If there are 25 students in a class in which 5 of the 11 guys wear glasses and 6 out of the 14 girls wear glasses- what is the probability that one of the students in the class is a guy that he wears glasses?

Answers

Answer 1

Answer:

6 out of 25

Step-by-step explanation:


Related Questions

Find the equation of a parabola that has a vertex (3,5) and passes through the point (1,13).
Oy= -27 - 3)' +5
Oy=2(x + 3) - 5
Oy=2(0 - 3)' + 5
Oy= -3(2 – 3) + 5
PLEASE HELP ME!!

Answers

Answer:

y = 2(x - 3)² + 5

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (3, 5), thus

y = a(x - 3)² + 5

To find a substitute (1, 13) into the equation

13 = a(1 - 3)² + 5 ( subtract 5 from both sides )

8 = 4a ( divide both sides by 4 )

a = 2, then

y = 2(x - 3)² + 5 ← equation of parabola in vertex form

please help with this

Answers

Answer:

[tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \rightt)+C[/tex]

Step-by-step explanation:

We are given the graph of r = cos( θ ) + sin( 2θ ) so that we are being asked to determine the integral. Remember that [tex]\:r=cos\left(\theta \right)+sin\left(2\theta \right)[/tex] can also be rewritten as [tex]\int \cos \left(\theta \right)+\sin \left(2\theta \right)d\theta \right[/tex].

Let's apply the functional rule [tex]\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx[/tex],

[tex]\int \cos \left(\theta \right)+\sin \left(2\theta \right)d\theta \right[/tex] = [tex]\int \cos \left(\theta \right)d\theta \right+\int \sin \left(2\theta \right)d\theta \right[/tex]

At the same time [tex]\int \cos \left(\theta \right)d\theta \right=\sin \left(\theta \right)[/tex] = [tex]sin( \theta \right ))[/tex], and [tex]\int \sin \left(2\theta \right)d\theta \right[/tex] = [tex]-\frac{1}{2}\cos \left(2\theta \right)[/tex]. Let's substitute,

[tex]\int \cos \left(\theta \right)d\theta \right+\int \sin \left(2\theta \right)d\theta \right[/tex] = [tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \right)[/tex]

And adding a constant C, we receive our final solution.

[tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \rightt)+C[/tex] - this is our integral

Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answers down to the nearest whole number.)a. m = 12, n = 15, s1 = 4.0, s2 = 6.0b. m = 12, n = 21, s1 = 4.0, s2 = 6.0c. m = 12, n = 21, s1 = 3.0, s2 = 6.0d. m = 10, n = 24, s1 = 4.0, s2 = 6.0

Answers

Answer:

Part a ) The degrees of freedom for the given two sample non-pooled t test is 24

Part b ) The degrees of freedom for the given two sample non-pooled t test is 30

Part c ) The degrees of freedom for the given two sample non-pooled t test is 30

Part d ) The degrees of freedom for the given two sample non-pooled t test is 25

Step-by-step explanation:

Degrees of freedom for a non-pooled two sample t-test is given by;

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Now given the information;

a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0

we substitute

Δf =  {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}

Δf  = 30184 / 1241

Δf  = 24.3223 ≈ 24 (down to the nearest whole number)

b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 56320 / 1871

Δf = 30.1015 ≈ 30 (down to the nearest whole number)

c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 29095 / 949

Δf = 30.6585 ≈ 30 (down to the nearest whole number)

d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}

Δf = 1044 / 41  

Δf = 25.4634 ≈ 25 (down to the nearest whole number).

Question 15 please and i will mark the brainliest!!! And thank you to whoever answers

Answers

Answer: C) 12

Explanation:

We have 4 options for the first choice and 3 options for the next. So there are 4*3 = 12 different combos possible. The tree diagram below shows 12 different paths to pick from. For instance, the right-most path has us pick the number 4 and the color yellow.

The double number lines show the ratio of cups to gallons. How many cups are in 333 gallons? _____ cups

Answers

Answer:

5328 cups.

