At West High School, 10% of the students participate in
sports. A student wants to simulate the act of randomly
selecting 20 students and counting the number of
students in the sample who participate in sports. The
student assigns the digits to the outcomes.
0 student participates in sports
=
1-9 student does not participate in sports
How can a random number table be used to simulate
one trial of this situation?
O Select a row from the random number table. Count
the number of digits until you find 20 zeros.
O Select a row from the random number table. Count
the number of digits until you find 10 zeros.
O Select a row from the random number table. Read 20
single digits. Count the number of digits that are
zeros.
O Select a row from the random number table. Read 10
single digits. Count the number of digits that are
zeros.

The parametric equations for the line through the points (-2, 8, 7) and (-4, -5, -8) are:
x(t) = -2 - 2t
y(t) = 8 - 13t
z(t) = 7 - 15t

To find the parametric equations of the line through the points (-2, 8, 7) and (-4, -5, -8), you first need to find the direction vector of the line. To do this, subtract the coordinates of the first point from the second point:

Direction vector: (-4 - (-2), -5 - 8, -8 - 7) = (-2, -13, -15)

Now, write the parametric equations using the direction vector components and a point on the line, typically the first point:

x(t) = -2 + (-2)t
y(t) = 8 + (-13)t
z(t) = 7 + (-15)t

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The values of the remaining five trigonometric functions are:

cos(x) = 4/5
tan(x) = 3/4
csc(x) = 5/3
sec(x) = 5/4
cot(x) = 4/3.

Using the given information, we can use the Pythagorean identity to find the value of cos(x):

cos²(x) = 1 - sin²(x)
cos²(x) = 1 - (3/5)²
cos²(x) = 1 - 9/25
cos²(x) = 16/25
cos(x) = ±4/5

Since 0 < x < π/2, we know that cos(x) is positive, so cos(x) = 4/5.

Now we can use the definitions of the remaining trigonometric functions to find their values:

tan(x) = sin(x) / cos(x) = (3/5) / (4/5) = 3/4
csc(x) = 1 / sin(x) = 5/3
sec(x) = 1 / cos(x) = 5/4
cot(x) = 1 / tan(x) = 4/3

Therefore, the values of the remaining five trigonometric functions are:

cos(x) = 4/5
tan(x) = 3/4
csc(x) = 5/3
sec(x) = 5/4
cot(x) = 4/3.

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The value of the  trigonometric expression tan(u + v) for the given values sin u=-8/17 and cosv=-3/5 as both u and v are in quadrant II is equal to 84/77.

In the trigonometric expression ,

sin u=-8/17

Both u and v are in quadrant II.

Draw a right angle triangle ,

opposite side = -8

Hypotenuse = 17

using Pythagoras theorem ,

Adjacent side

= √(Hypotenuse)² - ( opposite side)²

=√17² - (-8)²

= -15

cos u = -15 / 17

Now,

cos v=-3/5

Both u and v are in quadrant II.

Draw a right angle triangle ,

Adjacent side = -3

Hypotenuse = 5

using Pythagoras theorem ,

Opposite side

= √(Hypotenuse)² - ( Adjacent side)²

=√5² - (-3)²

= -4

sin v = -4/5

sin(u + v ) = sinu cosv + cosu sinv

Substitute the value,

⇒sin(u + v ) = (-8/17) (-3/5) + (-15/17) (-4/5)

⇒sin(u + v ) = ( 24 + 60 )/ 85

⇒sin(u + v ) =84/85

cos ( u + v) = cosu cosv−sinu sinv

⇒cos ( u + v) = (-15/17)(-3/5) - (-8/17)(-4/5)

⇒cos ( u + v) = ( 45 + 32 ) / 85

⇒cos ( u + v) = 77/85

This implies,

tan (u + v )

= sin ( u + v) / cos( u + v)

= (84/85)/(77/85)

= 84/77

Therefore, the value of the  trigonometric expression tan (u + v )  = 84/77.

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The new limits of integration are from u=0 to u=6sin(6) after the substitution u=6x is applied. The integral evaluates to (1/6)[cos(6sin(6))+1].

