I need help answering this ASAP

Answer:

0.0022 = 0.22% probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question, the elements are arranged, so we have to use the arrangements formula.

Arrangements formula:

The number of possible arrangements of n elements is:

[tex]A_{n} = n![/tex]

Desired outcomes:

Pine trees(6!) then the willows(6!) or

Willows(6!) then the pine trees(6!). So

[tex]D = 2*6!*6! = 1036800 [/tex]

Total outcomes:

12 trees, so:

[tex]T = 12! = 479001600 [/tex]

What is the probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other?

[tex]p = \frac{D}{T} = \frac{1036800 }{479001600 } = 0.0022[/tex]

0.0022 = 0.22% probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other.

Hey there!

First,  divide 5,000 by 10. You will get 500.

Now, 500 ÷ 10, and you will get your answer, 50.

Hope this helps! Have a great day!

Answer:

0.0022 = 0.22% probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other.

Step-by-step explanation:

The trees are arranged, so the arrangements formula is used to solve this question. Also, a probability is the number of desired outcomes divided by the number of total outcomes.

Arrangements formula:

The number of possible arrangements of n elements is given by:

[tex]A_n = n![/tex]

Desired outcomes:

Two cases:

6 redwoods(6! ways) then the 6 pine trees(6! ways)

6 pine trees(6! ways) then the 6 redwoods(6! ways)

So

[tex]D = 2*6!*6![/tex]

Total outcomes:

12 trees, so:

[tex]D = 12![/tex]

What is the probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other?

[tex]p = \frac{D}{T} = \frac{2*6!*6!}{12!} = 0.0022[/tex]

0.0022 = 0.22% probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other.

Answer:

I hope it help and I guess it is correct

Answer:

x is 5

Step-by-step explanation:

[tex] \frac{4x - 5}{9x - 5} = \frac{3}{8} \\ \\ 8(4x - 5) = 3(9x - 5) \\ 32x - 40 = 27x - 15 \\ 5x = 25 \\ x = \frac{25}{5} \\ \\ x = 5[/tex]

Step-by-step explanation:

as you can see as i solved above. all you need to do was to rationalize the both equations

Answer:

Slope is 5/2

y-intercept is 2

Step-by-step explanation:

Turn the equation into slope intercept form [ y = mx +  b ].

2y = 5x + 4

~Divide everything by 2

y = 5/2x + 2

Remember that in slope intercept form, m = slope and b = y-intercept.

Best of Luck!

Answer:

slope: 2.5

y-intercept: 2

Step-by-step explanation:

First isolate the y variable which changes the equation to y=2.5x+2

The equation of a line is mx + b where m is the slope and b and the

y-intercept. Leading us to conclude that 2.5 is the slope and 2 is the y-intercept.

You're looking for a solution of the form

[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n[/tex]

Differentiating twice yields

[tex]\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/tex]

[tex]\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n[/tex]

Substitute these series into the DE:

[tex]\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0[/tex]

[tex]\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

[tex]\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0[/tex]

which indicates that the coefficients in the series solution are governed by the recurrence,

[tex]\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}[/tex]

Use the recurrence to get the first few coefficients:

[tex]\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}[/tex]

You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the n = 1 position being the odd one out. So we have

[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots[/tex]

which looks a lot like the power series expansion for -7eˣ.

Fortunately, we can rewrite the linear term as

3x = 10x - 7x = 10x - 7/1! x

and in doing so, we can condense this solution to

[tex]\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}[/tex]

Just to confirm this solution is valid: we have

y = 10x - 7eˣ   ==>   y (0) = 0 - 7 = -7

y' = 10 - 7eˣ   ==>   y' (0) = 10 - 7 = 3

y'' = -7eˣ

and substituting into the DE gives

-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0

as required.

Answer:

a) 0.0081 = 0.81% probability that he receives no job offer

b) He expects to get 2.8 job offers.

c) 0.0837 = 8.37% probability that more than half of the firms he applied do not make him any offer.

d) Each job must be independent of other jobs. Additionaly, if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal approximation to the binomial distribution can be used.

e) 0.2401 = 24.01% probability of him securing more than 3 offers.

Step-by-step explanation:

For each application, there are only two possible outcomes. Either he gets an offer, or he does not. The probability of getting an offer for a job is independent of any other job, which means that the binomial probability distribution is used 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.

He can expect to receive a job offer from 70% of the firms to which he applies.

This means that [tex]p = 0.7[/tex]

The student decides to apply to only four firms.

This means that [tex]n = 4[/tex]

(a) What is the probability that he receives no job offer?

This is [tex]P(X = 0)[/tex]. So

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

[tex]P(X = 0) = C_{4,0}.(0.7)^{0}.(0.3)^{4} = 0.0081[/tex]

0.0081 = 0.81% probability that he receives no job offer.

(b) How many job offers he expects to get?

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

In this question:

[tex]E(X) = 4(0.7) = 2.8[/tex]

He expects to get 2.8 job offers.

(c) What is the probability that more than half of the firms he applied do not make him any offer?

Less than 2 offers, which is:

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

So

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

[tex]P(X = 0) = C_{4,0}.(0.7)^{0}.(0.3)^{4} = 0.0081[/tex]

[tex]P(X = 1) = C_{4,1}.(0.7)^{1}.(0.3)^{3} = 0.0756[/tex]

Then

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0081 + 0.0756 = 0.0837[/tex]

0.0837 = 8.37% probability that more than half of the firms he applied do not make him any offer.

(d) What assumptions do you need to make to find the probabilities? To increase the chance of securing more job offers, the student decides to apply to as many companies as possible, he sent out 60 applications to all different accounting firms.

Each job must be independent of other jobs. Additionaly, if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal approximation to the binomial distribution can be used.

(e) What is the probability of him securing more than 3 offers?

Between 4 and n, since n is 4, 4 offers, so:

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

[tex]P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401[/tex]

0.2401 = 24.01% probability of him securing more than 3 offers.

Answer:

5880 ways

Step-by-step explanation:

For selections like this, we solve using the combination theory. Recall that

nCr = n!/(n-r)!r!

Hence given to find the number of ways 6 gentleman and 4 ladies can be choosen out of 10 gentleman and 8 ladies,

= 10C6 * 8C4

= 10!/(10-6)!6! * 8!/(8-6)!6!

= 10 * 9 * 8 * 7 * 6!/4 *3 *2 * 6! * 8 * 7 * 6!/2 * 6!

= 210 * 28

= 5880 ways

The arrangement can be done in 5880 ways

Answer:40, 80 and 62

Step-by-step explanation:

182-22= 160

160/4 = 40 so,

Shortest side is 40

Longest is 80

Third side is 62

Answer:

Step-by-step explanation:

Answer:

The answer is "0.07404893".

Step-by-step explanation:

Applying the binomial distribution:

[tex]n = 22\\\\p= 82\%=0.82\\\\q = 1-0.82 = 0.18\\\\[/tex]

Calculating the probability for not enough seats:

[tex]=P(X>20)\\\\= P(21) + P(22)\\\\[/tex]

[tex]= \binom{22}{21} (0.82)^{21}(0.18)^1+ \binom{22}{22} (0.82)^{22}(0.18)[/tex]

[tex]=0 .06134598+ 0.01270295\\\\=0.07404893[/tex]

Answer:

50%

Step-by-step explanation:

35 is halve of 70 therefore it is 50%

hope it helps u...........

Answer:

384 cars

Step-by-step explanation:

To find the total number of spaces in the carpark, we must multiply the number of rows by how many cars they can accommodate:

34 ⋅ 40 = 1360

As you can see, we have 1360 total spaces. Since there are 976 cars parked, and we want to find out how many spaces are left, we have to subtract the amount of cars parked from the total spaces.

1360 - 976 = 384

Therefore, our answer is 384, specifically, 384 cars.

Answer:

384 cars.

Step-by-step explanation:

40 * 34 - 976

= 1360 - 976

= 384.

Answer:

y = 25x + 40

Step-by-step explanation:

The electrician charges $25 per hour.

The number of hours is x.

Therefore after x hours the electrician will charge $25x. (multiply the charge by the number of hours $25 * x)

Therefore fee(y) charged by the electrician = $40 + $25x

Hence y = 25x + 40

Answer:

Step-by-step explanation:

a) categorical

b) add all of the numbers and divide by how many numbers there were.

c) outliers means any that were far away from the rest of the data

d) not entirely, you can make an estimate based on it, but nat an exact answer.

Answer:

a) Everyone on the team talks until the entire team agrees on one decision.

Step-by-step explanation:

Option B consists of voting and not everyone would like the outcome. Option C is making an outsider the decision maker, which can't be helpful since he / she won't have as strong opinions as the team itself. Option D is just plain wrong as it defeats the purpose of team work and deciding as one team. So, I believe option A makes the most sense

Answer:

a. 16 slug b. 3.2 ft

Step-by-step explanation:

a. Total mass of the rod

Since the linear density at a point of the rod,λ varies directly as the third power of the measure of the distance of the point form the end, x

So, λ ∝ x³

λ = kx³

Since the linear density λ = 2 slug/ft at then center when x = L/2 where L is the length of the rod,

k = λ/x³ = λ/(L/2)³ = 8λ/L³

substituting the values of the variables into the equation, we have

k = 8λ/L³

k = 8 × 2/4³

k = 16/64

k = 1/4

So, λ = kx³ = x³/4

The mass of a small length element of the rod dx is dm = λdx

So, to find the total mass of the rod M = ∫dm = ∫λdx we integrate from x = 0 to x = L = 4 ft

M = ∫₀⁴dm

= ∫₀⁴λdx

= ∫₀⁴(x³/4)dx

= (1/4)∫₀⁴x³dx

= (1/4)[x⁴/4]₀⁴

= (1/16)[4⁴ - 0⁴]

= (256 - 0)/16

= 256/16

= 16 slug

b. The center of mass of the rod

Let x be the distance of the small mass element dm = λdx from the end of the rod. The moment of this mass element about the end of the rod is xdm =  λxdx = (x³/4)xdx = (x⁴/4)dx.

We integrate this through the length of the rod. That is from x = 0 to x = L = 4 ft

The center of mass of the rod x' = ∫₀⁴(x⁴/4)dx/M where M = mass of rod

= (1/4)∫₀⁴x⁴dx/M

= (1/4)[x⁵/5]₀⁴/M

= (1/20)[x⁵]₀⁴/M

= (1/20)[4⁵ - 0⁵]/M

= (1/20)[1024 - 0]/M

= (1/20)[1024]/M

Since M = 16, we have

x' =  (1/20)[1024]/16

x' = 64/20

x' = 3.2 ft

Answer:

B. 1 + ln 2 - ln x

General Formulas and Concepts:

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle ln(\frac{2e}{x})[/tex]

Step 2: Simplify

Answer:

[tex]56[/tex] choices

Step-by-step explanation:

We know that we'll have to solve this problem with a permutation or a combination, but which one do we use? The answer is a combination because the order in which the child picks the candy does not matter.

To further demonstrate this, imagine I have 4 pieces of candy labeled A, B, C, and D. I could choose A, then C, then B or I could choose C, then B, then A, but in the end, I still have the same pieces, regardless of what order I pick them in. I hope that helps to understand why this problem will be solved with a combination.

Anyways, back to the solving! Remember that the combination formula is

[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex], where n is the number of objects in the sample (the number of objects you choose from) and r is the number of objects that are to be chosen.

In this case, [tex]n=8[/tex] and [tex]r=3[/tex]. Substituting these values into the formula gives us:

[tex]_8C_3=\frac{8!}{3!5!}[/tex]

[tex]= \frac{8*7*6*5*4*3*2*1}{3*2*1*5*4*3*2*1}[/tex] (Expand the factorials)

[tex]=\frac{8*7*6}{3*2*1}[/tex] (Cancel out [tex]5*4*3*2*1[/tex])

[tex]=\frac{8*7*6}{6}[/tex] (Evaluate denominator)

[tex]=8*7[/tex] (Cancel out [tex]6[/tex])

[tex]=56[/tex]

Therefore, the child has [tex]\bf56[/tex] different ways to pick the candies. Hope this helps!

