What is the solution set for the following system of linear inequalities? Graph the solution set of the system of linear inequalities in the coordinate plane. First, select the Graph 1 button to graph the line and choose the line style. To graph a line, select two points in the coordinate plane. A line will connect the points. Then select the Graph 2 button to graph the line and choose the line style. Then select the Solution Set button to select the desired region.

The equation of the parabola in vertex form is: y = (1/4)x² - 5. We can calculate it in the following manner.

Since the parabola opens either up or down, its vertex form is given by:

y = a(x - h)² + k

where (h, k) is the vertex of the parabola and "a" is a constant that determines the direction and shape of the parabola.

Using the given vertex (0, -5), we have:

h = 0 and k = -5

Substituting the vertex coordinates into the vertex form, we get:

y = a(x - 0)² - 5

y = ax² - 5

To determine the value of "a", we use the given point (–2, –4) on the parabola. Substituting these coordinates into the equation, we get:

-4 = a(-2)² - 5

-4 = 4a - 5

1 = 4a

a = 1/4

Substituting this value of "a" into the equation for the parabola, we get:

y = (1/4)x² - 5

Therefore, the equation of the parabola in vertex form is:

y = (1/4)x² - 5

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Each data point's ones digit is displayed in the second column, while its tens digit is displayed in the first column. Each "1|1" in the key denotes the value "11" in the data collection.

The term "statistical data" refers to numerical data that is gathered from a sample or population in order to characterise or draw conclusions about a larger group. Observational studies, surveys, experiments, and other techniques can all be used to gather statistical data, as can data from already-existing records and databases. Many approaches and methodologies, including data visualisation, inferential statistics, and descriptive statistics, can be used to examine statistical data. Measures of central tendency (mean, median, mode), indicators of dispersion (standard deviation, range, interquartile range), and graphical representations are some examples of the key characteristics of a dataset that are summarised and described using descriptive statistics (histograms, box plots, scatter plots). A bigger population can be inferred from a sample of data using inferential statistics, on the other hand.

given

The stem-and-leaf plot that depicts the data set accurately is as follows:

Season Points for the 2021 Riverside Red Dragon Football

0 0 3

1 1 7 7 7 7

2 0 0 0 2 4 8 8

3 1 3

Key 1|1 = 11

This stem-and-leaf plot depicts the frequency distribution of the data set in an accurate manner.

Each data point's ones digit is displayed in the second column, while its tens digit is displayed in the first column. Each "1|1" in the key denotes the value "11" in the data collection.

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Answer:

h=

[tex] \frac{15}{4} [/tex]

Step-by-step explanation:

Divide both sides by 8:

[tex]2 - 7 + 4h = \frac{80}{8} [/tex]

Divide 80 by 8 to get 10:

[tex]2 - 7 + 4h = 10[/tex]

subtract 7 from 2 to get -5:

[tex] - 5 + 4h = 10[/tex]

add 5 to both sides:

[tex]4h = 10 + 5[/tex]

Add 10 and 5 to get 15:

[tex]4h = 15[/tex]

divide both sides by 4.:

[tex]h = \frac{15}{4} [/tex]

Hopefully this helps! :)

Answer:

x = - 2 [tex]\frac{1}{4}[/tex]

Step-by-step explanation:

10(5x + 12) = 2(5x + 15)  Distribute the 10 and the 2

10(5x) + 10(12)  = 2(5x) + 2(15)

50x + 120 = 10x + 30  Subtract 10x from both sides

50x - 10x +120 = 10x - 10x + 30

40x + 120 = 30    Subtract 120 from both sides

40x + 120 - 120 = 30 - 120

40x = -90  Divide both sides by 40

[tex]\frac{40x}{40}[/tex] = [tex]\frac{-90}{40}[/tex]

x = [tex]\frac{-9}{4}[/tex] or - 2[tex]\frac{1}{4}[/tex]

Helping in the name of Jesus.

For the first part, the number of license plates is: 65,780,800 and for the second part, the number of license plates with 3 consecutive digits is: 16,524,288.

If there are no restrictions on where the digits and letters are placed, the number of 8-place license plates consisting of 5 letters and 3 digits with no repetitions allowed can be found using the permutation formula:

nPr = n! / (n-r)!

where n is the number of available characters (26 letters and 10 digits) and r is the number of characters needed for the license plate (8).