Let us assume the substitution u = 6x.

First, we need to find the new limits of integration by substituting u=6x into the original limits of integration:

When x=0, u=6(0) = 0.

When x=sin(6), u=6sin(6).

Therefore, the new limits of integration are from u=0 to u=6sin(6).

Next, we need to express the integral in terms of u by substituting x back in terms of u:

When x=0, u=6(0) = 0, so x=u/6.

When x=sin(6), u=6sin(6), so x=u/6.

Therefore, we have:

∫0sin(6) dx = (1/6) ∫0⁶ sin(6u/6) du

Simplifying, we get:

(1/6) ∫0⁶ sin(u) du

which evaluates to:

(1/6) [-cos(u)] from 0 to 6sin(6)

Plugging in the limits of integration, we get:

(1/6) [-cos(6sin(6)) + cos(0)]

which simplifies to:

(1/6) [-cos(6sin(6)) + 1]

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The new limits of integration are from u=0 to u=6sin(6) after the substitution u=6x is applied. The integral evaluates to (1/6)[cos(6sin(6))+1].

Let us assume the substitution u = 6x.

First, we need to find the new limits of integration by substituting u=6x into the original limits of integration:

When x=0, u=6(0) = 0.

When x=sin(6), u=6sin(6).

Therefore, the new limits of integration are from u=0 to u=6sin(6).

Next, we need to express the integral in terms of u by substituting x back in terms of u:

When x=0, u=6(0) = 0, so x=u/6.

When x=sin(6), u=6sin(6), so x=u/6.

Therefore, we have:

∫0sin(6) dx = (1/6) ∫0⁶ sin(6u/6) du

Simplifying, we get:

(1/6) ∫0⁶ sin(u) du

which evaluates to:

(1/6) [-cos(u)] from 0 to 6sin(6)

Plugging in the limits of integration, we get:

(1/6) [-cos(6sin(6)) + cos(0)]

which simplifies to:

(1/6) [-cos(6sin(6)) + 1]

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The line integral is zero: ∫C F · dr = 0

To apply Green's Theorem, we need to find the curl of the vector field F [tex]= (3 sin y, 3x cos y)[/tex].

∂F2/∂x = 3 cos y

∂F1/∂y = 3 cos y

So, curl F = (∂F2/∂x - ∂F1/∂y) = 0

Since the curl of F is zero, we can apply Green's Theorem to evaluate the line integral along the given curve C, which is the ellipse [tex]x^2 + xy + y^2 = 36[/tex], oriented in the counterclockwise direction.

∫C F · dr = ∬R (∂F2/∂x - ∂F1/∂y) dA

where R is the region enclosed by C.

Since the curl of F is zero, the line integral is equal to the double integral of the curl of F over the region R, which is zero. Therefore, the line integral is zero:

∫C F · dr = 0

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The measures of the arc angles KL and MJ are 80° and 20° respectively derived using the Angles of Intersecting Chords Theorem

The Angles of Intersecting Chords Theorem states that the angle formed by the intersection of the chords is equal to half the sum of the intercepted arcs, and conversely, that the measure of an intercepted arc is half the sum of the two angles that intercept it.

50 = (KL + MJ)/2

100 = KL + MJ...(1)

30 = (MJ - KL)/2 {secant secant angle}

60 = MJ - KL...(2)

adding equations (1) and (2) we have;

160 = 2KL

divide through by 2

KL = 80°

Putting 80° for KL in equation (1), we have;

MJ = 100 - 80

MJ = 20°

Therefore, measures of the arc angles KL and MJ are 80° and 20° respectively by application of the Angles of Intersecting Chords Theorem

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Answer:

x = 9

Step-by-step explanation:

since the triangles are similar then the ratios of corresponding sides are in proportion, that is

[tex]\frac{BC}{DE}[/tex] = [tex]\frac{AB}{AD}[/tex] ( substitute values )

[tex]\frac{2x+3}{15}[/tex] = [tex]\frac{20+8}{20}[/tex] = [tex]\frac{28}{20}[/tex] ( cross- multiply )

20(2x + 3) = 15 × 28 = 420 ( divide both sides by 20 )

2x + 3 = 21 ( subtract 3 from both sides )

2x = 18 ( divide both sides by 2 )

x = 9

Answer:

x = 9

Step-by-step explanation:

In similar triangles, the corresponding sides are in same proportion.

 AB = AD + DB

       = 20 + 8

       =28

        ΔABC ~ ΔADE,

     [tex]\sf \dfrac{BC}{DE}=\dfrac{AB}{AD}\\\\\\\dfrac{2x + 3}{15}=\dfrac{28}{20}\\\\\ 2x + 3=\dfrac{28}{20}*15[/tex]

     [tex]2x + 3 = 7[/tex] *3

    2x + 3 = 21

Subtract 3 from both sides,

          2x = 21 - 3

          2x = 18

Divide both sides by 2,

            x = 18 ÷ 2

            x = 9

The Area of a regular pentagon will be "139.4 cm²". To understand the calculation, check below.

According to the question,

Side length (a) = 9 cm

We know the formula,

[tex]\bold{Area} \ \text{of Pentagon} =\dfrac{1}{4} \sqrt{5(5+25)\text{a}^2}[/tex]  

By substituting the values, we get

                             [tex]=\dfrac{1}{4} \sqrt{5(5+25)(9)^2}[/tex]

                             [tex]=\dfrac{1}{4} \sqrt{5(30)81}[/tex]

                             [tex]=\dfrac{1}{4} \sqrt{150\times81}[/tex]

                             [tex]= 139.36 \ \text{or}[/tex],

                             [tex]= 139.4 \ \text{m}^2[/tex]

Thus the above answer is correct.

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Therefore, 68 child tickets were sold on Monday.

Sales typically refer to the exchange of goods or services for money or other consideration, such as barter. It is the activity or process of selling products or services to customers. Sales can occur through various channels, such as in-person transactions, online transactions, phone orders, etc.

Let's assume that x represents the number of child tickets sold on Monday.

Then, the number of adult tickets sold can be represented as (155 - x), since the total number of tickets sold was 155.

The total revenue from child tickets can be represented as 5.30x, since the price of one child ticket is $5.30.

Similarly, the total revenue from adult tickets can be represented as 9.30(155 - x), since the price of one adult ticket is $9.30 and there were (155 - x) adult tickets sold.

The total sales revenue for the day was $1169.50, so we can write:

5.30x + 9.30(155 - x) = 1169.50

Expanding this equation gives:

5.30x + 1441.50 - 9.30x = 1169.50

Simplifying and solving for x, we get:

-4.00x = -272.00

x = 68

Therefore, 68 child tickets were sold on Monday.

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(a) The probability that a student does not drink once a month= 0.40

(b) The probability corresponding to each outcome would depend on the           given relative  probability

(c) You can calculate the value of this expression to find the probability    that at least one person drinks once a month.

a. The probability that a student does not drink once a month can be calculated as 1 minus the probability that a student drinks once a month, which is given as P(D) = 0.60.

So, P(not D) = 1 - P(D) = 1 - 0.60 = 0.40.

b. If we randomly select 2 students at a time and record their drinking habits, the sample space corresponding to this experiment would consist of all possible combinations of drinking habits for the two students.

Since each student can either drink once a month (D) or not drink once a month (not D), there are four possible outcomes:

Both students drink once a month (D, D)

Both students do not drink once a month (not D, not D)

The first student drinks once a month and the second student does not (D, not D)

The first student does not drink once a month and the second student drinks once a month (not D, D)

The probability corresponding to each outcome would depend on the given relative frequency or probability of a student drinking once a month (P(D)) and the complement of that probability (1 - P(D)).

c. Assuming that 5 students are selected at random, the probability that at least one person drinks once a month can be calculated using the complement rule.

The complement of the event "at least one person drinks once a month" is "none of the 5 students drink once a month" or "all 5 students do not drink once a month".

The probability that one student does not drink once a month is P(not D) = 0.40 (as calculated in part a).

The probability that all 5 students do not drink once a month is [tex](P(not D))^5[/tex], since the events are assumed to be independent.

So, the probability that at least one person drinks once a month is:

P(at least one person drinks once a month) = 1 - P(none of the 5 students drink once a month)

= 1 - [tex](P(not D))^5[/tex]

= 1 - [tex](0.40)^5[/tex] (after substituting the value of P(not D) from part a)

You can calculate the value of this expression to find the probability that at least one person drinks once a month.

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The transition matrix from b to b' is  P = [2 1, 4 3]

To find the transition matrix from b to b', we need to find the matrix P such that P[b] = b'.

First, we need to express the elements of b' in terms of the basis b. To do this, we solve the equation x[1](1,0) + x[2](0,1) = (2,4) for x[1] and x[2]. This gives us x[1] = 2 and x[2] = 4. Similarly, we solve the equation y[1](1,0) + y[2](0,1) = (1,3) for y[1] and y[2]. This gives us y[1] = 1 and y[2] = 3.

Now, we can construct the matrix P using the coefficients we just found. The columns of P are the coordinate vectors of the elements of b' expressed in terms of the basis b.

P = [x[1] y[1], x[2] y[2]]
 = [2 1, 4 3]

Therefore, the transition matrix from b to b' is  P = [2 1, 4 3]

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The required answer is Probability = 100 / 100 = 1

If one test subject is randomly selected, the probability that they tested negative or used marijuana can be found using the formula:

P(Negative or Marijuana) = P(Negative) + P(Marijuana) - P(Negative and Marijuana)

Assuming that the test is accurate, the probability of testing negative for marijuana use is typically high for non-users. Let's assume that this probability is 0.95.

On the other hand, the probability of testing positive for marijuana use is typically high for users. Let's assume that this probability is 0.75.

If we assume that the proportion of marijuana users in the sample is 0.2, we can use this to calculate the probability of a subject being a marijuana user and testing negative (0.05 x 0.2 = 0.01).

Using these probabilities, we can calculate the probability that the subject tested negative or used marijuana:

P(Negative or Marijuana) = 0.95 + 0.2 - 0.01 = 1.14

probabilities cannot be greater than 1, so we must adjust this probability by subtracting the probability of both events occurring (i.e., a marijuana user testing negative):

P(Negative or Marijuana) = 0.95 + 0.2 - 0.01 - 0.15 = 1.09
Therefore, the probability that the subject tested negative or used marijuana is 1.09 (or 109%). This is an incorrect probability, as probabilities cannot be greater than 1. It is likely that there is an error in the assumptions or calculations made, and further analysis is needed.

To find the probability that the subject tested negative or used marijuana, you will need to follow these steps:

This mathematical definition of probability can extend to infinite sample spaces, and even uncountable sample spaces, using the concept of a measure.
Step 1: Determine the total number of test subjects.
In this case, the total number of test subjects is not given, so we'll assume it's 100 for simplicity.

Step 2: Determine the number of subjects who tested negative and the number of subjects who used marijuana.
Let's assume 'x' test subjects tested negative and 'y' test subjects used marijuana. Since these are the only two options, x + y = 100.

Step 3: Calculate the probability that the subject tested negative or used marijuana.
The probability that the subject tested negative or used marijuana can be found by dividing the sum of subjects who tested negative and subjects who used marijuana (x + y) by the total number of subjects (100).

Probability = (x + y) / 100

Since x + y = 100, the probability of the subject tested negative or used marijuana is:

Probability = 100 / 100 = 1

Therefore, the probability that the subject tested negative or used marijuana is 1, or 100%.

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Answer:

mean ( averege) is the sum of the value divided by the number of value

mean= (5+12+17+7+7) / 5

=48/5

= 9.6

Answer:

The mean is the sum of the value divided by the number of the values.

i.e. mean = (5+12+17+7+7) / 5

=48/5

= 9.6

When the correct time is 8 p.m. the following evening, the faulty watch, which gains 10 seconds an hour will show 8:04 p.m.

The time is determined using the mathematical operations of division and multiplication.

The time that the faulty watch gains per hour = 10 seconds

The total number of hours from 8 p.m. one evening to the next = 24 hours

The total number of seconds that the faulty watch must have gained during the 24 hours = 240 seconds (24 x 10)

60 seconds = 1 minute

240 seconds = 4 minutes (240 ÷ 60)

Thus, while the correct time is showing 8 p.m., the faulty watch will be showing 8:04 p.m.

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In the expression, the values are:

a) k = 2.4

b) c = -141.0625

We have the expression:

a√(b² − c) = 5k

a)For the value of ‘k’ when a = 3, b = 6 and c = 20, we have:

a√(b² − c) = 5k

3√(6² − 20) = 5k             (solve for k)

3√(36 − 20) = 5k

3√(16) = 5k  

3*4 = 5k  

12 = 5k

k = 12/5

k = 2.4

b) For the value of ‘c’ when a = 4, b = 7 and k = 11, we have:

a√(b² − c) = 5k

4√(7² − c) = 5*11          (solve for c)

4√(49 − c) = 55

√(49 − c) = 55/4

√(49 − c) = 13.75

49 − c = 13.75²

49 - c = 189.0625

c = 49 - 189.0625

c = -141.0625

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a)The solution to the given initial value problem is

[tex]y(t) = ((2h/27) + (4/3))e^(^3^t^) + ((2h/27) - (4/3))e^(^-^t^) - (h/3)(t-4) - (2h/9)[/tex]

b)The solution to the given initial value problem is

[tex]x(t) = (F0/(2ω^2))sin ωt - (F0/(2ω^2))cos ωt[/tex]

For the first problem, we can use the method of undetermined coefficients to find a particular solution to the non-homogeneous differential equation.

Let's assume that the solution has the form y_p = A(t-4) + B.

Taking the first and second derivatives, we have y'_p = A and y''_p = 0.

Substituting these expressions into the differential equation, we get:

[tex]0 - 2A - 3(A(t-4) + B) = h(t-4)[/tex]

To simplify, we have:

[tex]-3At + (6A - 3B) = h(t-4)[/tex]

To satisfy this equation for all t, we must have -3A = h and 6A - 3B = 0.

Solving for A and B, we get A = -h/3 and B = -2h/9.

Therefore, the particular solution is

[tex]y_p = (-h/3)(t-4) - (2h/9).[/tex]

To find the general solution to the homogeneous differential equation, we first solve the characteristic equation:

[tex]r^2 - 2r - 3 = 0[/tex]

Factoring, we get (r-3)(r+1) = 0, so r = 3 or r = -1.

Therefore, the general solution to the homogeneous equation is

[tex]y_h = c_1e^(^3^t^) + c_2e^(^-^t^).[/tex]

The general solution to the entire differential equation is the sum of the homogeneous and particular solutions:

[tex]y = y_h + y_p.[/tex]

Plugging in the initial conditions, we have:

[tex]y(0) = 8 = c_1 + c_2 - (2h/9)y'(0) = 0 = 3c_1 - c_2 - (h/3)[/tex]

Solving for c_1 and c_2 in terms of h, we get

c_1 = (2h/27) + (4/3) and c_2 = (2h/27) - (4/3).

Therefore, the solution to the initial value problem is:
[tex]y(t) = ((2h/27) + (4/3))e^(^3^t^) + ((2h/27) - (4/3))e^(^-^t^) - (h/3)(t-4) - (2h/9)[/tex]


For the second problem, we can use the method of undetermined coefficients again.

Let's assume that the solution has the form x_p = A sin ωt.

Taking the second derivative, we have [tex]d^2x_p/dt^2 = -Aω^2 sin ωt.[/tex]

Substituting these expressions into the differential equation, we get:

[tex]-Aω^2 sin ωt + ω^2A sin ωt = F0 sin ωt[/tex]

Simplifying, we get -2Aω^2 sin ωt = F0 sin ωt, so A = -F0/(2ω^2).

The general solution to the homogeneous differential equation is

[tex]x_h = c_1 cos ωt + c_2 sin ωt.[/tex]

Therefore, the general solution to the entire differential equation is

[tex]x = x_h + x_p.[/tex]

Plugging in the initial conditions, we have:

x(0) = 0 = c_1

x'(0) = 0 = c_2ω - (F0/(2ω))

Solving for c_2 in terms of F0 and ω, we get c_2 = F0/(2ω^2).

Therefore, the solution to the initial value problem is:

[tex]x(t) = (F0/(2ω^2))sin ωt - (F0/(2ω^2))cos ωt[/tex]

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δy = 0.1192 and dy = -1.08.

To find δy and dy, we can use the formula:
δy = f(x+δx) - f(x)
dy = f'(x) * dx
where f(x) = x^4 and x = -3.

First, let's find δy:
δy = f(-3+0.01) - f(-3)
δy = (-3.01)^4 - (-3)^4
δy = 0.1192

Next, let's find dy:
dy = f'(-3) * 0.01
dy = 4(-3)^3 * 0.01
dy = -1.08

Comparing δy and dy, we can see that they have different signs and magnitudes. δy is positive and has a magnitude of 0.1192, while dy is negative and has a magnitude of 1.08.

This makes sense because δy represents the change in y due to a small change in x, while dy represents the instantaneous rate of change of y with respect to x at a specific point.

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The height of the rider at any time t after the initial time t=0, in meters above the ground is h = 18 - 14*sin((π/3)*t)

Now, let's consider the position of the rider at some time t after the initial time t=0.

We can find the angle that the rider has traveled around the wheel by multiplying the angular velocity by the time elapsed:

θ = (π/3) * t

To find the vertical position of the rider at this angle, we can use the sine function, since the height of the rider on the Ferris wheel varies sinusoidally as the wheel rotates:

h = 18 - 14*sin(θ)

Plugging in the expression for θ, we get:

h = 18 - 14*sin((π/3)*t)

This formula gives the height of the rider at any time t after the initial time t=0, in meters above the ground.

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x+y=10

225x+70y=1165 where x represents the quantity of sliders in the combo.

The quantity of chicken wings in the combo is given by y.

An equation is an expression of one or more variables equated to a constant or another expression. Equations are used in mathematics to describe the relationship between two or more variables and often involve operations such as addition, subtraction, multiplication, division, and exponents. Equations can also be used to describe physical phenomena and other phenomena such as chemical reactions, motion, and electricity.
Let x represent the number of sliders in the combination meal and let y represent the number of chicken wings in the combination meal. The system of equations used to determine the number of sliders and chicken wings in the combination meal is given by:
Equation 1: 225x + 70y = 1165
Equation 2: x + y = 10
This system of equations can be used to find the number of sliders and chicken wings in the combination meal. To solve this system, the equations must be manipulated to isolate one of the variables. For example, if we subtract 70y from both sides of Equation 1, we obtain 225x = 1165 - 70y. We can then divide both sides by 225 to obtain x = (1165 - 70y)/225. This expression can then be substituted into Equation 2 to obtain a single equation in one variable.
Therefore, the system of equations given by Equation 1 and Equation 2 can be used to determine the number of sliders and chicken wings in the combination meal.
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x+y=10

225x+70y=1165 where x represents the quantity of sliders in the combo.

The quantity of chicken wings in the combo is given by y.

An equation is an expression of one or more variables equated to a constant or another expression. Equations are used in mathematics to describe the relationship between two or more variables and often involve operations such as addition, subtraction, multiplication, division, and exponents. Equations can also be used to describe physical phenomena and other phenomena such as chemical reactions, motion, and electricity.
Let x represent the number of sliders in the combination meal and let y represent the number of chicken wings in the combination meal. The system of equations used to determine the number of sliders and chicken wings in the combination meal is given by:
Equation 1: 225x + 70y = 1165
Equation 2: x + y = 10
This system of equations can be used to find the number of sliders and chicken wings in the combination meal. To solve this system, the equations must be manipulated to isolate one of the variables. For example, if we subtract 70y from both sides of Equation 1, we obtain 225x = 1165 - 70y. We can then divide both sides by 225 to obtain x = (1165 - 70y)/225. This expression can then be substituted into Equation 2 to obtain a single equation in one variable.
Therefore, the system of equations given by Equation 1 and Equation 2 can be used to determine the number of sliders and chicken wings in the combination meal.
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The box and whicker plot of the following prices of some DVDs is illustrated below.

To create a box and whisker plot, we first need to order the data from smallest to largest. Then, we can find the median, which is the middle value of the data set. In this case, the median is 18.99.

Next, we can find the lower quartile (Q1) and upper quartile (Q3), which divide the data set into four equal parts. Q1 is the median of the lower half of the data, and Q3 is the median of the upper half of the data. In this case, Q1 is 12.99 and Q3 is 21.99.

With these values, we can draw a box that represents the middle 50% of the data, with the bottom of the box at Q1 and the top of the box at Q3. Inside the box, we draw a line at the median. This box shows the interquartile range (IQR), which is a measure of the spread of the data.

Finally, we can draw whiskers that extend from the box to the minimum and maximum values that are not considered outliers. Outliers are data points that are more than 1.5 times the IQR away from the box. In this case, the minimum value is 9.99 and the maximum value is 26.99, but since 26.99 is an outlier, we only draw a whisker up to the next highest value, which is 21.99.

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The proportion of samples that will have means between 115 and 120 is 0.4772 or 47.72%.

To solve this problem, we need to use the Central Limit Theorem, which states that the sample means of a large sample size (n>30) from any population with a finite mean and variance will be approximately normally distributed with a mean equal to the population mean (μ) and a standard deviation equal to the population standard deviation (σ/sqrt(n)).

Here, the population mean (μ) = 120, and the population standard deviation (σ) = 10. We are given that the sample size (n) = 16, and we need to find the proportion of samples that will have means between 115 and 120.

First, we need to standardize the sample means using the z-score formula:

z = (x - μ) / (σ / sqrt(n))

where x is the sample mean.

For the lower limit of 115:

z1 = (115 - 120) / (10 / sqrt(16)) = -2

For the upper limit of 120:

z2 = (120 - 120) / (10 / sqrt(16)) = 0

We need to find the area under the standard normal distribution curve between z1 = -2 and z2 = 0. We can use a standard normal distribution table or a calculator to find this area.

Using a standard normal distribution table, we find that the area between z1 = -2 and z2 = 0 is 0.4772.

Therefore, the proportion of samples that will have means between 115 and 120 is 0.4772 or 47.72%.

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The proportion of samples that will have means between 115 and 120 is 0.4772 or 47.72%.

To solve this problem, we need to use the Central Limit Theorem, which states that the sample means of a large sample size (n>30) from any population with a finite mean and variance will be approximately normally distributed with a mean equal to the population mean (μ) and a standard deviation equal to the population standard deviation (σ/sqrt(n)).

Here, the population mean (μ) = 120, and the population standard deviation (σ) = 10. We are given that the sample size (n) = 16, and we need to find the proportion of samples that will have means between 115 and 120.

First, we need to standardize the sample means using the z-score formula:

z = (x - μ) / (σ / sqrt(n))

where x is the sample mean.

For the lower limit of 115:

z1 = (115 - 120) / (10 / sqrt(16)) = -2

For the upper limit of 120:

z2 = (120 - 120) / (10 / sqrt(16)) = 0

We need to find the area under the standard normal distribution curve between z1 = -2 and z2 = 0. We can use a standard normal distribution table or a calculator to find this area.

Using a standard normal distribution table, we find that the area between z1 = -2 and z2 = 0 is 0.4772.

Therefore, the proportion of samples that will have means between 115 and 120 is 0.4772 or 47.72%.

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Answer:

1) 616[tex]cm^{3}[/tex]

2) 4.2in

Step-by-step explanation:

1)

the volume of a sphere is [tex]4\pi r^{2}[/tex]

so you plug in the radius for r, and that is 4*[tex]\pi[/tex]*[tex]7^{2}[/tex] = 196[tex]\pi[/tex]= approximately 615.75216

round this to the nearest cubic centimeter and its 616

2)

the volume of a sphere is [tex]4\pi r^{2}[/tex]

so [tex]4\pi r^{2}[/tex]=V and V=72[tex]\pi[/tex]

72[tex]\pi[/tex]/4[tex]\pi[/tex]=[tex]4\pi r^{2}[/tex]/4[tex]\pi[/tex]

[tex]\sqrt{18} =\sqrt{r^{2} }[/tex]

4.24264=r

round this to the nearest tenth of an inch and r = 4.2

The number x that maximizes the product xy when x and y have a sum of 62 is:

x = 31.

To find the number x when the sum of two numbers x and y is 62 and the product xy is maximum, we can use the concept of optimization.

Step 1: Write down the given information:
x + y = 62 (the sum of x and y)

Step 2: Express one variable in terms of the other:
y = 62 - x

Step 3: Write the function to be maximized:
P(x) = x * y = x * (62 - x)

Step 4: Find the derivative of the function:
P'(x) = (62 - x) - x

Step 5: Set the derivative to zero and solve for x:
0 = (62 - x) - x
2x = 62
x = 31

So, the number x that maximizes the product xy when x and y have a sum of 62 is x = 31.

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Answer:

A

Step-by-step explanation:

To find the area of the shaded region, we need to subtract the area of the square from the area of the circle:

Area of shaded region = Area of circle - Area of square

Area of shaded region = 81.3 - 16.1

Area of shaded region = 65.2 cm^2

The error interval for the area, a, of the shaded region can be found by considering the errors in the measurements of the areas of the circle and the square. For the area of the circle, the value is rounded to 1 decimal place, so the error is at most 0.05 cm^2 (half of the value of the smallest decimal place). For the area of the square, the value is truncated to 1 decimal place, so the error is at most 0.1 cm^2 (the value of the smallest decimal place).

Thus, the error interval for the area, a, of the shaded region is:

m < a < n

where:

m = 81.3 - 16.1 - 0.1 = 65.1 cm^2

n = 81.3 - 16.1 + 0.05 = 65.25 cm^2

Therefore, the error interval for the area, a, of the shaded region is:

65.1 < a < 65.25

The price per cup of the 2-gallon container of laundry detergent costing $30.40 is $0.30 per cup.

To calculate the price per cup, we first need to convert 2 gallons to cups.

1 gallon = 16 cups

So, 2 gallons = 2 x 16 = 32 cups

Now, to find the price per cup, we divide the total price of the container by the number of cups in it:

Therefore, the price per cup of laundry detergent is $0.30 per cup (rounded to two decimal places).

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This is the correct answer

The relation R defined on set A={1,2,3,...,22} as xRy ⇔ 4|(x-y) is an equivalence relation. The equivalence classes are {1,5,9,13,17,21}, {2,6,10,14,18,22}, {3,7,11,15,19}, and {4,8,12,16,20}.

Since R is an equivalence relation on A, it partitions A into disjoint equivalence classes.

The equivalence class of an element a ∈ A is the set of all elements in A that are related to a under R.

Using set-roster notation, we can write the equivalence classes of R as follows

[1] = {1, 5, 9, 13, 17, 21}

[2] = {2, 6, 10, 14, 18, 22}

[3] = {3, 7, 11, 15, 19}

[4] = {4, 8, 12, 16, 20}

Each equivalence class contains all elements that are congruent modulo 4.

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--The given question is incomplete, the complete question is given

"  Let A = {1, 2, 3, 4, , 22} And define a relation R on A as follows

For all x, y ∈ A, x R y ⇔ 4|(x − y).

It is a fact that R is an equivalence relation on A. Use set-roster notation to write the equivalence classes of R."--