Answer:

y=-1/7x + 12/7

Step-by-step explanation:

Start by finding the slope

m=(1-0)/(-5-2)

m=-1/7

next plug the slope and the point (-5,1) into point slope formula

y-y1=m(x-x1)

y1=1

x1= -5

m=-1/7

y- 1 = -1/7(x - -5)

y-1=-1/7(x+5)

Distribute -1/7 first

y- 1=-1/7x + 5/7

Add 1 on both sides, but since its a fraction add 7/7

y=-1/7x + (5/7+7/7)

y=-1/7x+12/7

Answer:

Step-by-step explanation:

(-5,1) (2,0)

m=(y-y)/(x-x)

m = (0-1)/2- -5)

m = -1/7

(2,0)

y-0= -1/7 (x-2)

y = -1/7x + 2/7

Answer:

BC^2=10^2+30^2

Step-by-step explanation:

Using pythagorean theorem

[tex]\\ \sf\longmapsto BC^2=10^2+30^2[/tex]

[tex]\\ \sf\longmapsto BC^2=100+300[/tex]

[tex]\\ \sf\longmapsto BC^2=400[/tex]

[tex]\\ \sf\longmapsto BC=\sqrt{400}[/tex]

[tex]\\ \sf\longmapsto BC=20[/tex]

Answer:

W=7 and L=11

Step-by-step explanation:

We have two unknowns so we must create two equations.

First the problem states that  length of a rectangle is 10 yd less than three times the width so: L= 3w-10

Next we are given the area so: L X W = 77

Then solve for the variable algebraically. It is just a system of equations.

3W^2 - 10W - 77 = 0

(3W + 11)(W - 7) = 0

W = -11/3 and/or W=7

Discard the negative solution as the width of the rectangle cannot be less then 0.

So W=7

Plug that into the first equation.

3(7)-10= 11 so L=11

Answer:

Step-by-step explanation:

Side=AC=9[tex]\sqrt{2}[/tex]

Side AB= x

Hypotenuse =CB= y

Side AB = 9[tex]\sqrt{2}[/tex]

Hypotenuse CB = 36

Answer:

0.1088 or 10.88%

Step-by-step explanation:

q = 1 - 0.35 = 0.65

P(X=2) = 12C2 × (0.35)² × (0.65)¹⁰

= 0.1088

Given:

The given expression is:

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

To find:

Part A: The expression by factoring out the greatest common factor.

Part B: Factor the entire expression completely.

Solution:

Part A:

We have,

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

Taking out the highest common factor 3y, we get

[tex]=3y(2x^2-x-8xy+4y)[/tex]

Therefore, the required expression is [tex]3y(2x^2-x-8xy+4y)[/tex].

Part B:

From part A, we have,

[tex]3y(2x^2-x-8xy+4y)[/tex]

By grouping method, we get

[tex]=3y(x(2x-1)-4y(2x-1))[/tex]

[tex]=3y(x-4y)(2x-1)[/tex]

Therefore, the required factored form of the given expression is [tex]3y(x-4y)(2x-1)[/tex].

Answer:

The unknown is 100

Step-by-step explanation:

A straight line is 180 degrees

We have two angles x, and 80

x+80 = 180

x = 180-80

x= 100

Answer:

distributive

Step-by-step explanation:

a(b + c)=ab + ac

it's distributive one

Answer:

Not a function

Domain: {3,4}

Range: {4,5}

Step-by-step explanation:

A function is a relation where each input has its own output. In other words if the x value has multiple corresponding y values then the relation is not a function

For the relation given {(3, 4), (3, 5), (4, 4), (4, 5)} the x value 3 and 4 have more than one corresponding y value therefore the relation shown is not a function

Now let's find the domain and range.

Domain is the set of x values in a relation.

The x values of the given relation are 3 and 4 so the domain is {3,4}

The range is the set of y values in a relation

The y value of the given relation include 4 and 5

So the range would be {4,5}

Notes:

The values of x and y should be written from least to greatest when writing them out as domain and range.

They should be written inside of brackets

Do not repeat numbers when writing the domain and range