The graphs for the functions f(x) = √x−3 and g(x) = √x−3 are shown in the graph below. Describe the transformations of a parent function to obtain the graphs of f(x) and g(x).

Answer:

What is the question? I might be able to help

Answer:

Can you please give the question because I might be able to help you.

Hope you have a great day, have a wonderful weekend and stay safe!

Also Happy Halloween! :) :D :3

Answer:

12 is the difference

Step-by-step explanation:

it has a range of 12˚ F

Answer:

12°F

Step-by-step explanation:

Use highest temp. - lowest temp. = difference.

Substitute in values: 3 - (-9) = 3 + 9

                                              = 12

Answer:

f(x) = 0.85x

g(x) = 0.85x - 100

Step-by-step explanation:

Given that :

Sale price = 15% off the item price

Let original price = x

Hence, sale price = (100 - 15)%

Let the sale price function = f(x)

f(x) = 0.85x

Where, x = price before 15% off

Customer also has a coupon discount of 100 pesos

Let coupon discount price = g(x)

g(x) = f(x) - 100

Hence,

g(x) = 0.85x - 100

x = price before 15% off

Answer:

Ratio of boys to girls,

=> 5:4

If there are 60 girls, then the number of boys

= 5 * 15 : 4 * 15

=> 75 : 60

Therefore, there were 75 boys in the choir

If my answer helped, kindly mark me as the brainliest!!

Thank You!!

The number of boys in the choir is 75 when there are 60 girls.

Given information:

The ratio of the number of boys in the choir to the number of girls is 5 to 4 [tex]\left( \dfrac{5}{4}\right)[/tex].

There are 60 girls in the choir.

Let the number of boys in the choir be x. We need to find the value of x.

The number of boys, x, can be calculated as,

[tex]\dfrac{x}{60}=\dfrac{5}{4}\\x=60\times \dfrac{5}{4}\\x=15\times5\\x=75[/tex]

Therefore, the number of boys in the choir is 75 when there are 60 girls.

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Answer:

-6

Step-by-step explanation:

f(x) = 2x2 + 5x - 4

Let x=-2

f(-2) = 2*(-2)^2 + 5(-2) - 4

      = 2 * 4 +5(-2) -4

      = 8-10-4

      = -6

Answer:

y=-6.6

hope this helps

Answer: 1. 5^3

2.-5/1

3. -1/3

Step-by-step explanation:

The other trigonometric ratios for the given cos(theta) = 5/13 are: sin(theta) = 5/13tan(theta) = 5/12cot(theta) = 12/5sec(theta) = 13/5csc(theta) = 13/5.

Based on the given trigonometric ratio cos(theta) = 5/13, we can determine the other trigonometric ratios for the angle theta.

1. Sine (sin) theta:

The sine ratio is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In this case, the opposite side is 5 and the hypotenuse is 13, so sin(theta) = 5/13.

2. Tangent (tan) theta:

The tangent ratio is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. In this case, the opposite side is 5 and the adjacent side is 12, so tan(theta) = 5/12.

3. Cotangent (cot) theta:

The cotangent ratio is the reciprocal of the tangent ratio, which means it is the ratio of the adjacent side to the opposite side. In this case, the adjacent side is 12 and the opposite side is 5, so cot(theta) = 12/5.

4. Secant (sec) theta:

The secant ratio is defined as the reciprocal of the cosine ratio. In this case, the cosine ratio is 5/13, so sec(theta) = 13/5.

5. Cosecant (csc) theta:

The cosecant ratio is defined as the reciprocal of the sine ratio. In this case, the sine ratio is 5/13, so csc(theta) = 13/5.

In summary, the other trigonometric ratios for the given cos(theta) = 5/13 are:

sin(theta) = 5/13

tan(theta) = 5/12

cot(theta) = 12/5

sec(theta) = 13/5

csc(theta) = 13/5.

These ratios represent the relationships between the sides of a right triangle in terms of the angle theta.

