Choose the correct answer.

The History Activities were about​ ___ and​ ___. They were both from​ ___. One of them had a​ ___ named after him. The other had a​ ___ named after him. The work they did together on a problem they got from a​ ___ laid the foundation for probability theory.

A. ​Pascal, Decartes,​ England, graph,​ book, relative
B. ​Luniz, Kolmogrov,​ Germany, church,​ square, teacher
C. ​Fermat, Luniz,​ England, law,​ diagram, priest
D. ​Pascal, Fermat,​ France, triangle,​ theorem, gambler

9514 1404 393

Answer:

Step-by-step explanation:

For continuous compounding the "rule of 69" applies. That is the doubling time can be found from ...

  t = 69.3147/r . . . . where r is the interest rate in percent.

Here, r=2, so ...

  t = 69.3147/2 = 34.6574 . . . years

That's 34 years and 240 days.

Answer:

can you say again please

Answer:

Step-by-step explanation:

Let length = y    width = x

y = 3x + 2

Perimeter = Sum of all sides (or sum of both lengths and both widths)

2y + 2x

2(3x + 2) + 2x

6x + 4 + 2x

8x + 4

Answer:

chocolate chips are $2.00 per pound.

nd walnuts must be $3.50 per pound.

Step-by-step explanation:

Let x be the price of walnuts and y the price of chocolate chips.

2x + 5y = 17 (i)

8x + 3y = 34 (ii)

Multiply (i) by 4 to get

8x + 20y = 68

Subtract (ii) to get

 17y = 34

Dividing by 17, we see that chocolate chips are $2.00 per pound.

Substituting y=2 in (i) or (ii), walnuts must be $3.50 per pound.

Answer:

The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.  

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Siberian Huskies:

Sample of 47, mean of 5.2 minutes, standard deviation of 1.4. So

[tex]\mu_1 = 5.2[/tex]

[tex]s_1 = \frac{1.4}{\sqrt{47}} = 0.2042[/tex]

Others:

Sample of 39, mean of 5.5 minutes, standard deviation of 1.1. So

[tex]\mu_2 = 5.5[/tex]

[tex]s_2 = \frac{1.1}{\sqrt{39}} = 0.1761[/tex]

Distribution of the difference:

[tex]\mu = \mu_1 - \mu_2 = 5.2 - 5.5 = -0.3[/tex]

[tex]s = \sqrt{s_1^2+s_2^2} = \sqrt{0.2042^2+0.1761^2} = 0.2692[/tex]

Confidence interval:

We have that 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 pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.

Now, find the margin of error M as such

[tex]M = zs[/tex]

In which s is the standard error. So

[tex]M = 1.96(0.2692) = 0.5276[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is -0.3 - 0.5276 = -0.8276.

The upper end of the interval is the sample mean added to M. So it is -0.3 + 0.5276 = 0.2276

The 95% confidence interval for the true difference between the mean times on this course for teams of Siberian Huskies and teams of other breeds of sled dogs is (-0.8276, 0.2276).

Here we want to solve a question involving fractions, we will find that:

3/4 gallon fils the complete bin.

Ok, so we know that 1/2 gallon of water, fills 2/3 of the bin.

We want to find the total amount of water that would fill the entire bin.

So we could write an equation like:

amount of water =  amount of the bin that it fills.

Then, using the above information, we have:

1/2 gal  = 2/3 of a bin

Now we want to get at 1 on the right side, this would mean "1 bin"

Then we multiply both sides by (3/2)

(3/2)*(1/2) gal = (3/2)*(2/3) of a bin

3/4 gal = 1 bin

From this, we can conclude that (3/4) gallons of water would fill the complete bin.

If you want to learn more about algebra, you can read:

https://brainly.com/question/4837080

x = 0,75 gallons      or       x  = 3/4   gallons      The volume of the bin

The volume of the bin is: In terms of a fraction

1  =  3/3      or any unitary fraction   5/5    7/7    9/9

We will take 3/3 since we have the information that 2/3 of the volume of the bin was filled with 2/3 of a gallon

If  2/3 of the volume of the bin was filled with 1/2 gallon then we make a rule of three according to:

If     0,5  gal.         fill    2/3 of the volume of the bin   then

          x   gal         fill     3/3  ( the volume of the bin)

solving

0,5 (gal) * 3/3    =   (2/3)*x       ( The equation)

0,5*3 = 2*x

x  =  (0,5*3)/2

x = 0,75 gallons      or       x  = 3/4   gallons

Answer:

The third choice down

Step-by-step explanation:

|-2x| = 4

There are two solutions, one positive and one negative

-2x = 4  and -2x = -4

Divide by -2

-2x/-2 = 4/-2    -2x/-2 = -4/-2

x = -2   and x = 2

Answer:

"A"

Step-by-step explanation:

a+b >c

a+c>b

b+c>a

~~~~~~~~~~~~

A. T,T,T

B. T,T,F

C. T,F,T

Answer:

rectangle

Step-by-step explanation:

Perimeter of 20 feet

rectangle (square is technically a rectangle):

sides 5 and 5

5*5 = 25ft²

Circle:

20/(2π) = 3.18309...

3.1809...²π = 31.831ft²

Max area of rectangle (i.e. square) has a smaller area than a circle.

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

Step-by-step explanation:

before Tax

Coast = 888

in 2ND Item = $21

• 888/21

= $42.28

Answer:

m=1/2

y-8=1/2(x-1)

y-8=1/2x-1/2

multiply through by 2

2y-16=x-1

2y-16+1-x=0

2y-15-x=0

2y-x-15=0

It looks like the limit you want to find is

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4}[/tex]

One way to compute this limit relies only on the definition of the constant e and some basic properties of limits. In particular,

[tex]e = \displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x[/tex]

The idea is to recast the given limit to make it resemble this definition. The definition contains a fraction with x as its denominator. If we expand the fraction in the given limand, we have a denominator of x - 1. So we rewrite everything in terms of x - 1 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\dfrac{x-1+5}{x-1}\right)^{x-1+5} \\\\ = \left(1+\dfrac5{x-1}\right)^{x-1+5} \\\\ =\left(1+\dfrac5{x-1}\right)^{x-1} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Now in the first term of this product, we substitute y = (x - 1)/5 :

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(1+\dfrac1y\right)^{5y} \times \left(1+\dfrac5{x-1}\right)^5[/tex]

Then use a property of exponentiation to write this as

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \left(\left(1+\dfrac1y\right)^y\right)^5 \times \left(1+\dfrac5{x-1}\right)^5[/tex]

In terms of end behavior, (x - 1)/5 and x behave the same way because they both approach ∞ at a proportional rate, so we can essentially y with x. Then by applying some limit properties, we have

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty} \left(\left(1+\dfrac1x\right)^x\right)^5 \times \left(1+\dfrac5{x-1}\right)^5 \\\\ = \lim_{x\to\infty}\left(\left(1+\dfrac1x\right)^x\right)^5 \times \lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)^5 \\\\ =\left(\lim_{x\to\infty}\left(1+\dfrac1x\right)^x\right)^5 \times \left(\lim_{x\to\infty}\left(1+\dfrac5{x-1}\right)\right)^5[/tex]

By definition, the first limit is e and the second limit is 1, so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = e^5\times1^5 = \boxed{e^5}[/tex]

You can also use L'Hopital's rule to compute it. Evaluating the limit "directly" at infinity results in the indeterminate form [tex]1^\infty[/tex].

Rewrite

[tex]\left(\dfrac{x+4}{x-1}\right)^{x+4} = \exp\left((x+4)\ln\dfrac{x+4}{x-1}\right)[/tex]

so that

[tex]\displaystyle \lim_{x\to\infty} \left(\frac{x+4}{x-1}\right)^{x+4} = \lim_{x\to\infty}\exp\left((x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ = \exp\left(\lim_{x\to\infty}(x+4)\ln\dfrac{x+4}{x-1}\right) \\\\ =\exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right)[/tex]

and now evaluating "directly" at infinity gives the indeterminate form 0/0, making the limit ready for L'Hopital's rule.

We have

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\ln\dfrac{x+4}{x-1}\right] = -\dfrac5{(x-1)^2}\times\dfrac{1}{\frac{x+4}{x-1}} = -\dfrac5{(x-1)(x+4)}[/tex]

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{x+4}\right]=-\dfrac1{(x+4)^2}[/tex]

and so