Step-by-step explanation:

Given that 333 gallons

We know that

1 gallons = 16 cups

1 cups = 0.0625 gallons

Therefore,from the above conversion we can say that

Now by putting the values in the above conversion

333 gallons = 16 x 333 cups

333 gallons = 5328 cups

So , we can say that 333 gallons is equal to 5328 cups.

Thus the answer will be 5328 cups.

Answer:

48 cups(BTW he meant 33 galons, IVE had this before). lol you need to put the double number line image. first u have to divide 64/4 to get 16, Then it says "How many cups are in 3 gallons". There fore, U multiply 16 to 3 to get ur answer "48".

Emily made a pot of cream of pumpkin soup for thanksgiving dinner. She put 5
cups of cream in the soup. She poured the soup into 24 small soup bowls. How
much cream (measured in oz.) is used for each small bowl of soup?

Answers

Answer:

1 2/3 ounces in each bowl

Step-by-step explanation:

We need to convert 5 cups to ounces

1 cup = 8 ounces

5 cups = 5*8 = 40 ounces

We divide the 40 ounces into 24 bowls

40 ounces / 24 bowl

5/3 ounces per bowl

1 2/3 ounces in each bowl

Answer:

each bowl can contain 5/3 oz. of soup.

Step-by-step explanation:

1 cup = 8 oz.

                   8 oz.

5 cups x --------------  =  40 oz.

                    1 cup

to get the measurement of each bowl,

40 oz. divided into 24 bowls.

therefore, each bowl can contain 5/3 oz. of soup.

When proving a statement using mathematical induction, part of the process is assuming that the statement is true for the nth case. (True or False).

Answers

Answer:

True

Step-by-step explanation:

We assume that is true for the nth case and prove it for the n+1 case

and show that it is true for the case when n=1

For a certain casino slot machine, the odds in favor of a win are given as 17 to 83. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.

Answers

Step-by-step explanation:

83P (E)=17-17P (E),

P (E)=17/100=0.17

Can somebody please solve this problem for me!

Answers

Answer:

x = 200.674

Step-by-step explanation:

tan∅ = opposite/adjacent

Step 1: Find length of z

tan70° = 119/z

ztan70° = 119

z = 119/tan70°

z = 43.3125

Step 2: Find length z + x (denoted as y)

tan26° = 119/y

ytan26° = 119

y = 119/tan26°

y = 243.986

Step 3: Find x

y - z = x

243.986 - 43.3125 = x

x = 200.674

What is the approximate area of the unshaded region under the standard normal curve below? Use the portion of the standard normal table given to help answer the question.

A normal curve with a peak at 0 is shown. The area under the curve shaded is 1 to 2.

z
Probability
0.00
0.5000
1.00
0.8413
2.00
0.9772
3.00
0.9987
0.14
0.16
0.86
0.98

Answers

Answer:

0.14

Step-by-step explanation:

The z score is a score used in statistics to determine by how many standard deviations ti the raw score above or below the mean. If the raw score is above the mean then the z score is positive while If the raw score is below the mean then the z score is negative, It is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

From the normal distribution table, The area under the curve shaded is 1 to 2 = P(1 < z < 2) = P(z < 2) - P(z < 1) = 0.9772 - 0.8413 = 0.1359 ≈ 0.14

The area under the curve shaded is 1 to 2 is 0.14

What are probabilities?

Probabilities are used to determine the chances of an event

The shaded region represents the probability of the z-scores

The shaded region 1 to 2 is represented as:

P(1 < z < 2) =

Using the probability of z-score, we have the formula

P(1 < z < 2) = P(z < 2) - P(z < 1)

From the given standard normal table:

P(z < 2) = 0.9772

P(z < 1) = 0.8413

So, we have:

P(1 < z < 2) = 0.9772 - 0.8413

P(1 < z < 2) = 0.1359

Approximate

P(1 < z < 2) = 0.14

Hence, the area under the curve shaded is 1 to 2 is 0.14

Read more about normal distribution at:

https://brainly.com/question/4079902

Identify each x-value at which the slope of the tangent line to the function f(x) = 0.2x^2 + 5x − 12 belongs to the interval (-1, 1).

