Which explanation could be used to verify weather the functions are inverse

Answer:

D. 28

Step-by-step explanation:

Given:

m∠AFD = 90°

m∠AFB = 31°

Required:

m∠DFE

Solution:

m<AFB = m<CFD (both angles are marked as congruent angles)

Since m<AFB = 31°, therefore,

m<CFD = 31°

m<AFB + m<CFD + m<BFC = m<AFD

Plug in the known values

31° + 31° + m<BFC = 90°

62° + m<BFC = 90°

Subtract 62° from each side

m<BFC = 90° - 62°

m<BFC = 28°

m<BFC = m<DFE = 28° (both angles are marked congruent to each other)

Therefore,

m<DFE = 28°

Answer:

$199.29

Step-by-step explanation:

Total payments = $7,174.24

Total interest = $1,174.24

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

Explanation:

You can use the AAS (angle angle side) theorem to prove that triangle ABD is congruent to triangle CBD.

From there, we can then say that AD and DC are the same length

AD = DC

3y+6 = 5y-18

3y-5y = -18-6

-2y = -24

y = (-24)/(-2)

y = 12

Answer:

12

Step-by-step explanation:

If f(x) = 5 - x

Then f(-7) = 5 - (-7)

f(-7) = 5 + 7

f(-7) = 12

Answer:

Each group has 1 fiction book and 2 nonfiction book(s).

Answer:

[tex]Pr= \frac{1}{21}[/tex]

Step-by-step explanation:

Given

[tex]n(S) = 7[/tex] --- number of games

Required

Probability of bike and movie in consecutive order

This probability is represented as:

[tex]Pr = P(Bike\ and\ Movie) \ or\ P(Movie\ or\ Bike)[/tex]

So, we have:

[tex]Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)[/tex]

The probability is an illustration of selection without replacement;

So, we have:

[tex]P(Bike\ and\ Movie) = P(Bike) * P(Movie)[/tex]

[tex]P(Bike\ and\ Movie) = \frac{n(Bike)}{n(S)} * \frac{n(Movie)}{n(S) - 1}[/tex] ---- without replacement

Bike and Movie appear in the game list 1 time.

So, the equation becomes

[tex]P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{7 - 1}[/tex]

[tex]P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{6}[/tex]

[tex]P(Bike\ and\ Movie) = \frac{1}{42}[/tex]

Similarly,

[tex]P(Movie\ and\ Bike) = \frac{1}{42}[/tex]

So, we have:

[tex]Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)[/tex]

[tex]Pr= \frac{1}{42}+\frac{1}{42}[/tex]

Take LCM

[tex]Pr= \frac{1+1}{42}[/tex]

[tex]Pr= \frac{2}{42}[/tex]

[tex]Pr= \frac{1}{21}[/tex]

The required probability of a couple having a baby boy when their third child is​ born is 1/2.

probability is the ratio of the number of favorable outcomes and the total number of possible outcomes. The chance that a particular event (or set of events) will occur expressed on a linear scale from 0 (impossibility) to 1 (certainty), also expressed as a percentage between 0 and 100%.

Given:

Assuming boys and girls are equally​ likely.

The first two children were both boys

According to given question we have

The probability of having a baby girl is an independent probability.

The first two children were both boys

So, it is not related to the previous child.

So required probability = 1/2

Therefore, the required probability of a couple having a baby boy when their third child is​ born is 1/2.

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Answer:

A.  5

Step-by-step explanation:

Parallel lines have the same slope.

Answer:

5

Step-by-step explanation:

Answer:

I think the answer is B. 65%

Answer:

option D is the answer

Step-by-step explanation:

using the HH,ll,ha and la,

where h is the hypotenuse and l is the leg

Answer:

x = 38.4

Step-by-step explanation:

tan(38) = 30/x

x = 30/tan(38)

x = 38.4

Answered by GAUTHMATH

[tex]\boxed{ \sf{Answer}} [/tex]

[tex] \sf \: 3(3x - 4) - 5(x + 3) \\ \sf =( 3 \times 3x) - (3 \times 4) + ( - 5 \times x) +( - 5 \times 3 ) \\ \sf = 9x - 12 - 5x - 15 \\ \sf = 9x - 5x - 12 - 15 \\ = \underline{ \bf 4x - 27}[/tex]

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐

[tex]\\ \sf\longmapsto 3(3x-4)-5(x+3)[/tex]