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Answer: z= 1.72

Step-by-step explanation:

Don’t ask me how I guessed and it was right

The standardized score for the student after the conversion is z=1.72. The correct option is C.

In statistics, standardization is the process of putting different variables on the same scale. This process allows you to compare scores between different types of variables. Typically, to standardize variables, you calculate the mean and standard deviation for a variable.

A student had a standardized score of z=1.72 before the conversion.

We know that, z=(x-μ)/σ

The standardized score is z=1.72

Therefore, option C is the correct answer.

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Step-by-step explanation:

This is a probability related question, let the event be E

We know that the likelihood of an event happening is given as

Pr(E)=1

if an event will not occur the probability is

Pr(E)=0

a. This event is impossible: Pr(E)=0

b.This event will occur more often than not, but is not extremely likely:  

Pr(E)=0<E>0.5

c.This event is extremely unlikely, but it will occur once in a while in a long sequence of trials:

Pr(E)=0<E<0.5

d.This event will occur for sure: Pr(E)=0

Answer:

1.99 for 24 ounces

Step-by-step explanation:

Answer:

A. A set is a group or a collection of objects.

Step-by-step explanation:

The correct choice is that a set is a group or a collection of objects. Each of the objects in a set are distinct and well defined.

Sets are used extensively to solve mathematical problems.

A set is not a collect of an object but distinct objects that are well defined.

Answer:

a c d e

Step-by-step explanation:

Answer:

Step-by-step explanation:

First, we have to figure out how many there are all together.

22

so it is going to be

5/22 to get a cherry candy

and to put it back and get a grape is

8/22

Answer:

-18

Step-by-step explanation:

Answer:

B. 55°

Step-by-step explanation:

they are alternate interior angles which mean they are congruent

Answer:

answer what?

Step-by-step explanation:

Answer:

40 girls

Step-by-step explanation:

50/5=10

10*4=40

40 girls

Answer:

Step-by-step explanation:

answer is b.

the domain may not be used more than once

The class width used for the frequency distribution is 6.

The class midpoint for the class 23-29 is 26.

The class midpoint for the class 30-36 is 33.

The class midpoint for the class 37-43 is 40.

To find the class width of the frequency distribution, we need to determine the range of each age class. The range is the difference between the upper and lower boundaries of each class. Looking at the table, we can see that the class boundaries are as follows:

23-29

30-36

37-43

For the class 23-29, the lower boundary is 23 and the upper boundary is 29. To find the class width, we subtract the lower boundary from the upper boundary:

Class width = 29 - 23 = 6

So, the class width for the frequency distribution is 6.

To find the class midpoint for each class, we take the average of the lower and upper boundaries of each class.

For the class 23-29:

Class midpoint = (23 + 29) / 2 = 52 / 2 = 26

For the class 30-36:

Class midpoint = (30 + 36) / 2 = 66 / 2 = 33

For the class 37-43:

Class midpoint = (37 + 43) / 2 = 80 / 2 = 40

So, the class midpoint for the class 23-29 is 26, for the class 30-36 is 33, and for the class 37-43 is 40.

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Answer:

a) player’s expected payoff is $ 240

b) probability the player loses $1000 or more is 0.1788

c)  probability the player wins is 0.3557

d) probability of going broke is 0.0594

Step-by-step explanation:

Given:

Since there are 60 hands per hour and the player plays for four hours then the sample size is:

n = 60 * 4 = 240

The player’s strategy provides a probability of .49 of winning on any one hand so the probability of success is:

p = 0.49

a)

Solution:

Expected payoff is basically the expected mean

Since the bet is $50 so $50 is gained when the player wins a hand and $50 is lost when the player loses a hand. So

Expected loss =  μ

                        = ∑ x P(x)

                        = 50 * P(win) - 50 * P(lose)

                        = 50 * P(win) + (-50) * (1 - P(win))

                         = 50 * 0.49 - 50 * (1 - 0.49)