Answers

Answer:

Step-by-step explanation:

Hello, the slope of the tangent is the value of the derivative.

f'(x) = 2*0.2x + 5 = 0.4x + 5

So we are looking for

[tex]-1\leq f'(x) \leq 1 \\ \\<=> -1\leq 0.4x+5 \leq 1 \\ \\<=> -1-5=-6\leq 0.4x \leq 1-5=-4 \\ \\<=> \dfrac{-6}{0.4}\leq 0.4x \leq \dfrac{-4}{0.4} \\\\<=> \boxed{-15 \leq x\leq -10}[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Using derivatives, it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval is (-15,-10).

What is the slope of the tangent line to a function f(x) at point x = x_0?

It is given by the derivative at x = x_0, that is:

m = f'(x_0)

In this problem, the function is:

f(x) = 0.2x^2 + 5x − 12

Hence the derivative is:

f'(x) = 0.4x + 5

For a slope of -1, we have that,

0.4x + 5 = -1

0.4x = -6

x = -15.

For a slope of 1, we have that,

0.4x + 5 = 1.

0.4x = -4

x = -10

Hence it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval is (-15,-10).

More can be learned about derivatives and tangent lines at;

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The size of a television is the length of the diagonal of its screen in inches. The aspect ratio of the screens of older televisions is 4:3, while the aspect ratio of newer wide-screen televisions is 16:9. Find the width and height of an older 35-inch television whose screen has an aspect ratio of 4:3.

Answers

Answer:

The Width = 28 inches

The Height = 21 inches

Step-by-step explanation:

We are told in the question that:

The width and height of an older 35-inch television whose screen has an aspect ratio of 4:3

Using Pythagoras Theorem

Width² + Height² = Diagonal²

Since we known that the size of a television is the length of the diagonal of its screen in inches.

Hence, for this new TV

Width² + Height² = 35²

We are given ratio: 4:3 as aspect ratio

Width = 4x

Height = 3x

(4x)² +(3x)² = 35²

= 16x² + 9x² = 35²

25x² = 1225

x² = 1225/25

x² = 49

x = √49

x = 7

Hence, for the 35 inch tv set

The Width = 4x

= 4 × 7

= 28 inches.

The Height = 3x

= 3 × 7

= 21 inches

Question 2 Rewrite in simplest radical form 1 x −3 6 . Show each step of your process.

Answers

Answer:

√(x)

Step-by-step explanation:

(1)/(x^-(1/2)) that's 3 goes into -3 leaving 1 and goes into 6 leaving 2

1/2 is same as 2^-1

so therefore we can simplify the above as

x^-(-1/2)

x^(1/2)

and 4^(1/2)

is same as √(4)

so we conclude as

√(x)

I need help on this question, can someone please answer it correctly?

Answers

Answer:the one area < with line underneath then -4

St-by-step explanation: I’m pretty sure this is correct

Answer:

[tex] \boxed{x \leqslant - 4}[/tex]

Step-by-step explanation:

[tex] \mathrm{16x - 7 \leqslant - 71}[/tex]

Move constant to Right hand side and change its sign

[tex] \mathrm{16x \leqslant - 71 + 7}[/tex]

Calculate

[tex] \mathrm{16x \leqslant - 64}[/tex]

Divide both sides of the equation by 16

[tex] \mathrm{ \frac{16x}{16} \leqslant \frac{ - 64}{16} }[/tex]

Calculate

[tex] \mathrm{x \leqslant - 4}[/tex]

Hope I helped!

Best regards!

You are starting a sock company. You must determine your costs to manufacture your product. The start-up cost is $2000 (which helps you purchase sewing machines). Material and labor is $2.50 per pair of socks.

a. Write an equation to model your company’s cost for manufacturing the socks. (i.e. y=mx+b)
b. Which variable represents the domain? Explain your answer.
c. What is the domain for this situation?
d. Which variable represents the range? Explain your answer.
e. What is the range for this situation?
f. Using your equation, what would be the cost of manufacturing 25 pairs of socks?
g. How many socks could you make with $2500?
h. Create a coordinate graph on a sheet of paper to represent this situation. Describe the graph. Include the dimensions you would use for the x and y axes.
PLS HELP ASAP!