[tex]\\ \sf\longmapsto 9x-12-5x-15[/tex]

[tex]\\ \sf\longmapsto 9x-5x-15-12[/tex]

[tex]\\ \sf\longmapsto (9-5)x-27[/tex]

[tex]\\ \sf\longmapsto 4x-27[/tex]

Answer:

$3792116

Step-by-step explanation:

that's the answer above

Answer:

i believe it'll cost 200 dollars

Answer:

x = 8/3

Step-by-step explanation:

Log₂(9x) – Log₂3 = 3

The value of x can be obtained as follow:

Log₂(9x) – Log₂3 = 3

Recall

Log M – Log N = Log (M/N)

Thus,

Log₂(9x) – Log₂3 = 3

Log₂(9x/3) = 3

Log₂3x = 3

3x = 2³

3x = 8

Divide both side by 3

x = 8/3

Answer:

2

Step-by-step explanation:

8 ^ (5/3) ^ 1/5

We know a^b^c = a^(b*c)

8^ (5/3*1/5)

8^ 1/3

Rewriting 8 as 2^3

2^3 ^1/3

2 ^(3*1/3)

2^1

2

Answer:

2

Step-by-step explanation:

((2^3)^5/3)^1/5

= (2^5)^1/5

= 2

Answered by Gauthmath

(1) D

[tex]L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt[/tex]

Integrate by parts, taking

[tex]u = t \implies \mathrm du=\mathrm dt[/tex]

[tex]\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}[/tex]

Then

[tex]L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}[/tex]

[tex]L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}[/tex]

(2) A

[tex]L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}[/tex]

[tex]L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}[/tex]

Answer:

diện tích đa giác trong hình là :

186 cm2

Step-by-step explanation:

hãy tách hình đa giác trên thành 4 hình chữ nhật và tính diện tích từng hình chữ nhật

Answer:

Horizontal Shift: Right 1

Vertical Shift: Down 5

Reflection: None

Explanation: To find the transformation, compare the function to the parent function (being in this case g(x)=1/x) and check to see if there is a horizontal or vertical shift or a reflection.

So, the answer would be Right 1, and down 5

Hope this helps you out :)

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

[tex]\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42[/tex]

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

So

[tex]s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772[/tex]

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is [tex]n = 10[/tex]

The significance level is [tex]\alpha = 1 - 0.05 = 0.95[/tex]

The estimate, which is the sample standard deviation, is of [tex]s = 0.8772[/tex].

Now, we have to find the critical values for the Pearson distribution. They are:

[tex]\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228[/tex]

[tex]\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004[/tex]

The confidence interval for the population variance is:

[tex]\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}[/tex]

[tex]\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}[/tex]

[tex]0.3641 < \sigma^2 < 2.5646[/tex]

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

[tex]\sqrt{0.3641} = 0.6034[/tex]

[tex]\sqrt{2.5646} = 1.6014[/tex]

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

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Answer:

The minimum sample size is 1,704.

Step-by-step explanation:

We have to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.95}{2} = 0.025[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a p-value of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Standard deviation of 21,059,637 shares

This means that [tex]\sigma = 21059637[/tex]

What is the minimum required sample size if you would like your sampling error to be limited to 1,000,000 shares?

This is n for which [tex]M = 1000000[/tex], so:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]1000000 = 1.96\frac{21059637}{\sqrt{n}}[/tex]

[tex]1000000\sqrt{n} = 1.96*21059637[/tex]

[tex]\sqrt{n} = \frac{1.96*21059637}{1000000}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.96*21059637}{1000000})^2[/tex]

[tex]n = 1703.8[/tex]

Rounding up:

The minimum sample size is 1,704.

Answer:

1,000 defects

Step-by-step explanation:

Find how many defects that should be made by finding 1% of 100,000:

100,000(0.01)

= 1000

So, there should be 1,000 defects

9514 1404 393

Answer:

  13 < √181 < 14

Step-by-step explanation:

Apparently, you're supposed to know that ...

  13² = 169

  14² = 196

so √181 will lie between 13 and 14.

  13 < √181 < 14

Answer:

12024.7

Step-by-step explanation:

Searched it up.

then rounded

Answer:

A. 5

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Algebra II

Step-by-step explanation:

Step 1: Define

Identify points from graph.