[tex]\displaystyle \exp\left(\lim_{x\to\infty}\frac{\ln\dfrac{x+4}{x-1}}{\dfrac1{x+4}}\right) = \exp\left(\lim_{x\to\infty}\frac{-\dfrac5{(x-1)(x+4)}}{-\dfrac1{(x+4)^2}}\right) \\\\ = \exp\left(5\lim_{x\to\infty}\frac{x+4}{x-1}\right) \\\\ = \exp(5) = \boxed{e^5}[/tex]

The Total mileage is "400" and the further solution can be defined as follows:

Let t become the time he spent commuting on the first day of his vacation.

It is then calculated as [tex]t + 2[/tex].

[tex]\to 40\times(t+2) = 60(t) + 20 \\\\\to 40t+80 = 60t + 20 \\\\\to 80-20 = 60t + 40t \\\\\to 60 = 20t \\\\\to t=\frac{60}{20} \\\\\to t=\frac{6}{2} \\\\\to t= 3\\\\[/tex]

It traveled [tex]40\times (3 + 2) + 20 = 40\times 5 + 20 = 200+20=220[/tex] miles on its first day of operation.

The car traveled [tex]180\ miles[/tex] on the second day, which was [tex]60 \ miles \times 3[/tex].

So,

Total mileage= first day traveled + second day traveled [tex]= 220+ 180= 400 \miles[/tex]

Learn more:

Total distance traveled: brainly.com/question/20670144

Answer:

The answer is "36.14%"

Step-by-step explanation:

The complete question is given in the attached file please find it.

[tex]\mu =41\\\\\sigma= 4\\\\P(42<\bar{x}<48)= p(\bar{x}<48)-p(\bar{x}<42)\\\\Z =\frac{(42-41)}{4} = \frac{1}{4} =0.25\\\\Z =\frac{(48-41)}{4} = \frac{7}{4} = 1.75\\\\[/tex]

Using z-table to find the value.

[tex]\to P(41<\bar{x}<48) = 0.9599- 0.5987 = 0.3614\times 100= 36.14\%[/tex]

This means that between 42 and 48 months, 36.14 % of scanners could be predicted will break down.

The answer is A and D

good luck

Answer:

2025 cm

Step-by-step explanation:

Given the length of pieces - 45 cm, 75 cm and 81 cm

To find the length of the rope we have to find the L.C.M. of 45, 75 and 81 :

3  |  45, 75, 81

    | ________________      

3  |  15, 25, 27

    |________________

3  |    5, 25, 9

    |________________

3  |    5, 25, 3

    |________________

5  |    5, 25, 1

    |________________

5  |     1, 5, 1

    |________________

    |      1, 1, 1

L.C.M. = 3 × 3 × 3 × 3 × 5 × 5

= 2025 cm

So, the least length of the rope should be 2025 cm which can be cut into a whole number of pieces of length 45 cm, 75 cm and 81 cm.

Answer:

9/25

Step-by-step explanation:

3 novels , 1 bio , 1 poetry = 5 books

P( novel) = novels / books

               = 3/5

Book is returned

3 novels , 1 bio , 1 poetry = 5 books

P( novel) = novels / books

               = 3/5

P(novel, return, novel) = 3/5 * 3/5 = 9/25

Answer:

0.2725

0.0431

Step-by-step explanation:

The distribution here is a poisson distribution :

λ = 1.3

The poisson distribution :

p(x) = [(e^-λ * λ^x)] ÷ x!

Expected probability of finding male with 0 accident ; x = 0

p(0) = [(e^-1.3 * 1.3^0)] ÷ 0!

p(0) = [0.2725317 * 1] ÷ 1

p(0) = 0.2725317

= 0.2725

2.)

P(x ≥ 4) = 1 - P(x < 4)

P(x < 4) = p(x = 0) + p(x. = 1) + p(x = 2) + p(x = 3)

p(x = 0) =  p(0) = [(e^-1.3 * 1.3^0)] ÷ 0! = 0.2725

p(x = 1) = p(1) = [(e^-1.3 * 1.3^1)] ÷ 1! = 0.35429

p(x = 2) = p(2) = [(e^-1.3 * 1.3^2)] ÷ 2! = 0.23029 p(x = 3) = p(3) = [(e^-1.3 * 1.3^3)] ÷ 0! = 0.09979

P(x < 4) = 0.2725 + 0.35429 + 0.23029 + 0.09979 = 0.95687

P(x ≥ 4) = 1 - 0.95687 = 0.0431

Answer:

C.    AA

Step-by-step explanation:

Since m<Y = 27°, then m<W = 27°.

We have two angles of one triangle (A and B) congruent to two angles of the other triangle (W and X).

Answer: C.   AA

Answer:

Step-by-step explanation:

In order to find out how long it takes for the rocket to hit the ground, we only need set that position equation equal to 0 (that's how high something is off the ground when it is sitting ON the ground) and factor to solve for t:

[tex]0=-4.9t^2+112t+395[/tex]

Factor that however you are factoring in class to get

t = -3.1 seconds and t = 25.9 seconds.

