(PLEASE ANSWER A, B, AND C)
Give each probability as a simplified fraction, decimal, and percent.
If the spinner below is spun once, find each probability.
A. P (4)
B. P (a number less than 10)
C. P (a prime number)

Answer:

1. He get get to school 10 minutes faster riding by hisself than ridding with Christina.

Step-by-step explanation:

Christina is 22 minutes - Luis 12 minutes = 10 minutes.

Step-by-step explanation:

wot. that's not even a qn bro

Step-by-step explanation:

the difference between the numbers are 5

so this is an arthimetic sequence with D=5

5n-7 is the formula

The amount that would be in the account kept with BPI bank in 3 years is P3,011.2640

It would 6.9422 years for the deposit to double if interest rate is 10,5%

What is future value of immediate amount?

The future value of deposit today in 3 years means its future equivalent when the amount has earned interest over the 3 years period

FV=PV*(1+r)^N

FV=future value after 3 years=unknown

PV=immediate deposit=3,000

r=annual interest rate=0.125%

N=number of years that the deposit lasted=3

FV=3000*(1+0.125%)^3

FV=P3,011.2640

Time taken to double:

FV=future value=3000*2=6000

PV=3000

r=annual interest rate=10.5%

N=number of years that it takes initial deposit to double=unknown

6000=3000*(1+10.5%)^N

6000/3000=(1.105)^N

2=(1.105)^N

take log of both sides

ln(2)=N*ln(1.105)

N=ln(2)/ln(1.105)

N=6.9422 years

Annuity:

FV=PMT*(1+r)^N-1/r

FV=future value of annuity=unknown

PMT=annual payment=20000

r=interest rate=3%

N=number of annual payments=7

FV=20000*(1+3%)^7-1/7%

FV= 153,249.2436

FV=PMT*(1+r)^N-1/r*(1+r)

FV=20000*(1+3%)^7-1/7%*(1+7%)

FV= 157,846.7209

20,000,000=1,250,000*(1+r)^20

20,000,000/1,250,000=(1+r)^20

16=(1+r)^20

(16)^1=(1+r)^20

divide indexes on both sides by 20

(16)^(1/20)=1+r

r=(16)^(1/20)-1

r=14.8698%

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The inverse of the function f(x) = (3x)^2 is f-1(x) = 1/3√x

The function is given as:

f(x) = (3x)^2

Remove the bracket in the above equation

So, we have:

f(x) = 9x^2

Express f(x) as y

So, we have

y = 9x^2

Swap the positions of x and y

So, we have

x = 9y^2

Make y the subject of the formula

y^2 = x/9

Take the square root of both sides

y = 1/3√x

Express as an inverse function

f-1(x) = 1/3√x

Hence, the inverse of the function f(x) = (3x)^2 is f-1(x) = 1/3√x

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A line has no endpoints (line FG), a line segment has two endpoints (line segment AB), and a ray has one endpoint (ray CD).

A line segment can be described as a line having two definite endpoints.

A ray is a part of a line that has just one fixed endpoint and extends in the opposite direction of the endpoint to infinity.

A line has no endpoint. It extends in opposite directions to infinity.

1. The figure that is a line segment is the green figure. (line segment AB).

2. The blue figure is a ray (ray CD)

3. The red figure is a line (line FG).

In summary, a line has no endpoints (line FG), a line segment has two endpoints (line segment AB), and a ray has one endpoint (ray CD).

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[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

Domain : All possible values of x for which a function is defined.

According to given graph, the curve extends to infinity on both sides on x - axis, so it is defined for all real values.

i.e : Domain : All real

Range : All possible values of y for x, as per the given function

As per the given graph, the graph extends to infinity on lower end, but has maximum value of 0.

i.e Range : [tex]\boxed{ \sf y < 0} [/tex]

If the interval is [0,3] then the second function whose graph is given has the greater average rate of change.

Given two functions, one is in the table and the other one is in the form of graph.

We are required to choose the function which has the greater average rate of change.

Function is basically the relationship between two or more variables that are expressed in equal to form. The values that we enter are known as part of domain and the values that we get from the function are known as part of codomain or range of the function.

If we observe the table then we will find that in the interval [0,3] there is not any change in the value of function, it is constant to be 4.

If we observe the graph then we will find that the value of function is continuously decreasing.

So, the second function has greater average rate of change over the interval [0,3].

Hence if the interval is [0,3] then the second function whose graph is given has the greater average rate of change.