Point (8, 5)

Point (4, 2)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Answer:

9/49

Step-by-step explanation:

that is the procedure above

Answer:

c

Step-by-step explanation:

The first four terms are 17, 15, 13, and 11.

A function is a relationship between inputs where each input is related to exactly one output.

Example:

f(x) = 2x + 1

f(1) = 2 + 1 = 3

f(2) = 2 x 2 + 1 = 4 + 1 = 5

The outputs of the functions are 3 and 5

The inputs of the function are 1 and 2.

We have,

f(n):

f(1) = 17

f(n) = f(n - 1) - 2 if n > 1

Now,

The first term is 17.

The second term.
f(2) = f(2 - 1) - 2

= f(1) - 2

= 17 - 2

= 15

The third term.

= f(3 - 1) - 2

= f(2) - 2

= 15 - 2

= 13

The fourth term.

f(4) = f(4 - 1) - 2
= f(3) - 2

= 13 - 2

= 11

Thus,

The terms are 17, 15, 13, 11

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Answer:

the first option because I took the test

Answer:

a. 168000 for Eyeconnect, 132000 for Topconnect

b. 100,000

Step-by-step explanation:

a.

Because the change in customers are only due to leaving companies, we can say that, after one year, Eyeconnect loses 10% of its customers to Topconnect and Topconnect loses 5% of its customers to Eyeconnect. This represents all changes in customers.

First, we can calculate how much Eyeconnect loses, which is 10% of 180,000 = 0.1 * 180,000 = 18,000 . They then have 180,000 - 18,000 = 162,000 employees

Next, Topconnect loses 120,000 * 5% = 120,000 * 0.05 = 6,000. They then have 120,000-6,000 = 114,000 employees

We can then add the customer amounts. Note that we are subtracting both sides before adding as both companies gain and lose customers simultaneously.

We can then add how much one company lost to the other company's customers.

Eyeconnect gains 6,000 customers, so they then have 162,000 + 6,000 = 168000 employees. Topconnect gains 18,000 customers so they then have 114,000 + 18,000 = 132,000 employees

b.

After many years, the number of customers Eyeconnect has will be less than the number of customers that Topconnect has. One way to find the end amount of customers that Eyeconnect has is to figure out when the customer bases even out, or when Eyeconnect loses the same amount of customers as Topconnect so the customer base stays the exact same. We know that no customers leave or join the companies except to leave/join the other, so the total amount of customers between the two companies stays the exact same. The amount of customers is 180,000 + 120,000 = 300,000. Therefore, at the end amount,

Eyeconnect customers (E) + Topconnect customers (T) =300,000

Furthermore, if the amount of customers that leave Eyeconnect is the same that leaves Topconnect, we can say

E * 0.1 = T * 0.05

divide both sides by 0.05 to isolate the T

E * 0.1 / 0.05 = T

2 * E = T

plug that into the first equation

E + 2 * E = 300,000

3 * E = 300,000

divide both sides by 3 to isolate E

E = 100,000 after many years

With some rewriting, you get

[tex]\displaystyle \sum_{k=0}^\infty (-1)^k\frac{x^{k+2}}{4^k} = x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k[/tex]

Recall that for |x| < 1, you have

[tex]\displaystyle \frac1{1-x} = \sum_{k=0}^\infty x^k[/tex]

So as long as |-x/4| = |x/4| < 1, or |x| < 4, your series converges to

[tex]\displaystyle x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k = \frac{x^2}{1-\left(-\frac x4\right)} = \frac{x^2}{1+\frac x4} = \boxed{\frac{4x^2}{4+x}}[/tex]

Based on known expressions from Taylor series, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex].

Taylor series are polynomic approximations used to estimate values both from trascendental and non-trascendental functions. It is commonly used in trigonometric, potential, logarithmic and even rational functions.

In this question we must use series properties and common Taylor series-derived formulas to infer the expression behind the given series. Now we proceed to find the expression:

[tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]

[tex]x^{2}\cdot \sum\limits_{k = 0}^{\infty} \left(-\frac{x}{4} \right)^{k}[/tex]

[tex]x^{2}\cdot \left(\frac{1}{1+\frac{x}{4} } \right)[/tex]

[tex]\frac{4\cdot x^{2}}{4+x}[/tex]

Based on power and series properties and most common Taylor series- derived formulas, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex] represents a Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex]. [tex]\blacksquare[/tex]

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