TRUE or FALSE: The regression equation is always the best predictor of a y value for a given value of x. Defend your answer.

Answer:

(C) The value of r indicates that the number of sales decreases as the price increases.

ED2021.

The best statement, given the correlation coefficient of -0.63 is: value of r indicates that the number of sales decreases as the price increases.

A negative correlation coefficient has a negative sign, and implies a negative relationship between two variables.

This means that, as one variable decreases, the other variable increases.

Thus, a correlation coefficient of -0.63 shows a negative relationship between prices of smartphones and the number of sales.

Therefore, the best statement, given the correlation coefficient of -0.63 is: value of r indicates that the number of sales decreases as the price increases.

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Answer:

f(x)=15x+5

thanks

Step-by-step explanation:

p=2(length +width)

p=f(x)=2(5x+2+2x)

=f(x)=10x+4+5x

collecting like terms,we get;

f(x)=15x+5

The expression that represents the perimeter P as a function of x is 14x + 14

The perimeter can simply be defined as the part or portion surrounding a shape or boundary. The formula for calculating the perimeter of the rectangular yard is expressed as:

The perimeter of the rectangular yard = 2(L+W) where:

L is the length

W is the width

Given the following parameters

Length = 5x + 2

Width = 2x

Substitute the given parameters into the formula above:

Perimeter = 2(5x+2+2x)

Perimeter = 2(7x + 2)

Perimeter = 14x + 4

Hence the expression that represents the perimeter P as a function of x is 14x + 14

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9514 1404 393

Answer:

  x^2/1 +y^2/81 = 1

Step-by-step explanation:

We know that the equation of a unit circle is ...

  x^2 +y^2 = 1 . . . . . equation of a unit circle

We also know that replacing x with x/a in a function will expand the graph by a factor of 'a'. Similarly, replacing y with y/b will do the same in the vertical direction.

An ellipse is a circle that has had different expansion factors applied along its different axes. Here, the given points tell us the center of the ellipse is (0, 0), and that it has been expanded by a factor of 9 in the y-direction and a factor of 1 in the x-direction This means the equation for it would be ...

  (x/1)^2 +(y/9)^2 = 1 . . . . . equation for desired ellipse

In the required form, this is ...

  [tex]\dfrac{x^2}{1}+\dfrac{y^2}{81}=1[/tex]

Answer:

-3

1 + 4 sqrt( 241 )

1 - 4 sqrt( 241 )

Step-by-step explanation:

We need minus lambda on the entries down the diagonal. I'm going to use m instead of the letter for lambda.

[-43-m 0 80]

[40 -3-m 80]

[24 0 45-m]

Now let's find the determinant

(-43-m)[(-3-m)(45-m)-0(80)]

-0[40(45-m)-80(24)]

+80[40(0)-(-3-m)(24)]

Let's simplify:

(-43-m)[(-3-m)(45-m)]

-0

+80[-(-3-m)(24)]

Continuing:

(-43-m)[(-3-m)(45-m)]

+80[-(-3-m)(24)]

I'm going to factor (-3-m) from both terms:

(-3-m)[(-43-m)(45-m)-80(24)]

Multiply the pair of binomials in the brackets and the other pair of numbers;

(-3-m)[-1935-2m+m^2-1920]

Simplify and reorder expression in brackets:

(-3-m)[m^2-2m-3855]

Set equal to 0 to find the eigenvalues

-3-m=0 gives us m=-3 as one eigenvalue

The other is a quadratic and looks scary because of the big numbers.

I guess I will use quadratic formula and a calculator.

(2 +/- sqrt( (-2)^2 - 4(1)(-3855) )/(2×1)

(2 +/- sqrt( 15424 )/(2)

(2 +/- sqrt( 64 )sqrt( 241 )/(2)

(2 +/- 8 sqrt( 241 )/(2)

1 +/- 4 sqrt( 241 )

Answer:

0.08y - 0.0144

Step-by-step explanation:

We need to solve the below expression i.e.

(y - .18) x .08

It can be done as follows :

Using distributive property to solve it.

(y - .18) x .08 = 0.08(y) - 0.18(0.08)

= 0.08y - 0.0144

So, the equivalent expression is 0.08y - 0.0144.