Answers

a. y = 2.5x + 2000

b. The variable x represents the domain because the domain is the range of the possible x values.

c. x ≥ 0

d. The variable y represents the range because the range is the range of the possible y values.

e. y ≥ 2000

f. y = 2.5(25) + 2000

  y = 62.5 + 2000

  y = $2062.50

g. 2500 = 2.5x + 2000

   2.5x = 500

   x = 200

h. I am sorry I cannot make the graph but hopefully you can figure out how to make it using the info I have given in the above parts of the problem :)

A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after submersion is 120°F. After 1 min submerged, the temperature has cooled to 100°F. (Let y be measured in degrees Fahrenheit, and t be measured in seconds.) (a) Determine the cooling constant k. k = s−1 (b) What is the differential equation satisfied by the temperature y(t)? (Use y for y(t).) y'(t) = (c) What is the formula for y(t)? y(t) = (d) Determine the temperature of the bar at the moment it is submerged. (Round your answer to one decimal place.)

Answers

Answer:

a.  k = -0.01014 s⁻¹

b.  [tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]

c.  [tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]

d.  y(t) = 130.485°F

Step-by-step explanation:

A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after submersion is 120°F. After 1 min submerged, the temperature has cooled to 100°F.

(Let y be measured in degrees Fahrenheit, and t be measured in seconds.)

We are to determine :

a.  Determine the cooling constant k. k = s−1

By applying the new law of cooling

[tex]\dfrac{dT}{dt} = k \Delta T[/tex]

[tex]\dfrac{dT}{dt} = k(T_1-T_2)[/tex]

[tex]\dfrac{dT}{dt} = k (T - 60)[/tex]

Taking the integral.

[tex]\int \dfrac{dT}{T-60} = \int kdt[/tex]

㏑ (T -60) = kt + C

T - 60 = [tex]e^{kt+C}[/tex]

[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]

After 20 seconds, the temperature of the bar submersion is 120°F

T(20) = 120

From equation (1) ,replace t = 20s and T = 120

[tex]120 = 60 + C_1 e^{20 \ k}[/tex]

[tex]120 - 60 = C_1 e^{20 \ k}[/tex]

[tex]60 = C_1 e^{20 \ k} --- (2)[/tex]

After 1 min i.e 60 sec , the temperature  = 100

T(60) = 100

From equation (1) ; replace t = 60 s and T = 100

[tex]100 = 60 + c_1 e^{60 \ t}[/tex]

[tex]100 - 60 =c_1 e^{60 \ t}[/tex]

[tex]40 =c_1 e^{60 \ t} --- (3)[/tex]

Dividing equation (2) by (3) , we have:

[tex]\dfrac{60}{40} = \dfrac{C_1e^{20 \ k } }{C_1 e^{60 \ k}}[/tex]

[tex]\dfrac{3}{2} = e^{-40 \ k}[/tex]

[tex]-40 \ k = In (\dfrac{3}{2})[/tex]

- 40 k = 0.4054651

[tex]k = - \dfrac{0.4054651}{ 40}[/tex]

k = -0.01014 s⁻¹

 

b. What is the differential equation satisfied by the temperature y(t)?

Recall that :

[tex]\dfrac{dT}{dt} = k \Delta T[/tex]

[tex]\dfrac{dT}{dt} = \dfrac{- In (\dfrac{3}{2})}{40}(T-60)[/tex]

Since y is the temperature of the body , then :

[tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]

(c) What is the formula for y(t)?