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Answer:

1.2 × 10⁻⁵

Step-by-step explanation:

The exponent is negative, so the decimal was moved -5 places back, making the number a decimal.

Answer:

1.2 x 10^5

Step-by-step explanation:

All work is shown in the attached screenshot! :)

The measure of  angle m∠PZQ is 63degrees.

A point where two or more line segments meet is called a vertex.

The vertex point also forms an angle.

Therefore,

point P is in the interior of ∠OZQ,

Hence,

<OZQ = <OZP +  m∠PZQ

The following angles are given:

m∠OZQ = 125°

m∠OZP = 62°

Substitute the given values into the expression above:

125 = 62 +  m∠PZQ

m∠PZQ = 125 - 62

m∠PZQ = 63°

Therefore, the measure of  the angle m∠PZQ is 63degrees

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Interest rate compounded annually for the amount of $5,000 invested would have been worth $23,125.59 after 10 years is equal to 16.55% per year.

As given in the question,

Principal (P) = $5,000

Time (t) = 10 years

Amount = $23,125.59

[tex]r = n[(A/P)^{\frac{1}{nt}}-1]\\\\\implies r = 1[(23125.29/5000)^{\frac{1}{10} }-1]\\\\\implies r = 0.1655\\[/tex]

Convert r into percentage

r = 0.1655 × 100

 = 16.55% compounded annually

Therefore, interest rate compounded annually for the amount of $5,000 invested would have been worth $23,125.59 after 10 years is equal to 16.55% per year.

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In the picture we have to solve the individual variables A=2πr²+2πrh. We got function with r as subject is[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]

Given that,

In the picture we have to solve the individual variables

A=2πr²+2πrh

We have to find function with r as subject.

Taking A to left side we get

2πr²+2πrh-A=0

We can see the equation is in the form of quadratic equation with variable r.

So, The factor we find by using the formula

That is [tex]\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}[/tex]

Here, a=2π,b=2πh and c=-A

r=[tex]\frac{-2\pi h\pm\sqrt{(2\pi h)^{2}-4(2\pi)(-a) } }{2(2\pi)}[/tex]

r=[tex]\frac{-2\pi h\pm\sqrt{(4\pi^{2}h^{2} +8\pi a) } }{4\pi}[/tex]

r=[tex]\frac{-h}{2} \pm\frac{\sqrt{(4\pi^{2}h^{2} +8\pi a )} }{4\pi}[/tex]

r=[tex]\frac{-h}{2} \pm\sqrt{\frac{4\pi^{2}h^{2}+8\pi a }{16\pi^{2} } }[/tex]

r=[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]

Therefore, We got  function with r as subject is[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]

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Answer:

$700

Step-by-step explanation:

Since the deal is 30% off, that means that the price is now 70% of the original price.

70% of x = 490

0.7x = 490

x = 490/0.7

x = 700

Answer: $700

Answer:

x =20°

because m and n is parallel

infinitely many solutions 

Example

Simplify left side using distributive property to 6x+16. Combine like terms (2x and 4x) on the right side to 6x+16. Both sides simplify to the same expression (left side = right side) so there are infinitely many solutions. You can plug in any real number for x and the left side will always equal the right side.

The length of the field with area 119 square feet is 17 feet and the width of the feet is 7 feet.

The area is the total region or space in two dimension that is covered by a figure , surface or object. Area of a rectangle is calculated by the product of its length and width.

A rectangle has 4 sides where each pair of  opposite side is equal .The length of a rectangle is normally the longer side and the width defines the shorter side.

Given the area of the barn is 119 square feet.

Let us consider the width of the barn to be x feet. As the length is 10 feet longer than the width then we will find the length to be ( x +10 ) feet.

Area = length × width

or, 119 = x × (x+10)

or, 119 = x² +10x

or, x² +10x - 119=0

Now we will solve the quadratic equation thus formed by middle term-factorization method:

or, x² +17x - 7x - 119 = 0

or, x ( x + 17 ) - 7 ( x + 17) = 0

or, ( x + 17 ) ( x - 7 ) = 0

Now either x + 17 = 0 or x - 7 =0

Therefore either x = 7  

or, x = -17 (The width cannot be a negative value)

∴Width = 7 feet and length = 7 + 10 = 17 feet.

Therefore the length of the field is 17 feet and the width of the feet is 7 feet.

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Answer:

x ≥ 3

Step-by-step explanation:

2(x - 3) ≥ -3(-3 + x)

2x - 6 ≥ 9 - 3x

5x - 6 ≥ 9

5x ≥ 15

x ≥ 3

Answer:

x ≥ 3

Step-by-step explanation:

write, calculate, divide both sides

Answer:

Step-by-step explanation:

32.1 gb is answer

77.56

-45.46

second file would be 32.10 gb

Answer:

6325

Step-by-step explanation:

[tex]6000 \\ \: \: 300 \\ \: \: \: \: 20 \\ \: \: \: + 5[/tex]

___________

6325

Answer:

Your answer would be [tex]6325[/tex]

Step-by-step explanation:

[tex]=6325[/tex]

[tex]6000 +300+20+5[/tex]

[tex]=6325[/tex]

hopefully this helps! TwT

Answer:

x > 4

Step-by-step explanation:

2x + 2 > 10

2x > 10 - 2

2x > 8

x > 8/2

x > 4

The range of scores on the fourth test to give David a B grade for the semester is at least 73 but less than or equal to 89.

A range refers to the difference between the lowest and the highest score.

A range of scores gives two values, the lowest and the highest scores.

Scores secured by David = 81, 92, and 74

Total scores in three exams = 247

The average score for a B grade is:

80  ≤  B grade  < 90      ["at least" means ≤]

Therefore,

80 ≤ (81 + 92 + 74 + x) / 4  < 89  [assume equal weights for exams]

80 ≤ (247 + x) / 4 ≤ 89  [add values]

320 ≤ (247 + x) ≤ 356   [multiply by 4, retains a sense of inequality]

73 ≤ x < 109                   [subtracting 247 retains a sense of inequality]

Since exams are graded on 100 points, David cannot score 109, so the upper limit is put at 89.

Thus, the exam grade in the fourth test must be at least 73 but less than or equal to 89 to get a B average.

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same here 2383882828x7767=?

Answer:

The answer is yes. Since there is no x^2

Using Venn sets, the cardinalities are given as follows:

a) n(A U B) = 15.

b) n(A' U C) = 16.

c) n(A ∩ B)' = 10.

Venn amounts relates the cardinality of sets that intersect with each other.

For this problem, the sets are the ones given in this problem, A, B and C, while U is the universal set.

For this problem, the cardinalities are given as follows:

Hence:

Hence:

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Let [tex]S[/tex] denote the sum. We can first resolve the sum in [tex]m[/tex] by factorizing and decomposing into partial fractions.

[tex]\displaystyle S = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+1}}{mn^2 + mn + m^2n} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n \sum_{m=1}^\infty \frac1{m(m+n+1)} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)} \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right)[/tex]

Rewrite the [tex]m[/tex]-summand as a definite integral. Interchange the integral and sum, and evaluate the resulting geometric sums.

[tex]\displaystyle \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right) = \sum_{m=1}^\infty \int_0^1 \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{m=1}^\infty \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \frac{1 - x^{n+1}}{1 - x} \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{\ell=0}^n x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \int_0^1 x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \frac1{\ell+1} \\\\ ~~~~~~~~ = \sum_{\ell=1}^{n+1} \frac1\ell = H_{n+1}[/tex]

where

[tex]H_n = \displaystyle \sum_{\ell=1}^n \frac1\ell = 1 + \frac12 + \frac 13 + \cdots + \frac1n[/tex]

is the [tex]n[/tex]-th harmonic number. The generating function will be useful:

[tex]\displaystyle \sum_{n=1}^\infty H_n x^n = -\frac{\ln(1-x)}{1-x}[/tex]

To evaluate the remaining sum to get [tex]S[/tex], let

[tex]\displaystyle f(x) = \sum_{n=1}^\infty \frac{H_{n+1}}{n(n+1)} x^{n+1}[/tex]

and observe that [tex]S=\lim\limilts_{x\to-1^+} f(x)[/tex], which I'll abbreviate to [tex]f(-1)[/tex]. Differentiating twice, we have

[tex]\displaystyle f'(x) = \sum_{n=1}^\infty \frac{H_{n+1}}n x^n[/tex]

[tex]\displaystyle f''(x) = \sum_{n=1}^\infty H_{n+1} x^n[/tex]

[tex]\displaystyle \implies f''(x) = -\frac{\ln(1-x)}{x^2(1-x)} - \frac1x[/tex]

By the fundamental theorem of calculus, noting that [tex]f(0)=f'(0)=0[/tex], we have

[tex]\displaystyle \int_{-1}^0 f'(x) \, dx = f(0) - f(-1) \implies f(-1) = -\int_{-1}^0 f'(x) \, dx[/tex]

[tex]\displaystyle \int_x^0 f''(x) \, dx = f'(0) - f'(x) \implies f'(x) = -\int_x^0 f''(t) \, dt[/tex]

[tex]\displaystyle \implies S = f(-1) = \int_{-1}^0 \int_x^0 \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dt \, dx[/tex]

Change the order of the integration, and substitute [tex]t=-u[/tex].

[tex]S = \displaystyle \int_{-1}^0 \int_{-1}^t \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dx \, dt \\\\ ~~~ = - \int_{-1}^0 \left(\frac{(1+t) \ln(1-t)}{t^2(1-t)} + \frac1t + 1\right) \, dt \\\\ ~~~ = -1 - \int_{-1}^0 \left(\left(\frac2{1-t} + \frac2t + \frac1{t^2}\right) \ln(1-t) + \frac1t\right) \, dt \\\\ ~~~ = -1 - \int_0^1 \left(\left(\frac2{1+u} - \frac2u + \frac1{u^2}\right) \ln(1+u) - \frac1u\right) \, du[/tex]

For the remaining integrals, substitute and use power series.

[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}{1+u} \, du = \int_0^1 \ln(1+u) d(\ln(1+u)) = \frac{\ln^2(2)}2[/tex]

[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}u \, du = - \int_0^1 \frac1u \sum_{k=1}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}k \int_0^1 u^{k-1} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = \frac{\pi^2}{12}[/tex]

[tex]\displaystyle \int_0^1 \frac{\ln(1+u) - u}{u^2} \, du = - \int_0^1 \frac1{u^2} \left(\sum_{k=1}^\infty \frac{(-u)^k}k + u\right) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\int_0^1 \frac1{u^2} \sum_{k=2}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=2}^\infty \frac{(-1)^k}k \int_0^1 u^{k-2} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\sum_{k=2}^\infty \frac{(-1)^k}{k(k-1)} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1)} = 1 - 2\ln(2)[/tex]

Tying everything together, we end up with

[tex]S = -1 - \left(2 \cdot \dfrac{\ln^2(2)}2 - 2 \cdot \dfrac{\pi^2}{12} + (1-2\ln(2))\right) \\\\ ~~~ = \boxed{\frac{\pi^2}6 - 2 + 2\ln(2) - \ln^2(2)}[/tex]

Answer: c

Step-by-step explanation:

Answer:

10.75

Step-by-step explanation:

$118.25 / 11

We're finding the cost of each pizza. So, If a Coach bought 11 Pizza's for $118.25, We need to find how much x is. x = amount of cost per pizza.  Divide $118.25 by 11 to get $10.75.

Each Pizza Costs $10.75.

Answer:

A graphing calculator can help you.

Here is the graph, it is in the photo.

Answer:

[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]

Step-by-step explanation:

Given function:

[tex]f(x)=\dfrac{e^x}{\sqrt{e^{2x}+1}}[/tex]

The domain of the given function is unrestricted:  {x : x ∈ R}

The range of the given function is restricted:  {f(x) : 0 < f(x) < 1}

To find the inverse of a function, swap x and y:

[tex]\implies x=\dfrac{e^y}{\sqrt{e^{2y}+1}}[/tex]

Rearrange the equation to make y the subject:

[tex]\implies x\sqrt{e^{2y}+1}=e^y[/tex]

[tex]\implies x^2(e^{2y}+1)=e^{2y}[/tex]

[tex]\implies x^2e^{2y}+x^2=e^{2y}[/tex]

[tex]\implies x^2e^{2y}-e^{2y}=-x^2[/tex]

[tex]\implies e^{2y}(x^2-1)=-x^2[/tex]

[tex]\implies e^{2y}=-\dfrac{x^2}{x^2-1}[/tex]

[tex]\implies \ln e^{2y}= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies 2y \ln e= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies 2y(1)= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies 2y= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies y= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

Replace y with f⁻¹(x):

[tex]\implies f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

The domain of the inverse of a function is the same as the range of the original function.  Therefore, the domain of the inverse function is restricted to {x : 0 < x < 1}.

Therefore, the inverse of the given function is:

[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]