A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 70 and a standard deviation of 10.How many different samples of size 5 can be taken from a finite population of size 10?

Answer:

volume of hemisphere =1912.26cm^3

Step-by-step explanation:

Let y = x + 22 and dy = dx, so the integral becomes

[tex]\displaystyle \int_3^9 \frac{\mathrm dx}{(x+21)\sqrt{x+22}} = \int_{25}^{31} \frac{\mathrm dy}{(y-1)\sqrt{y}}[/tex]

Now let z = √y, so that z ² = y. Then 2z dz = dy, and the integral becomes

[tex]\displaystyle \int_3^9 \frac{\mathrm dx}{(x+21)\sqrt{x+22}} = \int_{\sqrt{25}}^{\sqrt{31}} \frac{2z}{(z^2-1)z} \\\\ = \int_5^{\sqrt{31}} \frac{2}{z^2-1}\,\mathrm dz[/tex]

Expand the integrand into partial fractions:

[tex]\dfrac{2}{z^2-1} = \dfrac1{z-1}-\dfrac1{z+1}[/tex]

Then we have

[tex]\displaystyle \int_3^9 \frac{\mathrm dx}{(x+21)\sqrt{x+22}} = \int_5^{\sqrt{31}}\left(\frac1{z-1}-\frac1{z+1}\right)\,\mathrm dz \\\\ = \left(\ln|z-1|-\ln|z+1|\right)\bigg|_5^{\sqrt{31}} \\\\ =\left[\ln\left|\frac{z-1}{z+1}\right|\right]\bigg|_5^{\sqrt{31}} \\\\ =\ln\left(\frac{\sqrt{31}-1}{\sqrt{31}+1}\right) - \ln\left(\frac{4}{6}\right) \\\\ =\ln\left(32-2\sqrt{31}\right) - \ln\left(\frac23\right) \\\\ =\boxed{\ln\left(48-3\sqrt{31}\right)}[/tex]

Step-by-step explanation:

check it out.

Hope it helps.

Answer:

question is not proper

Step-by-step explanation:

question is

Answer:

D.

Step-by-step explanation:

D. This contains an x^2 and is called a parabola ( curved line like a U).

Answer:D

Step-by-step explanation: d is the correct answer

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Just a correction, the characteristic roots of the equation are [tex]y = 7[/tex] and [tex]y = -\frac{1}{2}[/tex], thus, we should test for:

[tex]y = Ae^{7x} + Be^{-\frac{x}{2}}[/tex]

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Question a:

[tex]y = Ae^{7x} + Be^{-\frac{x}{2}}[/tex]

[tex]y^{\prime} = 7Ae^{-7x} - \frac{1}{2}Be^{-\frac{x}{2}}[/tex]

[tex]y^{\prime\prime} = 49Ae^{-7x} + \frac{1}{4}Be^{-\frac{x}{2}}[/tex]

[tex]2y^{\prime\prime} - 13y^{\prime} - 7y = 0[/tex]

[tex]2(49Ae^{-7x} + \frac{1}{4}Be^{-\frac{x}{2}}) - 13(7Ae^{-7x} - \frac{1}{2}Be^{-\frac{x}{2}}) - 7(Ae^{7x} + Be^{-\frac{x}{2}}) = 0[/tex]

[tex]98Ae^{-7x} + \frac{1}{2}Be^{\frac{x}{2}} - 91Ae^{-7x} + \frac{13}{2}e^{-\frac{x}{2}} - 7Ae^{7x} - 7Be^{-\frac{x}{2}} = 0[/tex]

[tex]98Ae^{-7x} - 91Ae^{-7x} - 7Be^{-\frac{x}{2}} + \frac{1}{2}Be^{\frac{x}{2}}  + \frac{13}{2}e^{-\frac{x}{2}} - 7Be^{-\frac{x}{2}} = 0[/tex]

[tex]0A + 0B = 0[/tex]

[tex]0 = 0[/tex], thus, we found the identity, and for each constant A and B, the following is a solution.

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Question b:

[tex]y = Ae^{7x} + Be^{-\frac{x}{2}}[/tex]

[tex]A + B = 3 \rightarrow B = 3 - A[/tex]

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[tex]y^{\prime} = 7Ae^{-7x} - \frac{1}{2}Be^{-\frac{x}{2}}[/tex]

[tex]7A - \frac{1}{2}B = -5[/tex]

Using [tex]B = 3 - A[/tex]

[tex]7A - \frac{3}{2} + \frac{A}{2} = -5[/tex]

[tex]\frac{14A}{2} + \frac{A}{2} = -\frac{10}{2} + \frac{3}{2}[/tex]

[tex]\frac{15A}{2} = -\frac{7}{2}[/tex]

[tex]15A = -7[/tex]

[tex]A = -\frac{7}{15}[/tex]

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Then, B is given by:

[tex]B = 3 - A = 3 - (-\frac{7}{15}) = \frac{45}{15} + \frac{7}{15} = \frac{52}{15}[/tex]

Thus, the values are: [tex]A = -\frac{7}{15}, B = \frac{52}{15}[/tex]

A similar problem is given at https://brainly.com/question/2456414

Answer:

Step-by-step explanation:

Answer:

74

Step-by-step explanation:

for this question you have to use the formula of the nth term which is Tn=a+(n-1)d,

T21=-6+(21-1)4

=-6+(20)4

=-6+80

=74

I hope this helps

Answer:

The correct choice is option B. y+1/y-1

Step-by-step explanation:

Answer:

9z-2y-3x

Step-by-step explanation:

(6z-4y+2x)+(3z-5x+2y)

Combine like terms

6z+3z-4y+2y+2x-5x

9z-2y-3x

Answer:

-3x - 2y + 9z

Step-by-step explanation:

Since you are adding two polynomials, the parentheses are unnecessary, so you can just drop them. Then combine like terms. Like terms are terms that have the same variable part such as 6z and 3z.

(6z - 4y + 2x) + (3z - 5x + 2y) =             Original problem.

= 6z - 4y + 2x + 3z - 5x + 2y                 Drop unnecessary parentheses.

= 2x - 5x - 4y + 2y + 6z + 3z                 Rearrange terms to group like terms.

= -3x - 2y + 9z                                        Combine each set of like terms.

Answer: -3x - 2y + 9z

Answer:

A square is a retangle

Step-by-step explanation:

Because a squre is a rectangle

Answer:

first choice..

first choice

Answer:

FIRST CHOISE FIRST CHOISE

Step-by-step explanation:

Answer: Volume is 4.19

Step-by-step explanation:

The equation for the volume of a cone is πr^2(h/3)

r = 1

h = 4

Fill in values into the equation

π(1)^2(4/3) = 4.19

Answer:

y = 6x ^2 - 6x + 9

Step-by-step explanation:

hopefully it's visible

:)

Step-by-step explanation:

Any number that can be divided by 2 without having remainder is considered an even number.

I hope it helped U

stay safe stay happy

Answer:

x = 50

Step-by-step explanation:

When two secants intersect in the interior of a circle, the angles formed are the average of the arc an angle and its vertical intercept. In this case, our angle, 73, should be the average of x and 96. We can translate this to an equation and solve:

[tex]\frac{x+96}{2} = 73[/tex]

x + 96 = 146

x = 50

Answer:

2) - 12

3) - 1

4)y = 8

5)r = 4

6)x= -7/29

Hope it helps you

9514 1404 393

Answer:

  2^6 = 64

Step-by-step explanation:

Each hour the initial value (1) is multiplied by 2 (again). So the number of times 2 multiplies the initial value is the number of passing hours (t).

The number of cells after t hours is ...

  c(t) = 1×2^t = 2^t

After 6 hours, this is

  c(6) = 2^6 = 64 . . . . cells after 6 hours

_____

Additional comment

As you know, multiplication simplifies repeated addition.

  x + x + x = 3x

In a similar way, exponentiation simplifies repeated multiplication.

 x·x·x = x³ . . . . (written as x^3 in plain text)

In this problem, we're repeatedly multiplying by 2. After we do that t times, the initial value is multiplied by 2^t.

Answer: $ 39.60

Step-by-step explanation:

36 + (36*0.10) = 39.60

Answer:

x^4

Step-by-step explanation:

x^(2/5) ^10

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

x^(2/5*10)

x^4

Answer:

[tex]{x}^{4} [/tex]

Step-by-step explanation:

[tex] ({x}^{ \frac{2}{5} } ) ^{10} \\ {x}^{ \frac{2}{5} \times 10} \\ {x}^{ \frac{20}{5} } \\ {x}^{4} [/tex]

Answer:

B

Step-by-step explanation:

Answer:

Step-by-step explanation:

vxcvxcvxcvxcvxcvcxvxcbcvbcvnxcgjfgjfgjghjghjghjghj

The revenue function is y₁ = –4x² + 400x, the cost function is y₁ = x² – 100x + 8000, and the break-even number of units is 20 or 80.

