The student council is hosting a homecoming event for past graduates and current students. The treasurer determines that the event revenue from the event can be represented by R (z) = 0.05x³75, where x is the number of tickets sold. The cost to put on the event is represented by the function C(z) = 30x + 12,500. Which function describes the funds raised, F(x), as a function of the number of tickets sold? O F(z) F(x) F(x) 0.05³ +30 - 12, 425 F(x) 0.05z³ 30x - 12,425 O = 0.05³ +30 - = 12,575 = 0.05³ 30 - 12,575

Using the normal distribution, there is a 0.2148 = 21.48% probability that the sum of the 40 values is less than 7,100.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

For this problem, these parameters are given as follows:

[tex]\mu = 180, \sigma = 20, n = 40, s = \frac{20}{\sqrt{40}} = 3.1623[/tex]

A sum of 7100 is equivalent to a sample mean of 7100/40 = 177.5, which means that the probability is the p-value of Z when X = 177.5, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{177.5 - 180}{3.1623}[/tex]

Z = -0.79

Z = -0.79 has a p-value of 0.2148.

There is a 0.2148 = 21.48% probability that the sum of the 40 values is less than 7,100.

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The Width of plot B is 183 ft.

According to the statement

we have given that the Plot A has a square footage of 46,060, and Plot B has a square footage of 43,005. The common boundary they share is 235 feet long.

And we have to find the width of the Plot B

We know that

Area of plot A = 46,060 square footage.

Area of plot B = 43005 square footage.

Common area share by both Plot = 235 feet long.

So, Now

The width of Plot B = Area of plot B / Common area

Substitute the values in it then

width of Plot B = Area of plot B / Common area

width of Plot B = 43005 / 235

width of Plot B = 183 feet.

So, The Width of plot B is 183 ft.

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1/3 portion of the mixture, costing $\$6$ contains chocolate.

Assume that, in a mixture:

chocolate = x pound

nuts = y pound

mixture= x+y

The desired value is found by dividing the weight of the chocolate by the overall weight, or x/(x+y).

From the given data,

cost of x pound chocolate at $\$10$per pound = 10x

cost of y pound nuts at $\$4$ per pound = 4y

cost of x+y pound mixture at $\$6$ = 6(x+y)

As per condition,

⇒10x + 4y = 6(x+y)

⇒10x + 4y = 6x+ 6y

⇒4x = 2y

⇒y = 2x

The fraction of the chocolate in the mixture is

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

Therefore, chocolate makes  1/3 of the mixture's weight.

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Answer:

1/3

Step-by-step explanation:

The value of c is 5 when we simplify the  logarithmic equation              log(-c) + log8= log (7c-75). This can be obtained using the properties of the logarithm.

Given that,

log(-c) + log8 = log (7c-75)

log (-8c) = log (7c-75)  (∵ log a + log b = log ab)

- 8c = 7c - 75              

(8+7)c = 75

15c = 75

⇒ c = 5

Hence the value of c is 5 when we simplify the  logarithmic equation              log(-c) + log8= log (7c-75).

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The value of λ in the vector is 0.2

Vectors are the opposite of scalar quantities and they are quantities that have directions and magnitude

The vector a is given as:

a = (−6, 8)

The magnitude of vector a is calculated using

|a| = √(x^2 + y^2)

So, the equation becomes

|a| = √((-6)^2 + 8^2)

Evaluate the exponent

|a| = √(36 + 64)

Evaluate the sum

|a| = √100

Take the square root of both sides

|a| = 10

Given that

λa = 5

This means that

λ * 10 = 5

Divide both sides by 10

λ = 0.2

Hence, the value of λ is 0.2

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The average rate of change is -5 per month

The interval is given as:

July to December

From the graph, we have:

July = 110

December = 30

The average rate of a function over the interval (a, b) is calculated as:

Rate = [f(b) - f(a)]/[b - a]

So, we have:

Rate = (30 - 110)/(December- July)

This gives

Rate = (30 - 110)/(12 - 7)

Evaluate

Rate = -16

Hence, the average rate of change is -5 per month

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Answer:

Option (3)

Step-by-step explanation:

Since the base of the exponential is less than 1, the function is decreasing for all x.

Answer:

the value of the variable x is 16

Step-by-step explanation:

[tex]given \: expression \\ \frac{5}{8}x = \frac{1}{2} x + 2 \\ substract \: \frac{1}{2} x from \: both \: sides \\ \frac{5}{8}x - \frac{1}{2} x = \frac{1}{2} x - \frac{1}{2} x + 2 - \frac{1}{2} x \\ factor \: out \: common \: x \\ x( \frac{5}{8} - \frac{1}{2} ) = \frac{1}{8} \\ simplify \\ \frac{1}{2} x + 2 - \frac{1}{2} x \: \frac{1}{2} x - \frac{1}{2} x = 0 = 2 \\ \frac{1}{8} x = 2 \\ multiply \: both \: sides \: by \: 8 \\ 8. \frac{1}{8} x = 2.8 \\ simplify \\ x = 16.[/tex]

Answer:

$270

Step-by-step explanation:

18x15=$270

HOPE THAT HELPED:)

a) The length of p is 21.4 cm

b) The measure of angle P is 45.6°

From the question, we are to determine the length of p

From the Pythagorean theorem, we can write that

30² = 21² + p²

p² = 30² - 21²

p² = 900 - 441

p² = 459

p = √459

p = 21.4 cm

∴ The length of p is 21.4 cm

b)

Measure of angle P

Using SOH CAH TOA

[tex]cos P= \frac{21}{30}[/tex]

cos P = 0.7

P = cos⁻¹(0.7)

P = 45.6°

Hence, the measure of angle P is 45.6°

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The amount of medicine infused in 8 hours is 1000ml.

According to the statement

we have given that the medicine normal saline (ns) iv to infuse at 125 ml/hr.

and we have to find the total amount of medicine infuse in the 8 hours in the whole number value.

So,

An infusion is the creation of a new substance by steeping another substance in a liquid, usually water

we have given a rate of infusion of medicine per hour which is 125 ml/hr.

And

Amount of medicine infuse in 1 hour is 125 ml

And the Amount of medicine infuse in 8 hours = 125*8

Amount of medicine infuse in 8 hours = 1000 ml.

So, The amount of medicine infused in 8 hours is 1000ml.

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Answer:

[tex]\sf A) \quad \begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]

B)  see below

C)  points C, E and F

Step-by-step explanation:

Given points:

A system of inequalities is a set of two or more inequalities in one or more variables.

To create a system of inequalities that only contains C and F in the overlapping shaded region, create a linear equation where points C, F and E are to the right of the line and a linear equation where points C, F, A, B and D are to the left of the line.

The easiest way to do this is to find the slope of the line that passes through points C and F, then add values to move the lines either side of the points.

[tex]\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{y_F-y_C}{x_F-x_C}=\dfrac{-4-1}{3-2}=-5[/tex]

Therefore:

[tex]\sf y = -5x + 5[/tex]  →  points C, F and E are to the right of the line.

[tex]\sf y=-5x+12[/tex]  →  points C, F, A, B and D are the left of the line.

Therefore, the system of inequalities that only contains points C and F in the overlapping shaded regions is:

[tex]\begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]

To graph the system of inequalities:

To verify that the points C and F are solutions to the system of inequalities created in Part A, substitute the x-values of both points into the system of inequalities.  If the y-values satisfy both inequalities, then the points are solutions to the system.

Point C (2, 1)

[tex]\implies \sf x=2 \implies 1 > -5(2)+5 \implies 1 > -5\quad verified[/tex]

[tex]\implies \sf x=2 \implies 1 < -5(2)+12\implies 1 < 2 \quad verified[/tex]

Point F (3, -4)

[tex]\implies \sf x=3 \implies -4 > -5(3)+5 \implies -4 > -10\quad verified[/tex]

[tex]\implies \sf x=3 \implies -4 < -5(3)+12\implies -4 < -3 \quad verified[/tex]

Method 1

Graph the line y = 7x - 4 (making the line dashed since it is y < 7x - 4).

Shade below the dashed line.

Points that are contained in the shaded region are the houses in which Erica is interested in living:  points C, E and F.

Method 2

Substitute the x-value of each point into the given inequality y < 7x - 4.

Any point where the y-value satisfies the inequality is a house that Erica is interested in living.

[tex]\sf Point\: A: \quad x=-5 \implies 5 < 7(-5)-4 \implies -5 < -39 \implies no[/tex]

[tex]\sf Point\: B: \quad x=-4 \implies -2 < 7(-4)-4 \implies -2 < -32 \implies no[/tex]

[tex]\sf Point\: C: \quad x=2 \implies 1 < 7(2)-4 \implies 1 < 10 \implies yes[/tex]

[tex]\sf Point\: D: \quad x=-2 \implies 4 < 7(-2)-4 \implies 4 < -18 \implies no[/tex]

[tex]\sf Point\: E: \quad x=2 \implies 4 < 7(2)-4 \implies 4 < 10 \implies yes[/tex]

[tex]\sf Point\: F: \quad x=3 \implies -4 < 7(3)-4 \implies -4 < 17 \implies yes[/tex]

Answer:

since those 2 lines are parrelel that means the angles are the same so

8x+6=4x+38

minus 6

8x=4x+32

minus 4x

4x=32

divide by 4

x=8

Hope This Helps!!!

The value of the x is 11 and the measure of the angle B is 82 degrees.

It is defined as the angles when two lines intersect each other and at the intersecting point, some pair of angles are formed which we call vertically opposite angles, as the name describes that they have vertically opposite angles.

We have:

The angle measurements in the diagram are represented by the following expressions:

Angle A = 8x - 6

Angle B = 4x + 38

By using the vertically opposite angle property in angle B

As we know the alternate exterior angle are equal

8x - 6 = 4x + 38

4x = 44

x =11

Measure of angle B = 4(11) + 38 = 82 degrees

Thus, the value of the x is 11 and the measure of the angle B is 82 degrees.

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Answer:

Step-by-step explanation:

hello :

|x – 5| + 2 < 20  means :  |x – 5| + 2-2 < 20-2

|x – 5|  < 18

-18< x - 5 < 18

-18+5< x - 5+5 < 18+5

-13< x< 23   ( solutions)

Answer:

12ft

Step-by-step explanation:

It takes 10 minutes for each plant and she spent 1 hour in total or 60 minutes: 60/10 = 6 plants

If each of the plants is 2 feet apart and there are 6 plants: 2ft*6 = 12ft

Answer:

Step-by-step explanation:

1a+1b(a+b-c)+1b+1c(b=c-a)+ 1a+1c(c+a-b)

1a+1b-c+1b +1c-a+1a+1c-b

1a+2b-c+a+1c-b

2a+1b+c this is the answer I think lol

The ratio of lunch interval to the total times of class periods is 1:10

Given that timing of school opens is 10am , the timing of school closes is 3:30pm and the lunch interval is 30 minutes

We need to find the ratio of lunch interval to the total times of class periods.

Now,

Timing of a school(10.a.mto3.30p.m)=5hours30minutes

Timing for lunch interval=30minutes

Total time of the class periods=5hours30minutes−30minutes

=5hours

=60×5=300 minutes.

Ratio of lunch interval to total time of the class period=30minutes:300 minutes=1:10

Hence the ratio of lunch interval to the total times of class periods is 1:10

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The linear inequality of the graph is: -x + 2y + 1 > 0

First, we calculate the slope of the dashed line using:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Two points on the graph are:

(1, 0) and (3, 1)

The slope (m) is:

[tex]m = \frac{1 - 0}{3 - 1}[/tex]

This gives

m = 0.5

The equation of the line is calculated as:

[tex]y = m(x -x_1) + y_1[/tex]

So, we have;

[tex]y = 0.5(x -1) + 0[/tex]

This gives

[tex]y = 0.5x -0.5[/tex]

Multiply through by 2

[tex]2y = x - 1[/tex]

Now, we convert the equation to an inequality.

The line on the graph is a dashed line. This means that the inequality is either > or <.

Also, the upper region of the graph that is shaded means that the inequality  is >.

So, the equation becomes

2y > x - 1

Rewrite as:

-x + 2y + 1 > 0

So, the linear inequality is: -x + 2y + 1 > 0

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Complete question

Find a linear inequality with the following solution set. Each grid line represents one unit. (Give your answer in the form ax+by+c>0 or ax+by+c [tex]\geq[/tex] 0 where a, b, and c are integers with no common factor greater than 1.)

In a proportionate stratified sample of 120, there are 72 Whites, 24 Latinos, 18 Blacks and 6  others.

Proportionate Stratified Sample

A proportionate stratified sample is one in which the size of the strata in the sample is proportional to the size of the strata in the population; in other words, the chance of selecting a unit from a stratum depends on the relative size of that stratum in the population.

Calculating the Proportionate Stratified Sample

The given percentage of -

Whites = 60%

Latinos = 20%

Blacks = 5%

Strength of the sample = 120

⇒ Number of Whites = 60% of 120

= 0.6 × 120

=72

Number of Latinos = 20% of 120

= 0.2 × 120

=24

Proportionate Stratified Sample of Blacks and Others

Count of Blacks = 15 % of 120

= 0.15 × 120

= 18

Count of others = 5% of 120

= 0.05 × 120

6

Thus, in a Proportionate Stratified Sample of 120, 72 are Whites, 24 are Latinos, 18 are Blacks, and 6 are others.

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The sample is the randomly selected customers.

Sampling:  

Suppose I want to estimate the proportion of New York state residents who are Buffalo Bills fans. So i ask, lets say, 1000 randomly selected New York state residents whether they are Buffalo Bills fans, and expand this to the entire population of New York State residents.

Sampling is a process used in statistical analysis in which a predetermined number of observations are taken from a larger population. The methodology used to sample from a larger population depends on the type of analysis being performed, but it may include simple random sampling or systematic sampling.

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The value of the fund in 18 years as required by the task is; $1553.5.

Since, the interest rate is compounded annually as described, the value of the fund in 18 years can be calculated by means of the compound interest formula as;

Value = 500(1.065)^18.

Hence, we have; Value = $1553.5.

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(-4x + 2) / 4 >= 6

-4x + 2 >= 24

-4x >= 22

x <= 22/4 OR x <= 5 1/2 OR x <= 5.5

Hope this helps!

Answer:

1st quartile is 136

The second quartile is also the median 142

The third quartile is 162

The interquartile range is the difference between the 3rd and first quartile or 26

Step-by-step explanation:

The dimensions of the rectangle are: length= 18 in and width =1 in.

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle can be found for the formula : l*w, where b = length and w =width. The question gives that the area is 18 in².

For this question, the length exceeds its width by 17 inches - l=w+17. Thus,  from the value of area given, you can find the values of the length and width of the rectangle.

A=l*w

18=(w+17)*w

18=w²+17w

w²+17w-18=0

Next step will be solve the previous equation ( W²+17W-18=0)

[tex]w_{1,\:2}=\frac{-17\pm \sqrt{17^2-4\cdot \:1\cdot \left(-18\right)}}{2\cdot \:1}\\ \\ w_{1,\:2}=\frac{-17\pm \:19}{2\cdot \:1}[/tex]

Therefore,

[tex]w_1=\frac{-17+19}{2\cdot \:1}=1\\ \\ \\ w_2=\frac{-17-19}{2\cdot \:1}=-18[/tex]

For dimensions, only positive numbers must be used. Then, the width is equal to 1 inch.

As, the area (l*w) is 18 in², you have.

18=l*w

18=l*1

l=18 in

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Answer:

area of A'B'C'D' = 18.4 units²

Step-by-step explanation:

given 2 similar figures with

ratio of corresponding sides = a : b , then

ratio of areas = a² : b²

here ratio of sides = 5 : 2

ratio of areas = 5² : 2² = 25 : 4

let x be the area of A'B'C'D' , then by proportion

[tex]\frac{25}{115}[/tex] = [tex]\frac{4}{x}[/tex] ( cross- multiply )

25x = 460 ( divide both sides by 25 )

x = 18.4

area of A'B'C'D' = 18.4 units²

Answer: B

Step-by-step explanation:

When you evaluate f(-1)=2x^4+x^2-5 we will get -2 in basic notation (-1,-2). We need to find the root of the function to find the other roots or factors, where y=0.

Answer: 3/4

Step-by-step explanation:

Tangent is opp/adj

Therefore tan (x) = 18/24, which reduces to 3/4

The change in the scale factor from the old scale to the new scale is 3.

A scale drawing is a reduced form in terms of dimensions of an original image / building / object. The scale drawing is usually reduced at a constant dimension.

The change in the scale factor = larger scale / smaller scale

3.3 / 1.1 = 3

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The given function for the difference in length is presented as follows;

[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} [/tex]

When the pogo stick will be equal to its non compressed length, the difference is zero, therefore;

[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} = 0[/tex]

[tex] 2 \cdot cos( \theta) = - \sqrt{3} [/tex]

[tex]\theta= arccos \left( \frac{ - \sqrt{3} }{2} \right) [/tex]

Which gives;

[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]

[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]

Part B; If the angle was doubled, we have;

[tex]f( \theta) = 2 \cdot cos(2 \cdot \theta) + \sqrt{3} = 0[/tex]

Therefore;

[tex] 2 \cdot cos(2 \cdot \theta) = - \sqrt{3} [/tex]

Which gives;

[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]

[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]

Between 0 and 2•π, we have;

[tex] \theta = \frac{5\cdot \pi}{12} [/tex]

[tex] \theta = \frac{ 17\cdot \pi}{12} [/tex]

Part C;

The toddler's pogo stick is presented as follows;

[tex]g( \theta) = 1- son^2( \theta) + \sqrt{3} [/tex]

Integrating the original function between 0 and theta gives;

[tex]2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]

The original length =

Therefore, when the lengths are equal, we have;

[tex]1- son^2( \theta) + \sqrt{3} = 2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]