                        = 24.5 - 50 ( 0.51 )

                        = 24.5 - 25.5

                        = -1

Since n=240 and expected loss is $1 per hand then the expected loss in four hours is:

240 * 1 = $ 240

b)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 110.5 - 117.6 / √117.6(1-0.49)

  = −7.1/√117.6(0.51)

  = −7.1/√59.976

  = −7.1/7.744417

  =−0.916789

Here the player loses 1000 or more when he loses at least 130 of 240 hands so the wins is 240-130 = 110

Using normal probability table:

P(X≤110) = P(X<110.5)

             = P(Z<-0.916)

             = 0.1788

c)

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 120.5 - 117.6 / √117.6(1-0.49)

  = 2.9/√117.6(0.51)

  = 2.9/√59.976

  = 2.9/7.744417

  =0.374463

Here the player wins when he wins at least 120 of 240 hands

Using normal probability table:

P(X>120) = P(X>120.5)

              = P(Z>0.3744)  

             =  1 - P(Z<0.3744)

             = 1 - 0.6443

             = 0.3557

d)

Player goes broke when he loses $1500

Using normal approximation of binomial distribution:

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution

Compute z-score:

z = x - np / √(np(1-p))

  = 105.5 - 117.6 / √117.6(1-0.49)

  = -12.1/√117.6(0.51)

  = -12.1/√59.976

  = -12.1/7.744417

  =−1.562416

Here the player loses 1500 or more when he loses at least 135 of 240 hands so the wins is 240-135 = 105

Using normal probability table:

P(X≤105) = P(X<105.5)

             = P(Z<-1.562)

             = 0.0594

The Player’s expected payoff is $ 240.

Probability the player loses $1000 or more is 0.1788.

Probability the player wins is 0.3557.

Probability of going broke is 0.0594.

Given that,

There are 60 hands per hour and the player plays for four hours then the sample size is:

n = 60 × 4 = 240

The player’s strategy provides a probability of .49 of winning on any one hand so the probability of success is:

p = 0.49

Since the bet is $50 so $50 is gained when the player wins a hand and $50 is lost when the player loses a hand. So

Expected loss =  μ  = ∑ x P(x)

                        = 50 × P(win) - 50 P(loss)

                       = 50 × P(win) + (-50) × (1 - P(win)

                       = 50 × 0.49 - 50 × (1 - 0.49)

                      = 24.5 - 50 ( 0.51 )

                      = 24.5 - 25.5

                      = -1

Since n=240 and expected loss is $1 per hand then the expected loss in four hours is:

240 × 1 = $ 240

n = 240

p = 0.49

q = 1 - p = 1 - 0.49 = 0.51

np = 240 * 0.49 = 117.6

nq = 240 * 0.51 = 122.5

Both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution.

To Compute z-score:

[tex]z = \frac{x - np}{\sqrt{no (1-p )} } \\\\z = \frac{110.5 - 117.6}{\sqrt{117.6 ( 1-0.49)} }\\\\z = \frac{- 7.1}{\sqrt{117.6 (0.51) } } \\\\z = \frac{-7.1}{\sqrt{59.97} } \\\\z = \frac{-7.1}{7.74}\\\\z = - 0.916[/tex]

Here the player loses 1000 or more when he loses at least 130 of 240 hands so the wins is 240-130 = 110.

Using normal probability table:

P(X≤110) = P(X<110.5)

             = P(Z<-0.916)  = 0.1788

To compute z-score:

[tex]z = \frac{x - np}{\sqrt{no (1-p )} } \\\\z = \frac{120.5 - 117.6}{\sqrt{117.6 ( 1-0.49)} }\\\\z = \frac{2.9}{\sqrt{117.6 (0.51) } } \\\\z = \frac{2.9}{\sqrt{59.97} } \\\\z = \frac{2.9}{7.74}\\\\z = 0.34[/tex]

Here the player wins when he wins at least 120 of 240 hands.