It is a polynomial that is equal to zero. Polynomial of variable power 2, 1, and 0 terms are there. Any equation having one term in which the power of the variable is a maximum of 2 then it is called a quadratic equation.

Jua Kali Products Ltd has been in operation for the last 10 years.

Its annual revenue and cost functions take the form of quadratic functions.

The following data was obtained from the records of the company.

Year          Unit Sold            Revenue             Cost

2017              5                        1900                 7525

2018             10                       3600                7100

2019             15                        5100                6725

We know that the quadratic equation is given as

[tex]\rm y = ax^2 + bx + c[/tex]

Let y₁ be the revenue function, y₂ be the cost function and x be the unis sold.

Then the revenue function will be

1900 = 25a + 5b + c  ...i

3600 = 100a + 10b + c  ...ii

5100 = 225a + 15b + c  ...iii

From equations (i), (ii), and (iii), we have

a = –4, b = 400, and c = 0

Then the revenue function will be

y₁ = –4x² + 400x

Similarly, the cost function will be

7525 = 25a + 5b + c  ...1

7100 = 100a + 10b + c  ...2

6725 = 225a + 15b + c  ...3

From equations 1, 2, and 3, we have

a = 1, b = –100, and c = 8000

Then the cost function will be

y₁ = x² – 100x + 8000

For the break-even units, the cost function and the revenue function will be equal. Then we have

[tex]\begin{aligned} x^2 -100x + 8000 &= -4x^2 + 400x\\\\5x^2 -500x + 8000 &= 0\\\\x^2 - 100x + 1600 &= 0\\\\x^2 - 80 x - 20x + 1600 &= 0\\\\x(x-80) - 20 (x-80) &= 0\\\\(x-80)(x-20) &= 0\\\\x &= 20, 80 \end{aligned}[/tex]

More about the quadratic equation link is given below.

https://brainly.com/question/2263981

Answer:

9

Step-by-step explanation:

use Tangent-chord formula for finding the arc knowing E

x=1/2n

4x+19=1/2n

n=8x+38

both arcs = 360

28x-2+8x+38=360

x=9

Answer: 1,526.04 yd³ (this is what i got on my calculator others may vary use any rounding you need)

VOLUME of CONE is: V= Pir² h / 3

Pi= 3.14

r= 9

h= 18

V= (3.14) x (9)² x (18)

9²= 81

3.14 x 81= 254.34 x 18= 4,578.12

4,578.12 / 3= 1,526.04 <-- answer

Answer:

∠ A ≈ 79.0°

Step-by-step explanation:

Using the Cosine rule to find the angle

cosA = [tex]\frac{b^2+c^2-a^2}{2bc}[/tex]

        = [tex]\frac{15^2+29^2-30^2}{2(15)(29)}[/tex]

        = [tex]\frac{225+841-900}{870}[/tex]

        = [tex]\frac{166}{870}[/tex] , then

A = [tex]cos^{-1}[/tex] ([tex]\frac{166}{870}[/tex] ) ≈ 79.0° ( to the nearest tenth )

Answer:

16 once is the better one.

Answer: 12-ounce bag is better by $0.02 per ounce

Concept:

When coming across questions that ask for a comparison between prices, we should make the final unit [price per object].

In finding [price per object], simply do [Total price / number of objects].

Solve:

A 12-ounce bag of rice costs $4.08

Total price / number of objects = 4.08 / 12 = $0.34 per ounce

A 16-ounce bag of rice costs $5.76

Total price / number of objects = 5.76 / 16 = $0.36 per ounce

$0.36 - $0.34 = $0.02

$0.34 < $0.36, therefore, 12-ounce bag is better by $0.02 per ounce.

Hope this helps!! :)

Please let me know if you have any questions

Step-by-step explanation: Let's begin by finding the

greatest common factor for the numbers 65 and 39.

I would make a factor tree and break up 65 and 39.

So 65 is 13 x 5 and 39 is 13 x 3.

Since the 13's match up, the greatest

common factor between 65 and 39 is 13.

For the variables, we use the smallest power on each of them.

So we use a^3 and b^4 to get 13a^3b^4 as our GCF.