From equation (1) ;

where;

[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]

Let y be measured in degrees Fahrenheit

[tex]y(t) = 60 + C_1 e^{-\dfrac{In (\dfrac{3}{2})}{40}t}[/tex]

From equation (2)

[tex]C_1 = \dfrac{60}{e^{20 \times \dfrac{-In(\dfrac{3}{2})}{40}}}[/tex]

[tex]C_1 = \dfrac{60}{e^{-\dfrac{1}{2} {In(\dfrac{3}{2})}}}[/tex]

[tex]C_1 = \dfrac{60}{e^ {In(\dfrac{3}{2})^{-1/2}}}}[/tex]

[tex]C_1 = \dfrac{60}{\sqrt{\dfrac{2}{3}}}[/tex]

[tex]C_1 = \dfrac{60 \times \sqrt{3}}{\sqrt{2}}}[/tex]

[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]

(d) Determine the temperature of the bar at the moment it is submerged.

At the moment it is submerged t = 0

[tex]\mathbf{y(0) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ 0}{40}}}[/tex]

[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} }[/tex]

y(t) = 60 + 70.485

y(t) = 130.485°F

what is the distance between the first and third quartiles of a data set called?

Answers

Answer:

Interquartile range is the distance between the first and third of a data.

Step-by-step explanation:

Hope it will help you :)

Express the quotient of z1 and z2 in standard form given that [tex]z_{1} = -3[cos(\frac{-\pi }{4} )+isin(\frac{-\pi }{4} )][/tex] and [tex]z_{2} = 2\sqrt{2} [cos(\frac{-\pi }{2} )+isin(\frac{-\pi }{2} )][/tex]

Answers

Answer:

Solution : [tex]-\frac{3}{4}-\frac{3}{4}i[/tex]

Step-by-step explanation:

[tex]-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right][/tex]

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

[tex]\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}[/tex]

=[tex]-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)[/tex] ÷ [tex]2\sqrt{2}\left(0-1\right)i[/tex]

= [tex]3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right)[/tex] ÷ [tex]-2\sqrt{2}i[/tex]

= [tex]\frac{3\left(1-i\right)}{\sqrt{2}}[/tex]÷ [tex]2\sqrt{2}i[/tex] = [tex]-3-3i[/tex] ÷ [tex]4[/tex] = [tex]-\frac{3}{4}-\frac{3}{4}i[/tex]

As you can see your solution is the last option.

What is the value of 20 + 3 (7 + 4) + 5 + 2 (7 + 9)?

Answers

Answer:

90

Step-by-step explanation:

Answer:

90

Step-by-step explanation:

Here is the equation

[tex]20+3\times(7+4)+5+2\times(7+9)[/tex]

In the order of operations parentheses go first so we get

[tex]20+3\times11+5+2\times16[/tex]

Next we do the multiplication

[tex]20+33+5+32\\[/tex]

And finally we add them all up

[tex]20+33+5+32=90\\[/tex]

Thus, 90 is the answer of [tex]20+3\times(7+4)+5+2\times(7+9)[/tex] or [tex]20+3(7+4)+5+2(7+9)[/tex]

You drive 15 miles in 0.1hours . How fast did you travel if 8=d/t

Answers

Answer:

150

Step-by-step explanation:

[tex]distance = 15 miles\\time = 0.1 hours\\\\Speed = \frac{Distance}{time}\\ Speed = \frac{15}{0.1}\\ Speed =150[/tex]

Answer:

[tex]150mph[/tex]

Step-by-step explanation:

Given:

s=15miles

t=0.1hours

Required:

v=?

Formula:

[tex]v = \frac{s}{t} [/tex]

Solution:

[tex]v = \frac{s}{t} = \frac{15m}{0.1h} = \frac{150m}{1h} = 150mph[/tex]

Hope this helps ;) ❤❤❤

How many positive integers less than 1,000,000 have exactly one digit equal to 9 and have a sum of digits equal to 13

Answers

Answer:

10,000

Step-by-step explanation:

There are 2970 positive integers less than 1,000,000 have exactly one digit equal to 9 and have a sum of digits equal to 13

What is Number system?

A number system is defined as a system of writing to express numbers.

We need to find

positive integers less than 1,000,000 have exactly one digit equal to 9 and have a sum of digits equal to 13

Let all 9 numbers ae

a+b+c+d+e+f+g+h+9=13

a+b+c+d+e+f+g+h=13-9

a+b+c+d+e+f+g+h=4

Then we use combinations

(n+k-1)Ck

¹¹C₄

11!/(11-4)!4!

11!/7!4!

330

Three hundred thirty times of nine is two thousand nine hundred seventy.

Now 330 ×9=2970

Hence there are 2970  positive integers less than 1,000,000 have exactly one digit equal to 9 and have a sum of digits equal to 13

To learn more on Number system click:

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Suppose the radius of a circle is 5 units. What is its circumference?​

Answers

Answer:

C≈31.42

Step-by-step explanation:

C=2πr

C=2xπx5

C≈31.42

pls mark as brainliest

The areas of two similar octagons are 4 m² and 9 m². What is the scale factor of their side lengths? PLZ PLZ HELP PLZ

Answers

Answer:

[tex] \frac{2}{3} [/tex]

Step-by-step explanation:

Area of Octagon A = 4 m²

Side length of Octagon A = a

Area of Octagon B = 9 m²

Side length of Octagon B = b

The scale factor of their side lengths = [tex] \frac{a}{b} [/tex]

According to the area of similar polygons theorem, [tex] \frac{4}{9} = (\frac{a}{b})^2 [/tex]

Thus,

[tex] \sqrt{\frac{4}{9}} = \frac{a}{b} [/tex]

[tex] \frac{\sqrt{4}}{\sqrt{9}} = \frac{a}{b} [/tex]

[tex] \frac{2}{3} = \frac{a}{b} [/tex]

Scale factor of their sides = [tex] \frac{2}{3} [/tex]

Answer:

3:5

Step-by-step explanation:

square root of 9 is 3.

square root if 25 is 5.

therefore, 3:5.

Of the three properties, reflexive, symmetric, and transitive that define the relation "is equal to," which one could also apply to "is less than" and "is greater than?" transitive reflexive symmetric

Answers

Answer: Transitive property.

Step-by-step explanation:

First, for the equality we have:

Reflexive:

  For all real numbers x, x = x.

Symmetric:  

 For all real numbers x, y

 if x= y, then y = x.

Transitive:

 For reals x, y and z.

 if x = y, and y = z, then x = z.

Now, let's talk about inequalities.

first, the reflexive property will say that:

x > x.

This has no sense, so this property does not work for inequalities.

Now, the reflexive.

If x > y, then y > x.

Again, this has no sense, if x is larger than y, then we can never have that y is larger than x. This property does not work for inequalities.

Not, the transitive property.

if x > y, and y > z, then x > z.

This is true.

x is bigger than y, and y is bigger than z, then x should also be bigger than z.

x > y > z.

And this also works for the inverse case:

x < y and y < z, then x < z.

So the correct option is transitive property.

Simplify to create an equivalent expression.
-k-(-8k+7)
a=7k−7
b=-7k-7
c=7k+7
d=-7k+7
choose one

Answers

Answer:

a. 7k - 7

Step-by-step explanation:

Step 1: Write out expression

-k - (-8k + 7)

Step 2: Distribute negative

-k + 8k - 7

Step 3: Combine like terms

7k - 7

And we have our answer!

The balances in two separate bank accounts that grow each month at different rales are represented by the functions f(x) and gix) In what month do the funds in the f(x) bank account exceed those in the glx)
bank account?
Month (x) f(x) = 2* g(x) = 4x + 12
1
2
16
2.
4
20
O Month 3
O Month 4
O Month 5
O Month 6​

Answers

Answer:

The balance in two separate bank accounts grows each month at different rates. the growth rates for both accounts are represented by the functions f(x) = 2x and g(x) = 4x 12. in what month is the f(x) balance greater than the g(x) balance?

Answer:

6 months

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

x = 5,

f(5) = [tex]2^5\\[/tex] = 32

g(5) = 4 x 5 + 12 = 20 + 12 = 32

x = 6,

f(6) = [tex]2^6[/tex] = 64

g(6) = 4 x 6 + 12 = 24 + 12 = 36

At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.

What is a function?

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(x) = [tex]2^{x}[/tex]

g(x) = 4x + 12

x = number of months

Now,

x = 3,

f(3) = 2³ = 8

g(3) = 4 x 3 + 12 = 12 + 12 = 24

x = 4,

f(4) = [tex]2^4[/tex] = 16

g(4) = 4 x 4 + 12 = 16 + 12 = 28

x = 5,

f(5) = [tex]2^5\\[/tex] = 32

g(5) = 4 x 5 + 12 = 20 + 12 = 32

x = 6,

f(6) = [tex]2^6[/tex] = 64

g(6) = 4 x 6 + 12 = 24 + 12 = 36

We see that,

At x = 6,

f(5) = 64

g(5) = 36

Thus,

At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.

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20 points!
Please help.

Answers

Man this is a hard one!

Write an expression to represent the given statement. Use n for the variable. Three times the absolute value of the sum of a number and 6

Answers

Answer:

3 · |x+6|

Step-by-step explanation:

Write out what you see. "Three times" is 3 · something; "the absolute value of the sum of a number and 6" is |number + 6|. We'll use x for our number. Put it all together and you get 3 · |x+6|

The expression of the statement, Three times the absolute value of the sum of a number and 6 is  [tex]\[3\left| n+6 \right|\][/tex] .

Representation of statement:Let n be the number.The sum of the numbers n and 6 is n+6.The absolute value of the sum of the numbers n and 6 is  [tex]\[\left| n+6 \right|\][/tex].Hence, three times the absolute value of the sum of a number and 6 is [tex]\[3\left| n+6 \right|\][/tex].

 

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Calculate how many different sequences can be formed that use the letters of the given word. Leave your answer as a product of terms of the form C(n, r). HINT [Decide where, for example, all the s's will go, rather than what will go in each position.]
georgianna
A) C(10, 7)
B) C(2, 10)C(1, 8)C(1, 7)C(1, 6)C(1, 5)C(2, 4)C(2, 2)
C) C(10, 2)C(8, 1)C(7, 1)C(6, 1)C(5, 1)C(4, 1)C(3, 1)C(2, 1)C(1, 1)
D) 10 · C(10, 2)C(8, 1)C(7, 1)C(6, 1)C(5, 1)C(4, 2)C(2, 2)
E) C(10, 2)C(8, 1)C(7, 1)C(6, 1)C(5, 1)C(4, 2)C(2, 2)

Answers

Answer: E) C(10, 2)C(8, 1)C(7, 1)C(6, 1)C(5, 1)C(4, 2)C(2, 2)

Step-by-step explanation:

According to the combinations: Number of ways to choose r things out of n things = C(n,r)

Given word: "georgianna"

It is a sequence of 10 letters with 2 a's , 2 g's , 2 n's , and one of each e, o,r, i.

If we think 10 blank spaces, then in a sequence we need 2 spaces for each of g.

Number of ways = C(10,2)

Similarly,

1 space for 'e' → C(8,1)

1 space for 'o' → C(7,1)

1 space for 'r' → C(6,1)

1 space for 'i' → C(5,1)

1 space for 'a' → C(4,2)

1 space for 'n' → C(2,2)

Required number of different sequences  = C(10,2) ×C(8,1)× C(7,1)× C(6,1)×C(5,1)×C(2,2).

Hence, the correct option is E) C(10, 2)C(8, 1)C(7, 1)C(6, 1)C(5, 1)C(4, 2)C(2, 2)

Use spherical coordinates. Evaluate e x2 + y2 + z2 dV, E where E is enclosed by the sphere x2 + y2 + z2 = 25 in the first octant.

Answers

Answer:

[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \frac{\pi (17e^5 - 2)}{2}[/tex]

General Formulas and Concepts:
Calculus

Integration

Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Integration Method [Integration by Parts]:
[tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]

[IBP] LIPET: Logs, Inverses, Polynomials, Exponentials, Trig

Multivariable Calculus

Triple Integrals

Cylindrical Coordinate Conversions:

[tex]\displaystyle x = r \cos \theta[/tex][tex]\displaystyle y = r \sin \theta[/tex][tex]\displaystyle z = z[/tex][tex]\displaystyle r^2 = x^2 + y^2[/tex][tex]\displaystyle \tan \theta = \frac{y}{x}[/tex]

Spherical Coordinate Conversions:

[tex]\displaystyle r = \rho \sin \phi[/tex][tex]\displaystyle x = \rho \sin \phi \cos \theta[/tex][tex]\displaystyle z = \rho \cos \phi[/tex][tex]\displaystyle y = \rho \sin \phi \sin \theta[/tex][tex]\displaystyle \rho = \sqrt{x^2 + y^2 + z^2}[/tex]

Integral Conversion [Spherical Coordinates]:
[tex]\displaystyle \iiint_T {f( \rho, \phi, \theta )} \, dV = \iiint_T {\rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]

Step-by-step explanation:

*Note:

Recall that φ is bounded by 0 ≤ φ ≤ 0.5π from the z-axis to the x-axis.

I will not show/explain any intermediate calculus steps as there isn't enough space.

Step 1: Define

Identify given.

[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV[/tex]

[tex]\displaystyle \text{Region E:} \ x^2 + y^2 + z^2 = 25 \ \text{bounded by first octant}[/tex]

Step 2: Integrate Pt. 1

Find ρ bounds.

[Sphere] Substitute in Spherical Coordinate Conversions:
[tex]\displaystyle \rho^2 = 25[/tex]Solve:
[tex]\displaystyle \rho = 5[/tex]Define limits:
[tex]\displaystyle 0 \leq \rho \leq 5[/tex]

Find θ bounds.

[Sphere] Substitute in z = 0:
[tex]\displaystyle x^2 + y^2 = 25[/tex][Circle] Graph [See 2nd Attachment][Graph] Identify limits [Unit Circle]:
[tex]\displaystyle 0 \leq \theta \leq \frac{\pi}{2}[/tex]

Find φ bounds.

[Circle] Substitute in Cylindrical Coordinate Conversions:
[tex]\displaystyle r^2 = 25[/tex]Solve:
[tex]\displaystyle r = 5[/tex]Substitute in Spherical Coordinate Conversions:
[tex]\displaystyle \rho \sin \phi = 5[/tex]Solve:
[tex]\displaystyle \phi = \frac{\pi}{2}[/tex]Define limits:
[tex]\displaystyle 0 \leq \phi \leq \frac{\pi}{2}[/tex]

Step 3: Integrate Pt. 2

[Integrals] Convert [Integral Conversion - Spherical Coordinates]:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex][dρ Integrand] Rewrite [Spherical Coordinate Conversions]:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \iiint_E {e^{\rho} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex][Integrals] Substitute in region E:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^5_0 {e^{\rho} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]

We evaluate this spherical integral by using the integration rules, properties, and methods listed above:

[tex]\displaystyle \begin{aligned} \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^5_0 {e^{\rho} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta \\ & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {\bigg[ (\rho^2 - 2 \rho + 2) e^{\rho} \sin \phi \bigg] \bigg| \limits^{\rho = 5}_{\rho = 0}} \, d\phi \, d\theta\end{aligned}[/tex]

[tex]\displaystyle \begin{aligned}\iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {(17e^5 - 2) \sin \phi} \, d\phi \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {\bigg[ -(17e^5 - 2) \cos \phi \bigg] \bigg| \limits^{\phi = \frac{\pi}{2}}_{\phi = 0}} \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {17e^5 - 2} \, d\theta \\& = (17e^5 - 2) \theta \bigg| \limits^{\theta = \frac{\pi}{2}}_{\theta = 0} \\& = \frac{\pi (17e^5 - 2)}{2}\end{aligned}[/tex]

∴ the given integral equals [tex]\displaystyle \bold{\frac{\pi (17e^5 - 2)}{2}}[/tex].

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Topic: Multivariable Calculus

Unit: Triple Integrals Applications

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