Answer:

x = 14

Step-by-step explanation:

Vertical angles are equal

5x+2 = 6x-12

Subtract 5x from each side

5x+2-5x = 6x-12-5x

2 = x-12

Add 12 to each side

2+12 = x-12+12

14 =x

Answer:

x = 15.92

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp / adj

tan 53 = x / 12

12 tan 53 = x

x=15.92453

Rounding to the nearest hundredth

x = 15.92

Answer:

15.92

Step-by-step explanation:

Answer:

-7

Step-by-step explanation:

63 divided by -9 = y

63 divided by -9 = -7

Check answers :

-9 x -7 = a

a= 63

Answer:

Perimeter: 3/4

Area: 9/16

Step-by-step explanation:

The ratio of the perimeters is equal to the ratio of the sides so:

18/24 = 3/4

Ratio of Area = (Ratio of Sides)^2

(3/4)^2 = 9/16

I wasn't sure about the answer so I used Gauthmath

Answer:

R x 1= price of 1 locker

Step-by-step explanation:

it would continue the same way. Just multiply R and the number of lockers.

Is there a certain situation it was asking about?

Answer:

dư 0... 7^300 chia 7 đc 7^299 mà

Answer:

3*sqrt(2) and 3pi/4

Step-by-step explanation:

tan(theta)=y/x and r^2=y^+x^2.

tan(theta)=-3/3=-1, theta=3pi/4

r=sqrt(3^2+3^2)=3*sqrt(2)

Answer:

a

Step-by-step explanation:

Statements? At least look for one that says it is equivalent to positive 5.

Answer:

12

Step-by-step explanation:

2²×3

I hope its correct

Answer:

4

[tex] {( - 2 + 1)}^{2} + 5(12 \div 3) - 9 \\ 2 + 1 + 5 \times 4 - 9 \\ 3 + 20 - 9 \\ 23 - 9 \\ 14[/tex]

Answer:

30 new residents

Step-by-step explanation:

Find how many new residents they will have by finding 1.2% of the current population.

2500(0.012)

= 30

So, Hawkville will have 30 new residents

Answer:

A) 220.45%

B) 0%

Step-by-step explanation:

A) 220.45 / 100 = 2.2045 * 100 = 220.45%

B) Costs exceeded her paycheck, so 0% of her pay was leftover and put in the bank.

Answer: A) 22.045% , B) 75.16%

I guess its a typo and weekly salary was $1000

So weekly salary = 1000

Spent on shirt = $220.45

Spent on CD = $12.95

Spent on Gasoline = $15

A) 220.45/1000×100

= 22.045%

B) Total spent = 220.45 + 12.95 + 15

= $248.4

Total left with her = 1000 - 248.4

= $751.6

Total percent of pay she put in bank = 751.6/1000×100

= 75.16%

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Answer:

17.62%

Step-by-step explanation:

z→   ρ

0.93 (L)0.82381 (R)0.17619

17.62% / 82.38%

The area in the right tail more extreme than z=0.93 in standard normal distribution is 17.62% rounded to three decimal places.

A normal distribution of data occurs when the majority of data points are relatively similar and the data set has a small range of values.

We need to find the area in the right tail more extreme than z=0.93 in standard normal distribution.

Here, z=0.93

z→  ρ

The mean ( the population mean, at 0.93 standard deviation units, and it represents an area that is, 17.619% of the total area of the standard distribution).

In this normal distribution, every given score is a z-score. A z-score explain us the distance from the mean in standard deviation units.

A = 17.62% / 82.38%

Hence, The area in the right tail more extreme than z=0.93 in standard normal distribution is 17.62% rounded to three decimal places.

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#SPJ2

Answer:

a. The average number of hashtags used in a social media post from a marketing agency is different than 7 hashtags.

Step-by-step explanation:

A social media platform states that a social media post from a marketing agency has 7 hashtags, on average.

This means that at the null hypothesis, we test if the mean is 7, that is:

[tex]H_0: \mu = 7[/tex]

A digital marketing specialist studying social media advertising believes the average number of hashtags used in a post from a marketing agency is different than the number stated by the social media platform.

Keyword is different, so at the null hypothesis, we test if the mean is different of 7, that is:

[tex]H_1: \mu \neq 7[/tex]

Thus, the correct answer is given by option a.

Answer:

n ≥  3

Step-by-step explanation:

Applying the remainder term in evaluating the minimum order of the Taylor polynomial

absolute error ≤ 10^-2

[tex]\sqrt{1.06}[/tex]

∴ n ≥   ?

The remainder term is the leftover term after computation ( dividing one polynomial with another )

attached below is the detailed solution

The minimum order of the Taylor polynomial, n≥3

Taylor polynomial is a series of functions that has an infinite sum of terms that are expressed in terms of the function's derivatives.

[tex]\rm f(a)+\frac{f'(a)}{1!} (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +....,[/tex]

Applying the Taylor series polynomial, the minimum order of the Taylor polynomial, centered at 0

[tex]\rm f(0)+\frac{f'(0)}{1!} (x-0)+\frac{f''(0)}{2!} (x-0)^{2} +....,[/tex]

f(x) = [tex]\sqrt{x+1}[/tex]

[tex]f'(x)=\frac{1}{2}[/tex]

[tex]f''(x)=3/8[/tex]

substituting in the Taylors series

T(x) = [tex]1+\frac{x}{2} -\frac{x^{2} }{8} +\frac{x^{3} }{16}[/tex]......

T(0.06) = [tex]1+\frac{0.06}{2} -\frac{0.06^{2} }{8} +\frac{0.06^{3} }{16}...[/tex]

T(0.06) =1.03

f(0.06) =

[tex]\sqrt{0.06+1}\\= 1.03[/tex]

Therefore, the minimum order of the Taylor polynomial, n≥3

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First concert mixed fractions to normal Form

[tex]\\ \sf\longmapsto 1\dfrac{1}{4}=\dfrac{5}{4}[/tex]

[tex]\\ \sf\longmapsto 1\dfrac{3}{4}=\dfrac{7}{4}[/tex]

[tex]\\ \sf\longmapsto \dfrac{3}{4}-\dfrac{1}{4}=\dfrac{5}{4}-\dfrac{3}{4}[/tex]

[tex]\\ \sf\longmapsto \dfrac{3-1}{4}=\dfrac{5-3}{4}[/tex]

[tex]\\ \sf\longmapsto \dfrac{2}{4}=\dfrac{2}{4}[/tex]

Hence it's true

Answer: True

Explanation:

1/4, 3/4, 1 1/4, 1 3/4

1 1/4 = 5/4

1 3/4 = 7/4

1/4, 3/4, 5/4, 7/4

= 0.25, 0.75, 1.25 , 1.75

This is arithmetic

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Answer:

There are 28 beads

Step-by-step explanation:

Total ratio:

[tex]{ \sf{ (4 + 7) = 11}}[/tex]

let total beads be x:

[tex]{ \sf{ \frac{4}{11} \times x = 16 }} \\ \\ { \sf{x = \frac{11 \times 16}{4} }} \\ x = 44 \: beads[/tex]

Blue beads:

[tex] = 44 - 16 \\ = 28 \: \: beads[/tex]

First let's compute dx/dt

[tex]x = t - \frac{1}{t}\\\\x = t - t^{-1}\\\\\frac{dx}{dt} = \frac{d}{dt}\left(t - t^{-1}\right)\\\\\frac{dx}{dt} = 1-(-1)t^{-2}\\\\\frac{dx}{dt} = 1+\frac{1}{t^{2}}\\\\\frac{dx}{dt} = \frac{t^2}{t^{2}}+\frac{1}{t^{2}}\\\\\frac{dx}{dt} = \frac{t^2+1}{t^{2}}\\\\[/tex]

Now compute dy/dt

[tex]y = 2t + \frac{1}{t}\\\\y = 2t + t^{-1}\\\\\frac{dy}{dt} = \frac{d}{dt}\left(2t + t^{-1}\right)\\\\\frac{dy}{dt} = 2 - t^{-2}\\\\\frac{dy}{dt} = 2 - \frac{1}{t^2}\\\\\frac{dy}{dt} = \frac{2t^2}{t^2}-\frac{1}{t^2}\\\\\frac{dy}{dt} = \frac{2t^2-1}{t^2}\\\\[/tex]

From here, apply the chain rule to say

[tex]\frac{dy}{dx} = \frac{dy*dt}{dx*dt}\\\\\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\\\\\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\\\\\frac{dy}{dx} = \frac{2t^2-1}{t^2} \div \frac{t^2+1}{t^{2}}\\\\\frac{dy}{dx} = \frac{2t^2-1}{t^2} \times \frac{t^{2}}{t^2+1}\\\\\frac{dy}{dx} = \frac{2t^2-1}{t^2+1}\\\\[/tex]

We could use polynomial long division, or we could add 2 and subtract 2 from the numerator and do a bit of algebra like so

[tex]\frac{dy}{dx} = \frac{2t^2-1}{t^2+1}\\\\\frac{dy}{dx} = \frac{2t^2-1+2-2}{t^2+1}\\\\\frac{dy}{dx} = \frac{(2t^2+2)-1-2}{t^2+1}\\\\\frac{dy}{dx} = \frac{2(t^2+1)-3}{t^2+1}\\\\\frac{dy}{dx} = \frac{2(t^2+1)}{t^2+1}-\frac{3}{t^2+1}\\\\\frac{dy}{dx} = 2-\frac{3}{t^2+1}\\\\[/tex]

This concludes the first part of 4b

=======================================================

Now onto the second part.

Since t is nonzero, this means either t > 0 or t < 0.

If t > 0, then,

[tex]t > 0\\\\t^2 > 0\\\\t^2+1 > 1\\\\\frac{1}{t^2+1} < 1 \ \text{ ... inequality sign flip}\\\\\frac{3}{t^2+1} < 3\\\\-\frac{3}{t^2+1} > -3 \ \text{ ... inequality sign flip}\\\\-\frac{3}{t^2+1}+2 > -3 + 2\\\\2-\frac{3}{t^2+1} > -1\\\\-1 < 2-\frac{3}{t^2+1}\\\\-1 < \frac{dy}{dx}\\\\[/tex]

note the inequality signs flipping when we apply the reciprocal to both sides, and when we multiply both sides by a negative value.

You should find that the same conclusion happens when we consider t < 0. Why? Because t < 0 becomes t^2 > 0 after we square both sides. The steps are the same as shown above.

So both t > 0 and t < 0 lead to [tex]-1 < \frac{dy}{dx}[/tex]

We can say that -1 is the lower bound of dy/dx. It never reaches -1 itself because t = 0 is not allowed.

We could say that

[tex]\displaystyle \lim_{t\to0}\left(2-\frac{3}{t^2+1}\right)=-1\\\\[/tex]

---------------------------------------

To establish the upper bound, we consider what happens when t approaches either infinity.

If t approaches positive infinity, then,

[tex]\displaystyle L = \lim_{t\to\infty}\left(2-\frac{3}{t^2+1}\right)\\\\\\\displaystyle L = \lim_{t\to\infty}\left(\frac{2t^2-1}{t^2+1}\right)\\\\\\\displaystyle L = \lim_{t\to\infty}\left(\frac{2-\frac{1}{t^2}}{1+\frac{1}{t^2}}\right)\\\\\\\displaystyle L = \frac{2-0}{1+0}\\\\\\\displaystyle L = 2\\\\[/tex]

As t approaches infinity, the dy/dx value approaches L = 2 from below.

The same applies when t approaches negative infinity.

So we see that [tex]\frac{dy}{dx} < 2[/tex]

---------------------------------------

Since [tex]-1 < \frac{dy}{dx} \text{ and } \frac{dy}{dx} < 2[/tex], those two inequalities combine into the compound inequality [tex]-1 < \frac{dy}{dx} < 2[/tex]

So dy/dx is bounded between -1 and 2, exclusive of either endpoint.

The correct answer to the given question is "[tex]\bold{392.47\ < \mu <\ 427.53}[/tex],[tex]\bold{0.28 \ < P <\ 0.34}[/tex], and for Interpret results go to the C part.

Following are the solution to the given parts:

A)

[tex]\to \bold{(n) = 16}[/tex]

[tex]\to \bold{(\bar{X}) = 410}[/tex]

[tex]\to \bold{(\sigma) = 40}[/tex]

In the given question, we calculate [tex]90\%[/tex] of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

[tex]\to \bold{\bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}}[/tex]

[tex]\to \bold{C.I= 0.90}\\\\\to \bold{(\alpha) = 1 - 0.90 = 0.10}\\\\ \to \bold{\frac{\alpha}{2} = \frac{0.10}{2} = 0.05}\\\\ \to \bold{(df) = n-1 = 16-1 = 15}\\\\[/tex]

Using the t table we calculate [tex]t_{ \frac{\alpha}{2}} = 1.753[/tex]  When [tex]90\%[/tex] of the confidence interval:

[tex]\to \bold{410 \pm 1.753 \times \frac{40}{\sqrt{16}}}\\\\ \to \bold{410 \pm 17.53\\\\ \to392.47 < \mu < 427.53}[/tex]

So [tex]90\%[/tex] confidence interval for the mean weight of shipped homemade candies is between [tex]392.47\ \ and\ \ 427.53[/tex].

B)

[tex]\to \bold{(n) = 500}[/tex]

[tex]\to \bold{(X) = 155}[/tex]

[tex]\to \bold{(p') = \frac{X}{n} = \frac{155}{500} = 0.31}[/tex]

Here we need to calculate [tex]90\%[/tex] confidence interval for the true proportion of all college students who own a car which can be calculated as

[tex]\to \bold{p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}}[/tex]

[tex]\to \bold{C.I= 0.90}[/tex]

[tex]\to\bold{ (\alpha) = 0.10}[/tex]

[tex]\to\bold{ \frac{\alpha}{2} = 0.05}[/tex]

Using the Z-table we found [tex]\bold{Z_{\frac{\alpha}{2}} = 1.645}[/tex]

therefore [tex]90\%[/tex] the confidence interval for the genuine proportion of college students who possess a car is

[tex]\to \bold{0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}}\\\\ \to \bold{0.31 \pm 0.034}\\\\ \to \bold{0.276 < p < 0.344}[/tex]

So [tex]90\%[/tex] the confidence interval for the genuine proportion of college students who possess a car is between [tex]0.28 \ and\ 0.34.[/tex]

C)

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9514 1404 393

Answer:

  perimeter ratio = dilation factor

Step-by-step explanation:

In general, all linear dimensions are scaled by the dilation factor. This includes lengths of sides, perimeter, circumference, diameter, radius, or any other 1-dimensional (length) measure of a figure.

The ratio of perimeters is equal to the ratio of side lengths.

Answer:

Hypotenuse = XY = 17 cm

Base = YZ = 15 cm

Perpendicular = XZ = 8 cm

Answer:

1.09c

Step-by-step explanation:

The clothes cost c before tax.

The tax is 9% of the price of the clothes, or 9% of c.

To find a percent of a number, change the percent to a decimal and multiply by the number.

9% as a decimal is 0.09

9% of c = 9% * c = 0.09c

The amount of tax is 0.09c.

Now we add the amount of tax to the original price, c.

c + 0.09c = 1.09c

Answer: 1.09c

Answer:

Test Statistic = 10

Critical Value = 9.488

Step-by-step explanation:

Given :

Grade A _ B _ C _ D _ F

_____14 _ 18 _32_ 20_16

H0 : distribution of grade is uniform

H1 : Distribution of grade is not uniform

Using the Chisquare statistic :

χ² = (observed - Expected)² / Expected

The expected value :

(14+18+32+20+16) / 5 = 20

χ² = (14-20)^2 / 20 + (18-20)^2 / 20 + (32-20)^2 / 20 + (20-20)^2 / 20 + (16-20)^2 / 20

χ² statistic = 10

The χ² critical at df = (n - 1) = 5 - 1 = 4

χ² Critical(10, 4) = 9.488

The product that provides the better value per sheet is Angel Soft toilet paper.

Each sheet of Angel Soft costs $0.20 whereas each sheet of Charmin costs $0.36.

Data and Calculations:

Charmin has 18 rolls, each with 200 sheets, for $12.97

Angel Soft has 24 rolls, each with 425 sheets, for $19.98

Product      Rolls  Sheets   Total Sheets      Total Cost    Cost per Sheet

                           in a Roll  (rolls x sheet)                   (Total cost/No. Sheets)

Charmin      18     200      3,600 (18 x 200)   $12.97  $0.36 ($12.97/3,600)

Angel Soft  24    425     10,200 (24 x 425)  $19.98  $0.20 ($19.98/10,200)

Thus, Rebecca should go for Angel Soft toilet paper because it provides a better value per sheet than Charmin toilet paper.

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