Since time can NEVER be negative, it takes the rocket approximately 26 seconds to hit the ground.

Answer:

354°15'38.99''

Step-by-step explanation:

Answer:

8 units^2

Step-by-step explanation:

The area of a tringle is 1/2 bh. The base, LK, measures 4 while the height is also 4(you can get these values by counting the squares). This means the area is:

1/2 * (4)(4) = 1/2 * 16 = 8 units^2

Answer:

Kindly check explanation

Step-by-step explanation:

H0 : μ = 15.7

H1 : μ < 15.7

This is a one sample t test :

Test statistic = (xbar - μ) ÷ (s/√(n))

n = sample size = 33

Using calculator :

The sample mean, xbar = 15.41

The sample standard deviation, s = 0.419

Test statistic = (15.41 - 15.70) ÷ (0.419/√(33))

Test statistic = - 3.976

Using the Pvalue calculator :

Degree of freedom, df = n - 1 ; 33 - 1 = 32

Pvalue(-3.976, 32) = 0.000187

Decison region :

Reject H0 if Pvalue < α

Since Pvalue < α ; we reject H0

There is significant evidence to conclude that the true average height of the seedlings treated with an extract from Vinca minor roots is less than 15.7 cm.

Answer:

[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]

Step-by-step explanation:

Given

[tex]\sqrt{128x^8y^3}[/tex] --- the complete expression

Required

The equivalent expression

We have:

[tex]\sqrt{128x^8y^3}[/tex]

Expand

[tex]\sqrt{128x^8y^3} = \sqrt{128* x^8 * y^3}[/tex]

Further expand

[tex]\sqrt{128x^8y^3} = \sqrt{64 * 2* x^8 * y^2 * y}[/tex]

Rewrite as:

[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2* 2 * y}[/tex]

Split

[tex]\sqrt{128x^8y^3} = \sqrt{64 * x^8 * y^2} * \sqrt{2 * y}[/tex]

Express as:

[tex]\sqrt{128x^8y^3} = (64 * x^8 * y^2)^\frac{1}{2} * \sqrt{2y}[/tex]

Remove bracket

[tex]\sqrt{128x^8y^3} = (64)^\frac{1}{2} * (x^8)^\frac{1}{2} * (y^2)^\frac{1}{2} * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 * x^\frac{8}{2} * y^\frac{2}{2} * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 * x^4 * y * \sqrt{2y}[/tex]

[tex]\sqrt{128x^8y^3} = 8 x^4 y \sqrt{2y}[/tex]

Answer:

2:3 is equal to 8:12

Step-by-step explanation:

2:3

To get the first number to 8

8/2 = 4

Multiply by all terms 4

2*3 : 3*4

8:12

2:3 is equal to 8:12

8:12 = 8/12

= 2/3

= 2:3

Therefore 2:3 and 8:12 are equalent to each other.

Answered by Gauthmath must click thanks and mark brainliest

Answer: C

Step-by-step explanation:

Answer:

Confidence Interval

Lower Limit = $4,233.3

Upper Limit = $4,766.7

With 95% confidence, the mean profit per customer that SoftBus can expect to obtain is between $4,233.30 and $4,766.7 based on the given sample data.

Step-by-step explanation:

The z-score of 95% = 1.96

             Profit         Mean      Square Root

                          Difference    of MD

P1        $5,000       $500        $250,000

P2         4,500          0              0

P3         4,000       -500         $250,000

Total $13,500                        $500,000

Mean $4,500 ($13,500/3)    $166,667 ($500,000/3)

Standard Deviation = Square root of $166,667 = 408.2

Margin of error = (z-score * standard deviation)/n

= (1.96 * 408.2)/3

= 266.7

= $266.7

Confidence Interval = Sample mean +/- Margin of error

= $4,500 +/- 266.7

Lower Limit = $4,233.3 ($4,500 - $266.7)

Upper Limit = $4,766.7 ($4,500 + $266.7)

Answer: 18π

okokok gg

Step-by-step explanation:

Here angle is given in degree.We have convert it into radian.

[tex] {1}^{\circ} =( { \frac{\pi}{180} } )^{c} \\ \therefore \: {80}^{\circ} = ( \frac{80\pi}{180} ) ^{c} = {( \frac{4\pi}{9} })^{c} \: = \theta ^{c} [/tex]

Area of green shaded regions = A

[tex] \sf \: A = \frac{1}{2} { {r}^{2} }{ { \theta}^{ c} } \\ = \frac{1}{2} \times {9}^{2} \times \frac{4\pi}{9} \\ = 18\pi \: {cm}^{2} [/tex]