Therefore, the number of license plates is:

(26 P 5) x (10 P 3) = (26!/21!) x (10!/7!) = 65,780,800

If the 3 digits must be consecutive, there are 8 possible positions for the block of digits (either the first 3, second 3, or last 3). Once the position of the block is chosen, the number of license plates can be found by counting the number of ways to arrange the letters and the block of digits. The number of ways to arrange the letters is 26 P 5, and the number of ways to arrange the block of digits is 10 (since there are only 10 possible sets of consecutive digits).

Therefore, the number of license plates with 3 consecutive digits is:

8 x (26 P 5) x 10 = 16,524,288

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The complete question goes thus:

If there are no restrictions on where the digits and letters are placed, how many 8

-place license plates consisting of 5

letters and 3

digits are possible if no repetitions of letters or digits are allowed? What if the 3

digits must be consecutive?

Answer:

The secant function is defined as the reciprocal of the cosine function. So if `sec(theta) = √3`, then `cos(theta) = 1/√3`. Since `sin^2(theta) + cos^2(theta) = 1`, we can solve for `sin(theta)`:

`sin^2(theta) + cos^2(theta) = 1`

`sin^2(theta) + (1/√3)^2 = 1`

`sin^2(theta) + 1/3 = 1`

`sin^2(theta) = 2/3`

`sin(theta) = ±√(2/3)`.

Since it is given that `sin theta > 0`, we can conclude that `sin(theta) = √(2/3)`.

Now that we have the values of sine and cosine, we can find the remaining trigonometric functions:

`tan(theta) = sin(theta)/cos(theta)`

`= (√(2/3)) / (1/√3)`

`= √(6)/3`

`cot(theta) = cos(theta)/sin(theta)`

`= (1/√3)/(√(2/3))`

`= √6 / 6`

`csc theta = 1/sin theta`

`= 1/(√(2/3))`

`= √6 / √4`

So, in summary:

- sec theta = √3

- cos theta = 1 / √3

- sin theta = √(2 / 3)

- tan theta = √6 / 3

- cot theta = √6 / 6

- csc theta = √6 / √4

Is there anything else you would like to know?

Answer:

FH = 6.0

FG = 3.6

m<H = 31°

Step-by-step explanation:

sin Θ = opp/hyp

cos Θ = adj/hyp

sin 59° = FH/GH

sin 59° = FH/7

FH = 7 × sin 59°

FH = 6.0

cos 59° = FG/GH

cos 59° = FG/7

FG = 7 × cos 59°

FG = 3.6

m<F + m<G + m<H = 180°

90° + 59° + m<H = 180°

m<H = 31°

Answer:

Refers to the attachment given below!

Step-by-step explanation:

Answer:

Step-by-step explanation:

(a) The probability that at least two of the five points selected from the interval (0,1) are less than 1/3 is (161/243).  (b) The probability that exactly two of the points have the first decimal point as 3 is (3645/100000).

(a) The probability that at least two points out of five will be less than 1/3 can be found by subtracting the probabilities that no points and one point are less than 1/3 from 1.

Let us assume that each point is selected independently and at random.

P(no point less than 1/3) = (2/3)^5

P(one point less than 1/3) = 5C1(1/3)(2/3)^4 = (5/3)(2/3)^5

So the probability that at least two points are less than 1/3 is:

P(at least two points less than 1/3) = 1 - P(no point less than 1/3) - P(one point less than 1/3)

= 1 - (2/3)^5 - (5/3)(2/3)^5

= (161/243)

(b) To find the probability that the first decimal digit of exactly two of the five selected points is 3, we use a combination of counting and probability reasoning. There are five ways to choose two points out of five, which can each start with 3:

(0.3**, **.3**, **3.*, **.33, 33..)

The probability that the first decimal place of any point is 3 is 1/10, since there are 10 equally likely numbers in each of the five intervals. The probability that the first decimal point of each of the other three points is not 3 is 9/10.

Therefore, the probability of choosing two points from five where exactly the first decimal digit of both is 3 is:

P(exactly two points have first decimal digit of 3) = 5(1/10)^2(9/10)^3= (3645/100000)

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Katie will pay $1.10 for shipping and handling.

The cost of shipping refers to the amount of money that is charged for the transportation of goods or products from one location to another.

To calculate how much Katie will pay for shipping and handling, we first need to find 10% of $11.

To do this, we multiply $11 by 0.10 (which is the decimal equivalent of 10%).

$11 x 0.10 = $1.10

Therefore, Katie will pay $1.10 for shipping and handling on top of the $11 she paid for the set of spoons.

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The value of [tex]$\tan A$[/tex] in its simplest radical form is [tex]8/3$.[/tex]

What is radical form?

Radical form refers to a mathematical expression that includes radicals, which are symbols that indicate a root of a number.

To find the value of [tex]$\tan A$[/tex], we first need to determine the value of [tex]\cos A$.[/tex]

Using the Pythagorean identity, we have:

[tex]$$\sin^2 A + \cos^2 A = 1$$[/tex]

Substituting the given value of [tex]$\sin A$[/tex], we get:

[tex]$$\left(\frac{8}{\sqrt{73}}\right)^2 + \cos^2 A = 1$$[/tex]

Simplifying the left-hand side, we get:

[tex]$\frac{64}{73} + \cos^2 A = 1$$$$\cos^2 A = \frac{9}{73}$$$$\cos A = \frac{3}{\sqrt{73}}$$[/tex]

Since angle [tex]$A$[/tex] is in the first quadrant, both [tex]$\sin A$[/tex] and [tex]$\cos A$[/tex] are positive.

Therefore, we have:

[tex]$$\tan A = \frac{\sin A}{\cos A} = \frac{8/\sqrt{73}}{3/\sqrt{73}} = \frac{8}{3}$$[/tex]

So the value of [tex]$\tan A$[/tex] in its simplest form is [tex]8/3$.[/tex]

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The prοbability that the prοpοrtiοn οf red marbles in the sample frοm bin a is greater than the prοpοrtiοn οf red marbles frοm bin b is 0.1573.

Prοbability is a way οf calculating hοw likely sοmething is tο happen. It is difficult tο prοvide a cοmplete predictiοn fοr many events. Using it, we can οnly fοrecast the prοbability, οr likelihοοd, οf an event οccurring. The prοbability might be between 0 and 1, where 0 denοtes an impοssibility and 1 denοtes a certainty.

Here by using probability calculator for sampling distribution of [tex]p_1-p_2[/tex].

[tex]z=\frac{\hat p_1-\hat p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}[/tex]

Here [tex]\hat p_1=p_1[/tex] = 40% = 0.40 and [tex]\hat {p_2}=p_2[/tex] = 52% = 0.52 and [tex]n_1=0 , n_2=40[/tex]

=> [tex]z=\frac{0.40-0.52}{\sqrt{\frac{0.40(1-0.40)}{0}+\frac{0.52(1-0.52)}{40}}}[/tex]

=> z = -1.00561

Here [tex]\hat p_1 > \hat p_2[/tex]

=> [tex]\hat p_1 - \hat p_2 > 0[/tex]

Then , P(x<z) = 0.1573.

Hence  the prοbability that the prοpοrtiοn οf red marbles in the sample frοm bin a is greater than the prοpοrtiοn οf red marbles frοm bin b is 0.1573.

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In terms of journey times, Bus 14 is more reliable than Bus 47, as seen by its lower IQR and narrower range of travel times based on the line plot.

We must examine the data's variability to ascertain which bus is the most reliable. The interquartile range (IQR), which is the range of the middle 50% of the data, is one way to measure variability.

The line plot for Bus 47 reveals a range of trip times of 4 to 28 minutes, with an IQR of 8 minutes (from 10 to 18 minutes). This reflects a reasonable degree of consistency in journey times, as it shows that half of the students on Bus 47 have trip times that are within 8 minutes of one another.

With an IQR of six minutes, the line plot for Bus 14 reveals a travel time range of 8 to 28 minutes (from 14 to 20 minutes). This demonstrates that journey times on Bus 14 are more consistent than on Bus 47, with half of the students' travel times being within 6 minutes of one another.

We may also have a look at the data range, which is the range of numbers between the maximum and minimum. Bus 14 has a range of 20 minutes, while Bus 47 has a range of 24 minutes (from 4 to 28 minutes) (from 8 to 28 minutes). The conclusion that Bus 14 is more reliable is further supported by the suggestion that its range of travel times is lower.

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Three-digit addition and subtraction with regrouping involves adding or subtracting numbers with three digits (numbers between 100 and 999) while carrying or borrowing values from one place value to another.

Addition is a basic arithmetic operation that involves combining two or more numbers or quantities to find the total amount or sum. It is often denoted by the symbol "+". For example, if we add 2 and 3, we get a sum of 5.

For example, let's say we want to add 357 and 468:

In this example, we first add the ones place (7 + 8 = 15), which gives us a sum of 5 and carry-over of 1. We then add the tens place (5 + 6 = 11) and add the carry-over from the ones place, which gives us a sum of 2 and carry-over of 1. Finally, we add the hundreds place (3 + 4 = 7) and add the carry-over from the tens place, which gives us a final sum of 825.

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The volume of the region outside the cone phi = pi /4 and inside the sphere rho = 4 cos phi. The volume is 63.717

The limits of integration for ρ are 0 and 4 cos∅, since the sphere has a radius of 4 cos phi. The limits for phi are π/4 and π/2 because we want to consider the region outside the cone but inside the sphere, and phi is the angle between the positive z-axis and the position vector of the point. Finally, the limits for theta are 0 and 2π because theta is the azimuthal angle and we want to integrate over the entire azimuthal angle.

The volume of the region is therefore given by:

V = 8 x ∫(0 to 2π) ∫(π/4 to π/2) ∫(0 to 4 cos∅) ρ²

Simplifying the integral using the base number, we get:

V = 8 x (64/3 - 16/3√2π) = 63.717

Therefore, the volume of the region outside the cone phi = π/4 and inside the sphere ρ = 4 cos∅ is 8 (64/3 - 16/3√2π), which is the exact answer using π as needed.

Complete Question:

Use spherical coordinates to find the volume of the region outside the cone phi = pi /4 and inside the sphere rho = 4 cos phi. The volume is?

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Answer:

Step-by-step explanation:

To find the volume of a figure, you need to multiply the length, width, and height of the figure.

Answer: 0.09m > 56.60 + 0.07m, where m > 2830.

Step-by-step explanation:

Let's assume that Plan A costs more than Plan B for monthly phone use greater than some value m, where m is the number of minutes of phone use in a month.

For Plan A, the monthly cost can be expressed as:

Cost_A = 0 + 0.09m = 0.09m

For Plan B, the monthly cost can be expressed as:

Cost_B = 56.60 + 0.07m

To find the value of m for which Plan A costs more than Plan B, we need to set the two costs equal to each other and solve for m:

0.09m = 56.60 + 0.07m

0.02m = 56.60

m = 56.60 / 0.02

m = 2830

Therefore, for monthly phone use greater than 2830 minutes, Plan A will cost more than Plan B.

In mathematical notation, we can express this as:

0.09m > 56.60 + 0.07m, where m > 2830.

The experimental probabilities have their values to be P(3) = 1/12, P(6) = 1/4 and P(Less than 4) = 1/2

Experimental probability of 3

From the table of values, we have

n(3) = 1

Total = 12

So, we have

P(3) = 1/12

Experimental probability of 6

From the table of values, we have

n(6) = 3

Total = 12

So, we have

P(6) = 3/12

P(6) = 1/4

Experimental probability of less than 4

From the table of values, we have

n(Less than 4) = 6

Total = 12

So, we have

P(Less than 4) = 6/12

P(Less than 4) = 1/2

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Answer:

B) x=12

Step-by-step explanation:

theres a 50/50 shot at getting heads, half of 24 is 12

A card is drawn from a standard deck of 52 playing cards. the probability that the card will be a spade or a jack is 17/52 or 0.3269 (rounded to four decimal places).

The total number of playing cards in a deck is 52. Out of these 52 playing cards, there are 4 suits, and each of these suits includes 13 cards.

Hence, there are 4 types of cards, and each of these types includes 13 cards. Therefore, there are 52/4 = 13 cards per type.A card is drawn from the deck of 52 playing cards.

The probability that it will be a spade or a jack is given below:

The probability of a spade card is 13/52 or 1/4, since there are 13 spade cards in the deck of 52 playing cards.

The probability of a jack card is 4/52 or 1/13, since there are 4 jacks in the deck of 52 playing cards.

Therefore, the probability that the card will be a spade or a jack is (13/52) + (4/52) = 17/52 or 0.3269 (rounded to four decimal places).

Thus, the probability that the card will be a spade or a jack is 17/52 or 0.3269 .

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Answer: geometric because it has a steady rate of multiplying by 4.5.

Step-by-step explanation:

Answer:

To find the area, we will use formula of "area of circle" which is πr^2

Given:

Radius: 78cm

To find:

Area of circle: ?

Formula:

πr^2

Solution:

Area: π×r^2

: 3.14× 78^2

Area: 19,113cm

Answer:

19113cm is the area of circle

Hope you understand

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The match can be scheduled in 900 different ways.

Here, the number of players from each school = 3.

Let us assue that the players of the first school be A, B, and C.

And the players of the second school be X, Y, and Z.

Here, each player from the first school has to play twice with each player from the second school.

We can organize the schedule into six different rounds, i.e.,

R1 : AX BY CZ

R2 : AX BZ CY

R3 : AY BX CZ

R4 : AY BZ CX

R5 : AZ BX CY

R6 : AZ BY CX

So, the number of possible ways to do this would be : 6! = 720 ways.

Now, three rounds are played simultaneously in each round.

R1 : AX BZ CY

R2 : AX BZ CY

R3 : AY BX CZ

R4 : AY BX CZ

R5 : AZ BY CX

R6 : AZ BY CX

And R1 : AX BY CZ

R2 : AX BY CZ

R3 : AY BZ CX

R4 : AY BZ CX

R5 : AZ BX CY

R6 : AZ BX CY

The total number of possible ways for the second case would be twice of  6! / (2! 2! 2!), which is equal to 90+ 90 = 180.

Therefore, the total number of ways in which the match can be scheduled is 720 + 180 = 900.

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The complete question is:

central high school is competing against northern high school in a backgammon match. each school has three players, and the contest rules require that each player play two games against each of the other school's players. the match takes place in six rounds, with three games played simultaneously in each round. in how many different ways can the match be scheduled?

Answer:

  -1239/1961

Step-by-step explanation:

You want cos(A+B) given that tan(A) = 45/28 and cos(B) = 12/37.

Here, we'll use the tangent relations ...

The tangent of angle B can be found from ...

  tan(B)² = 1/cos(B)² -1

  tan(B)² = 1/(12/37)² -1 = 1225/144

  tan(B) = 35/12

Now the tangent of the angle sum is ...

  tan(A+B) = (tan(A) +tan(B))/(1 -tan(A)tan(B))

  = (45/28 +35/12)/(1 -(45/28)(35/12)) = (95/21)/(1 -75/16) = -1520/1239

Note that the tangent of this sum of two first-quadrant angles is negative. That means the result is a second-quadrant angle, so the cosine will also be negative.

  cos(A+B) = -1/√(tan(A+B)² +1)

  cos(A+B) = -1/√((1520/1239)² +1)

  cos(A+B) = -1239/1961

__

Additional comment

Some calculators maintain enough internal accuracy that you can obtain the answer directly from ...

  cos(arctan(45/28) +arccos(12/35))

The one shown in the attachment is not able to provide the ratio of integers equal to the floating point value it computes for this.

The best unit of measure for the liquid in family-sized container of apple cider is the liter - True.

The metric unit of volume is known as the litre (international spelling) or liter (American English spelling) (SI symbols L and l,[1] with another symbol also used: ℓ). It is equivalent to 1 cubic decimetre (dm3), 1000 cubic centimetres (cm3) or 0.001 cubic metre (m3). A litre or cubic decimetre fills a space of 10 cm × 10 cm × 10 cm (see figure) and is therefore equal to one-thousandth of a cubic metre. One litre of liquid water has a weight of almost precisely one kilogram, because the kilogram was initially defined in 1795 as the weight of one cubic decimetre of water at the temperature of melting ice (0 °C). However, subsequent redefinitions of the metre and kilogram mean that this relationship is no longer exact.

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I don't actually know but you could ask a professional.

This gives a result of t = 10, meaning that it will take 10 days for the amount of money in the box to be reduced to 25 cents.

If there are R125 in a box and half of the total is removed each day, it will take 10 days to reduce the total to 25 cents. This is because the total amount is being halved each day, and the amount remaining at the end of each day is the amount left at the start of the day, divided by two. This can be expressed mathematically using the formula[tex]R125/(2^t)[/tex] = 25, where t is the number of days.If we solve the equation for t, we get t = log2(R125/25). This gives a result of t = 10, meaning that it will take 10 days for the amount of money in the box to be reduced to 25 cents.To calculate this process over the 10 days, we can use the formula [tex]R125/(2^t)[/tex] for each day, where t is the day number. Starting at day 0, the amount remaining would be R125. At the end of day 1, the amount remaining would be R125/2 = R62.50. At the end of day 2, the amount remaining would be R62.50/2 = R31.25. This process continues until at the end of day 10, the amount remaining would be R0.25.

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Mr. Nelson took 35 students on the trip.

The equation that can be used to solve this problem is:

100 + 6.50n = 327.50

Where n is the number of students. To solve this equation, we need to subtract 100 from both sides of the equation:

6.50n = 227.50

Next, we need to divide both sides by 6.50:

n = 35

Therefore, Mr. Nelson took 35 students on the trip.

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