By using normal probability table:

P(X>120) = P(X>120.5)

               =  1 - P(Z<0.3744)

               = 1 - 0.6443

              = 0.37

Both np and nq are greater than 5 so the binomial distribution can be approximated by normal distribution.

To Compute z-score:

[tex]z = \frac{x - np}{\sqrt{no (1-p )} } \\\\z = \frac{105.5 - 117.6}{\sqrt{117.6 ( 1-0.49)} }\\\\z = \frac{-12.1}{\sqrt{117.6 (0.51) } } \\\\z = \frac{-12.1}{\sqrt{59.97} } \\\\z = \frac{-12.1}{7.74}\\\\z = -1.56[/tex]

Here the player loses 1500 or more when he loses at least 135 of 240 hands so the wins is 240-135 = 105

Using normal probability table:

P(X≤105) = P(X<105.5)

               = P(Z<-1.562)

              = 0.54

For the more information about Probability distribution click the link given below.

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Answer:

b would be your answer

Step-by-step explanation:

what about it you need help with?

Answer:

[tex]\sqrt{500}[/tex] inches

Step-by-step explanation:

a^2+b^2=c^2

10^2+20^2=c^2

100+400=c^2

c^2=500

[tex]\sqrt{500}[/tex]

Answer:

see below

Step-by-step explanation:

Part I: Use scientific notation to estimate the number of hours in one year

Using the conversion factors

Estimate using scientific notation

31,556,952 seconds = 3.2 *  10^7

3600 seconds = 3.6 * 10 ^2

1 year = 3.2 *  10^7 seconds * 1 hour/ 3.6 * 10 ^2 seconds

3.2 /3.6 =.88888 (repeating)    

10^7 / 10^2 = 10^5

8.88888 * 10 ^5

8.888 * 10 ^4 hours

Part II: Use scientific notation to calculate the exact number of hours in one year.

 31,556,952 seconds = 3.1556952 *  10^7

3600 seconds = 3.6 * 10 ^2

1 year = 3.1556952 *  10^7 seconds * 1 hour/ 3.6 * 10 ^2 seconds

3.1556952 /3.6 =.876582    

10^7 / 10^2 = 10^5

.876582 * 10 ^5      

8.76582*10^4 hours

Part III: In two or more complete sentences, compare your answers to parts I and II. In your comparison, discuss similarities and differences in your answers.

Part II is an exact answer, where Part I is an estimate.  Both Parts give values in the thousands of hours,  and are both  between 8700 and 8900 hours.  Part I can be done without a calculator where Part II is much harder to do without a calculator.

Answer:

see below

Step-by-step explanation:

Part I: Use scientific notation to estimate the number of hours in one year

Using the conversion factors

Estimate using scientific notation

31,556,952 seconds = 3.2 *  10^7

3600 seconds = 3.6 * 10 ^2

1 year = 3.2 *  10^7 seconds * 1 hour/ 3.6 * 10 ^2 seconds

3.2 /3.6 =.88888 (repeating)    

10^7 / 10^2 = 10^5

8.88888 * 10 ^5

8.888 * 10 ^4 hours

Part II: Use scientific notation to calculate the exact number of hours in one year.

31,556,952 seconds = 3.1556952 *  10^7

3600 seconds = 3.6 * 10 ^2

1 year = 3.1556952 *  10^7 seconds * 1 hour/ 3.6 * 10 ^2 seconds

3.1556952 /3.6 =.876582    

10^7 / 10^2 = 10^5

.876582 * 10 ^5      

8.76582*10^4 hours

Part III: In two or more complete sentences, compare your answers to parts I and II. In your comparison, discuss similarities and differences in your answers.

Part II is an exact answer, where Part I is an estimate.  Both Parts give values in the thousands of hours,  and are both  between 8700 and 8900 hours.  Part I can be done without a calculator where Part II is much harder to do without a calculator.

Answer:

44

Step-by-step explanation: