + \frac{1}{s(s+3)})[/tex].......................(3)

Since y(t) is the solution we are looking for we need to find the Inverse Laplace Transform of equation (3) by first breaking every  fraction into partial fraction:

[tex]\frac{1}{s^2 (s+3)} = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)}[/tex]

[tex]\frac{1}{s (s+3)} = \frac{1}{3s} + \frac{1}{3(s+3)}[/tex]

We can rewrite equation (3) by representing the fractions by their partial fractions.

[tex]Y(s) = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} - e^{-s} [\frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} + \frac{1}{3s} + \frac{1}{3(s+3)}]\\Y(s) = \frac{-1}{9s} + \frac{1}{3s^2} + \frac{1}{9(s+3)} - e^{-s}[\frac{2}{9s} + \frac{1}{3s^2} - \frac{2}{9(s+3)}][/tex]................(4)

step 5: Take the inverse Laplace transform of equation (4)

[tex]y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - u(t-1)[\frac{2}{9} + \frac{1}{3}(t-1) - \frac{2}{9}e^{-3(t-1)}][/tex]

Simplifying the above equation:

[tex]y(t) = \frac{-1}{9} + \frac{1}{3}t + \frac{1}{9}e^{-3t} - [\frac{-1}{9} + \frac{1}{3}t - \frac{2}{9}e^{-3(t-1)}]u(t-1)[/tex]

The Laplace transform is use to solve the differential equation problem.

The solution for the given initial-value problem is,

[tex]y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)[/tex]

Given:

The given initial value problem is [tex]y' + 3y = f(t)[/tex].

Consider the left hand side of the given equation.

[tex]y'+3y[/tex]

Take the Laplace transform.

[tex]L(y' + 3y) = sY(s) + 3Y(s) \\L(y' + 3y) = Y(s)[s + 3][/tex]

Consider the right hand side and get the expression for [tex]f(t)[/tex].

[tex]f(t) = t[/tex]  when 0 ≤ t < 1

From time shift of the unit step

[tex]f_1(t) = t (1 - u(t - 1))[/tex]

For f(t) = 0     when t ≥ 1

Now,

[tex]f_2(t) = 0(u(t-1))f(t) = f_1(t) + f_2(t)f(t) = t - t u(t-1)[/tex]

Take the Laplace for above expression.

[tex]F(s)=\dfrac{1}{s^2}-e^{-s}\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right)[/tex]

Now, the equate the above two equation.

[tex]Y(s)\left[s+3\right ]=\dfrac{1}{s^2}-e^{-s}\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right)\\Y(s)=\dfrac {1}{(s^2(s+3))}-e^{-s}\left(\dfrac{1}{(s^2(s+3))}+\dfrac{1}{s(s+3)\right)}[/tex]

Find the inverse Laplace for the above equation.

[tex]\dfrac{1}{(s^2(s+3))}=\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}\\\dfrac{1}{(s(s+3))}=\dfrac{1}{3s}+\dfrac{1}{3(s+3)}[/tex]

Calculate the partial fraction of above equation.

[tex]Y(s)=\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}-e^{-s}\left[\dfrac{-1}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}+\dfrac{1}{3s}+\dfrac{1}{3(s+3)}\right]\\Y(s)=\dfrac{2}{9s}+\dfrac{1}{3s^2}+\dfrac{1}{9(s+3)}-e^{-s}\left[\dfrac{2}{9s}+\dfrac{1}{3s^2}-\dfrac{2}{9(s+3)}\right][/tex]

Take the inverse Laplace of the above equation.

[tex]y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)[/tex]

Thus, the solution for the given initial-value problem is,

[tex]y(t)=\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{1}{9}e^-3t-\left[\dfrac{-1}{9}+\dfrac{-1}{3}t+\dfrac{2}{9}e^{-3(t-1)}\right]u(t-1)[/tex]

Learn more about what Laplace transformation is here:

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Answer:

C. G

Step-by-step explanation:

Answer:

It is better for the warriors to use man-to-man defense.

Step-by-step explanation:

The complete question is: The Westwood Warriors basketball team wants to score more points. To get better at scoring points the  team is trying to improve its offensive strategies. Some opponents primarily use a zone defense, while  others primarily use a man-to-man defense. When the Warriors play against teams that use a zone defense they score an average of 67 points per game with a standard deviation of 8 points per game. When they play against teams that use a  man-to-man defense they score an average of 62 points per game with a standard deviation of 5 points per game.

Since the Warriors started using their improved offensive strategies they have played two  games with the following results.

