Please Help!!! Thank you!
find x

9514 1404 393

Answer:

  x = A/y

Step-by-step explanation:

Divide both sides of the equation by y

  A = xy

  A/y = (xy)/y . . . . . . divide by y

  A/y = x . . . . . . . . . .simplify

  x = A/y . . . . . . . . . write with x on the left

Answer:

w = -46/5

Step-by-step explanation:

Answer:

w=-9.2

Step-by-step explanation:

7(w+8)=2w+10

7w+56=2w+10

7w-2w=10-56

5w=-46

w=-9.2

Answer: (7, -4)

Concept:

For each rotation angle, there is a specified rule that could be applied. Please refer to the attachment below for the list of rotation rules.

Solve:

Given point = (-7, 4)

Given rotation rule = (x , y) ⇒ (-x, -y)

(-7, 4) ⇒ (7, -4)

Hope this helps!! :)

Please let me know if you have any questions

no te puedo contestarte yo no hablo inglés

Answer:

a

Step-by-step explanation:

1/3 x 3/5 x 3/4 =7/12 so therefore that's what the answer isn't

Answer:

20kg

Step-by-step explanation:

Let the weight of one duck be x and the weight of one duckling be y

ATQ, 3x+2y=32 and 4x+3y=44, solving for x and y we get, weight of one duck is 8kg and one duckling is 4kg. The weight of two ducks and one duckling is 20kg

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:

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)

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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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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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

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.

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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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:

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]

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:

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

Answer:

a

Step-by-step explanation:

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:

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:

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:

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:

dư 0... 7^300 chia 7 đc 7^299 mà

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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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.

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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Answer:

X = 64

Step-by-step explanation:

All of the angles are right angles (because of the square at one of the angles shown above). This means each angle equals 90 degrees. If X + 26 = 90, then X = 64 because 90 - 26 = 64. I hope this helps!

Answer: X = 64

Step-by-step explanation:

Answer:

Length:8

Width:5.5

Step-by-step explanation:

We're given area = 44m^2, and the formula for the area of a rectangle is the length multiplied by the width. So,

A = L * w = 44

We're given that the length is 3m shorter than 2 times the width, which is 2w - 3. "2w" is the same as "2 times the width", and the 3 is subtracted because it says 3m shorter than 2 times the width. So L = 2w - 3, and we can substitute that into our equation above.

(2w - 3)(w) = 44

2w^2 - 3w - 44 = 0

Use the quadratic formula here.

x = {3 ± √(-3)^2 - 4(-44)(2)}/2(2)

= {3 ± √9 + 352}/4

= (3 ± 19)/4

You'll get two answers, but remember, we're measuring the length of the sides of shapes, so it has to be positive. It's impossible to have negative lengths, so we're going to stick with the (3 + 19)/4 answer, which is 22/4, which is 5.5. However, we are not finished yet. This is just the width. Now we need to plug it into the equation for length, which was 2w - 3

2(5.5) - 3 = 11 - 3 = 8

The length is 8m and the width is 5.5m.

Answer:

Step-by-step explanation:

Slope of the given line: m=4/3

Slope of the perpendiclar : m'=-3/4 (the inverse of the opposed of m)

Equation of the perpendiclar line: (passing through (-7,2))

[tex]y-2=(x+7)*\dfrac{-3}{4} \ or\\\\ y=-\dfrac{3x}{4} -\dfrac{13}{4}[/tex]

Given:

The five number summary of two data sets are given as:

a) 0, 4, 12, 14, 20

b) 2, 8, 14, 18, 20

To find:

The range for the outliers.

Solution:

We know that,

An observation is considered an outlier if it is below [tex]Q_1-1.5(IQR)[/tex]

An observation is considered an outlier if it is above [tex]Q_3+1.5(IQR)[/tex]

Where, IQR is the interquartile range and [tex]IQR=Q_3-Q_1[/tex].

The five number summary of two data sets are given as:

0, 4, 12, 14, 20

Here, [tex]Q_1=4[/tex] and [tex]Q_3=14[/tex].

Now,

[tex]IQR=14-4[/tex]

[tex]IQR=10[/tex]

The range for the outliers is:

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[4-1.5(10),14+1.5(10)][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[4-15,14+15][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[-11,29][/tex]

An observation is considered an outlier if it is below -11.

An observation is considered an outlier if it is above 29.

The five number summary of two data sets are given as:

2, 8, 14, 18, 20

Here, [tex]Q_1=8[/tex] and [tex]Q_3=18[/tex].

Now,

[tex]IQR=18-8[/tex]

[tex]IQR=10[/tex]

The range for the outliers is:

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[8-1.5(10),18+1.5(10)][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[8-15,18+15][/tex]

[tex][Q_1-1.5(IQR),Q_3+1.5(IQR)]=[-7,33][/tex]

An observation is considered an outlier if it is below -7.

An observation is considered an outlier if it is above 33.