Against the McNeil Mavericks

Maverick defense: zone

Warrior points: 77

Against the Round Rock Dragons

Dragon defense: man-to-man

Warrior points: 71

What is the Z-score of these values?

We are given that when the Warriors play against teams that use a zone defense they score an average of 67 points per game with a standard deviation of 8 points per game. When they play against teams that use a  man-to-man defense they score an average of 62 points per game with a standard deviation of 5 points per game.

We have to find the z-scores.

Let X = points score by warriors when they use zone defense

The z-score probability distribution for the normal distribution is given by;

                            Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = mean score = 67 points

            [tex]\sigma[/tex] = standard deviation = 8 points

It is stated that the Warriors scored 77 points when they used zone defense, so;

   z-score for 77 =  [tex]\frac{X-\mu}{\sigma}[/tex]

                            =  [tex]\frac{77-67}{8}[/tex]  = 1.25

Let X = points score by warriors when they use man-to-man defense

The z-score probability distribution for the normal distribution is given by;

                            Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = mean score = 62 points

            [tex]\sigma[/tex] = standard deviation = 5 points

It is stated that the Warriors scored 71 points when they used man-to-man defense, so;

   z-score for 71 =  [tex]\frac{X-\mu}{\sigma}[/tex]

                            =  [tex]\frac{71-62}{5}[/tex]  = 1.8

So, it is better for the warriors to use man-to-man defense.

Full Question:

The function [tex]y =\sqrt[3]{-x} - 3[/tex] is graphed only over the domain of {x | –8 < x < 8}. what is the range of the graph?

Answer:

[tex]-5 < y < -1[/tex]

Step-by-step explanation:

Given

Function:[tex]y =\sqrt[3]{-x} - 3[/tex]

Range: {x | –8 < x < 8}

Required

Find the range of the graph

To calculate the range of the graph; we simply substitute the value of x (the domain) at both ends to the given function;

In other words, solve for y when x = -8 and when x = 8

To start with;

When x = -8

[tex]y =\sqrt[3]{-x} - 3[/tex]

[tex]y =\sqrt[3]{-(-8)} - 3}[/tex]

[tex]y =\sqrt[3]{8} - 3}[/tex]

[tex]y =2 - 3[/tex]

[tex]y = -1[/tex]

When x = 8

[tex]y =\sqrt[3]{-8} - 3[/tex]

[tex]y =-2 - 3[/tex]

[tex]y = -5[/tex]

Converting both values of y to inequalities

[tex]-5 < y < -1[/tex]

Hence, the range of the graph is [tex]-5 < y < -1[/tex]

Answer:

a)Parametric equations are

X= -10t

Y= -10t and

z= 25+25t

b) Symmetric equations are

(x/-10) = (y/-10) = (z- 25)/25

Step-by-step explanation:

We were told to fin two things here which are ; a) the parametric equations and b) the symmetric equations

The given two points are (0, 0, 25)and (10, 10, 0)

The direction vector from the points (0, 0, 25) and (10, 10, 0)

(a,b,c) =( 0 -10 , 0-10 ,25-0)

= < -10 , -10 ,25>

The direction vector is

(a,b,c) = < -10 , -10 ,25>

The parametric equations passing through the point (X₁,Y₁,Z₁)and parallel to the direction vector (a,b,c) are X= x₁+ at ,y=y₁+by ,z=z₁+ct

Substitute (X₁ ,Y₁ ,Z₁)= (0, 0, 25), and (a,b,c) = < -10 , -10 ,25>

and in parametric equations.

Parametric equations are X= 0-10t

Y= 0-10t and z= 25+25t

Therefore, the Parametric equations are

X= -10t

Y= -10t and

z= 25+25t

b) Symmetric equations:

If the direction numbers image and image are all non zero, then eliminate the parameter image to obtain symmetric equations of the line.

(x-x₁)/a = (y-y₁)/b = (z-z₁)/c

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

Answer:

Step-by-step explanation:

$1,250,000 in current assets (cr) and $500,000 in current liabilities(cl)

current ratio (cr)=1250000/500000=2.5

Δ note payable ( change in note payable NP)

minimum current=2.2

2.2=(1250000+ΔNP)/(500000+ΔNP)

2.2(500000+ΔNP)=1250000+ΔNP

1100000+2.2ΔNP=1250000+ΔNP

2,2ΔNP-ΔNP=1250000-1100000

1.2ΔNP=150000

ΔNP=150000/1.2=125000

Nelson's short-term debt (notes payable) increase without pushing its current ratio below 2.2=125000

assuming this amount used to increase the inventory

new inventory=335000+125000=360000

current asset= 1250000+125000=1375000

the new ratio=(1375000-360000)/500000+125000

new ratio = 1.624

Answer:

0.80+/- 0.016

(0.784, 0.816)

The 95% confidence interval on the proportion of people who regularly have a dental checkup is = (0.784, 0.816)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 2000/2500 = 0.80

Number of samples n = 2500

Confidence interval = 95%

z-value(at 95% confidence) = 1.96

Substituting the values we have;

0.80+/- 1.96√(0.80(1-0.80)/2500)

0.80+/- 1.96√(0.000064)

0.80+/- 0.01568

0.80+/- 0.016

(0.784, 0.816)

The 95% confidence interval on the proportion of people who regularly have a dental checkup is = (0.784, 0.816)

Answer:

B is the answer

Step-by-step explanation:

-3(2w+6)-4

-6w-18-4

-6w-22

Answer:

Step-by-step explanation:

-3(2w + 6) - 4

1. Distribute

 = -3*2w = -6w

 = -3 * 6 = -18

 = -6w -18

2. Simplify like terms

 = -18 - 4

 = -22

3. Place variables and numbers together

 = -6w - 22

Explanation:

2 * -3w = -6w

2*-11 = -22

Place them together and you get the answer!

Answer:

B

Step-by-step explanation:

In order to combine like terms, they must share the same variable. We can't combine things 9y and 3p because they contain two different variables. On the other hand, 7r and r work because there is only one. 7r and r combine to 8r+2

Answer:

the answer is b

Step-by-step explanation:

Answer:

Step-by-step explanation:

Surface Area of the rectangular box = 2(LW+LH+WH)

L is the length of the box

W is the width of the box

H is the height of the box

let dL, dW and dH be the possible error in the dimensions L, W and H respectively.

Since there is a possible error of 0.2cm in each dimension, then dL = dW = dH = 0.2cm

The surface Area of the rectangular box using the differentials is expressed as shown;

S = 2{(LdW+WdL)+(LdH+HdL)+(WdH+HdW)]

Also given L = 96cm W = 58cm and H = 48cm, on substituting this given values and the differential error, we will have;

S = 2{(96*0.2+58*0.2) + (96*0.2+48*0.2)+(58*0.2+48*0.2)}

S = 2{19.2+11.6+19.2+9.6+11.6+9.6}

S = 2(80.8)

S = 161.6 cm²

Hence, the surface area of the box is 161.6 cm²

Answer:

No. At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the bags are underfilled (population mean significantly less than 418 g.)

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=418\\\\H_a:\mu< 418[/tex]

The significance level is 0.1.

The sample has a size n=9.

The sample mean is M=413.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=20.

The estimated standard error of the mean is computed using the formula:

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{20}{\sqrt{9}}=6.6667[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{413-418}{6.6667}=\dfrac{-5}{6.6667}=-0.75[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=9-1=8[/tex]

This test is a left-tailed test, with 8 degrees of freedom and t=-0.75, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t<-0.75)=0.237[/tex]

As the P-value (0.237) is bigger than the significance level (0.1), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.1, there is not enough evidence to support the claim that the bags are underfilled (population mean significantly less than 418 g.)

Answer: $0.30 per dozen.

Step-by-step explanation:

Given: The whole cost of 13 dozen pencils = $9

If 3 dozen broke in transit, remaining dozens of pencils = 13-3 = 10 dozens

Also, Selling price of theses 10 dozens = [tex]\dfrac{1}{3}\times\text{whole cost}[/tex]  [given]

[tex]=\dfrac{1}{3}\times9=\$3[/tex]

Then, the selling price of each dozen = [tex]\dfrac{\$3}{10}=\$0.30[/tex]

Hence, the shopkeeper sells the remaining pencils at $0.30 per dozen.

Answer:

There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute.

Step-by-step explanation:

i have the test

There is a 95% chance that the population mean is between 73.9 and 76.1 beats per minute.

Since

The average heart rate was 75 beats per minute.

The standard deviation is 8 beats per minute

And, there is the study of 205 adults

Now the following formula is to be used

Since

[tex]x \pm z \frac{\sigma}{\sqrt{n} }[/tex]

Here

z = 1.96 at 95% confidence interval

So,

[tex]= 75 \pm 1.96 \frac{8}{\sqrt{205} } \\\\= 75 - 1.96 \frac{8}{\sqrt{205} } , 75 + 1.96 \frac{8}{\sqrt{205} }[/tex]

= 73.9 ,76.1

Hence, the above statement should be true.

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Answer:

True mathematical statement.

"If x = 0, then for any real number y, we have: y*x = 0."

This is true, and we can prove it with the axioms of the real set.

A false mathematical statement can be:

"if n and x are integer numbers, then n/x is also an integer number."

And a counterexample of this is if we took n = 1 and x = 2, both are integer numbers, so the first part is true, but:

n/x = 1/2 = 0.5 is not an integer number, then the statement is false,

Answer:

True mathematical statement.

"If x = 0, then for any real number y, we have: y*x = 0."

This is true, and we can prove it with the axioms of the real set.

A false mathematical statement can be:

"if n and x are integer numbers, then n/x is also an integer number."

And a counterexample of this is if we took n = 1 and x = 2, both are integer numbers, so the first part is true, but:

n/x = 1/2 = 0.5 is not an integer number, then the statement is false,

Step-by-step explanation:

Answer:

$81

Step-by-step explanation:

We have:

20x (20 dollars per ski lesson)

60y (60 dollars per lift pass)

70z (70 dollars for ski lesson and pass)

We are given:

42 lift tickets sold

23 lessons sold

50 people bought 1 option

We need to find how many people bought both the ski lesson and pass:

42 + 23 = 65 things sold total

Only 50 people bought 1 item

65 - 50 = 15 people bought both ski lesson and pass

Now we find the total costs of everything combined:

70(15) + 60(34) + 60(16) = 4050

Mean: 4050/50 people = $81

Answer:

x = -3h

x = -12

Step-by-step explanation:

Given expression is,

[tex]\frac{x}{h}+1=-2[/tex]

By adding 2 on both the sides of the equation,

[tex]\frac{x}{h}+1+2=-2+2[/tex]

[tex]\frac{x}{h}+3=0[/tex]

Now subtract 3 form both the sides,

[tex]\frac{x}{h}+3-3=0-3[/tex]

[tex]\frac{x}{h}=-3[/tex]

Multiply the equation by 'h'

x = -3h

If h = 4,

By substituting h = 4 in the equation,

x = -3(4)

x = -12

Answer:

(a) [tex]V = (-8W^3 + 800W^2)/3[/tex]

(b) [tex]W > 100[/tex]

Step-by-step explanation:

Let's call the length of the box L, the width W and the height H. Then, we can write the following equations:

"A rectangular box has a base that is 4 times as long as it is wide"

[tex]L = 4W[/tex]

"The sum of the height and the girth of the box is 200 feet"

[tex]H + (2W + 2H) = 200[/tex]

[tex]2W + 3H = 200 \rightarrow H = (200 - 2W)/3[/tex]

The volume of the box is given by:

[tex]V = L * W * H[/tex]

Using the L and H values from the equations above, we have:

[tex]V = 4W * W * (200 - 2W)/3[/tex]

[tex]V = (-8W^3 + 800W^2)/3[/tex]

The domain of V(W) is all positive values of W that gives a positive value for the volume (because a negative value for the volume or for the width doesn't make sense).

So to find where V(W) > 0, let's find first when V(W) = 0:

[tex](-8W^3 + 800W^2)/3 = 0[/tex]

[tex]-8W^3 +800W^2 = 0[/tex]

[tex]W^3 -100W^2 = 0[/tex]

[tex]W^2(W -100) = 0[/tex]

The volume is zero when W = 0 or W = 100.

For positive values of W ≤ 100, the term W^2 is positive, but the term (W - 100) is negative, then we would have a negative volume.

For positive values of W > 100, both terms W^2 and (W - 100) would be positive, giving a positive volume.

So the domain of V(W) is W > 100.

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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

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

And p is the probability of X happening.

The true proportion of defects is 0.15

This means that [tex]p = 0.15[/tex]

Sample of 6:

This means that [tex]n = 6[/tex]

What is the probability that there will be 0 or 1 defects in a sample of 6?

[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771[/tex]

[tex]P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993[/tex]

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764[/tex]

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Answer:

[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=5-1=4[/tex]  

The p value wuld be given by:

[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X=46[/tex] represent the mean height for the sample  

[tex]s=5.148[/tex] represent the sample standard deviation

[tex]n=5[/tex] sample size  

[tex]\mu_o =40[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 40[/tex]  

Alternative hypothesis:[tex]\mu \neq 40[/tex]  

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

Replacing we got:

[tex]t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=5-1=4[/tex]  

The p value wuld be given by:

[tex]p_v =2*P(t_{(4)}>2.606)=0.060[/tex]  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Answer:

width of a river = 45m

Step-by-step explanation:

ration and proportion

let x = width of a river

x            20.9 m

------  =    --------

56 m       26 m

x = (20.9 * 56) / 26

x = 45 m

therefore the width of a river is 45 m