Find the measure of one interior angle in the regular polygon hexagon

The quotient and remainder are x³ -5x² + 10x - 17 and 55.

What is Synthetic division?

When the divisor is a linear factor, synthetic division is a technique used to carry out the division operation on polynomials.

Here, (x+2) is a linear factor which indicates that synthetic division can be applied.

So, we will divide the x^4 - 3x^3 - 7x + 1 by x+2.

(refer the attached solution)

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Answer:

Option B is true

Step-by-step explanation:

The formula for Cos is Adjacent/hypotenuse. 12 is Adjacent to angle e, and 17 is the hypotenuse.

Answer:

B: cos e = 12/17

Step-by-step explanation:

When using the SOHCAHTOA method you can see that Cos is adjacent over hypotenuse and in the triangle diagram they give you the measurements of the hypotenuse and the adjacent only, so you know that they would be using the cos equation.

0.66 litres will be used if she wants to use equal amounts of the solution in 3 experiments.

In mathematics, division is the process of breaking a number into equal parts and determining how many equal parts can be made. For example, dividing 15 by 3 divides 15 into three equal groups of five.

The symbol '÷' or '/' is used to represent division.

Any positive number with no decimal or fractional part is referred to as a whole number. Because these criteria are met, the numbers 1,2,15,500, and 1546 are all whole numbers.

Children will practise dividing whole numbers as they begin to learn their multiplication tables. This is due to the fact that all of the numbers in the times tables are whole numbers!

They will also learn that division is the inverse operation to multiplication, which simply means that division and multiplication perform opposite functions in mathematics.

We have 2 litres of saline water

and want to divide them in 3 experiment equally, i.e.

2 ÷ 3 = 0.6667

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Answer:

Its c

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

A is. You can only have the "x" used once to be a function. It must pass the vertical line test.

To construct a 90% confidence interval for the average speed of cars on the highway with a margin of error no more than ±0.5 mph and a standard deviation of 7 mph, you must clock at least 339 cars.

Step 1: Identify the critical value (z-score) for a 90% confidence interval. This can be found in a standard z-table or using a calculator. The critical value for a 90% confidence interval is 1.645.

Step 2: Use the margin of error (E) formula to calculate the sample size (n):
E = (z * σ) / √n

Where E is the margin of error (0.5 mph), z is the critical value (1.645), σ is the standard deviation (7 mph), and n is the sample size.

Step 3: Rearrange the formula to solve for n:
n = (z * σ / E)^2

Step 4: Plug in the values and calculate the sample size:
n = (1.645 * 7 / 0.5)^2
n ≈ 338.89

Since the sample size must be a whole number, round up to the nearest whole number.

To construct a 90% confidence interval for the average speed of cars on the highway with a margin of error no more than ±0.5 mph and a standard deviation of 7 mph, you must clock at least 339 cars.

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Answer:

The large hose can fill 1/8 more of the tank in on hour than the small hose

Step-by-step explanation:

To understand this, we need to make 7/8 and 3/4 have the same denominator

3/4 and 6/8 are the same in decimal form (0.75)

So we change 3/4 to 6/8

So now we subtract what the small hose fills per hour from what the large hose fills per hour

7/8 - 6/8 = 1/8

So the large hose fills 1/8 more of the tank per hour than the small hose

Answer:

below

Step-by-step explanation:

A triangle can only be a triangle if the two shortest sides add up to be greater than the longest side.

For example on the first one, the segments 1 2 and 3 will not make a triangle as 1 + 2 = 3 and if the two smallest sides of a triangle are equal to the longest side it will not be a triangle at all.

So to make this easy, out of the three numbers given: add the two smallest numbers. This number has to be greater than the third number given, and it cannot be equal.

For question three: 3+9 = 12

And 12 is greater than 11, therefore this can make a triangle.

Answer:

x = -3/4

Step-by-step explanation:

Because you have the variable x and you do not know the number of fruit snacks he ate and you need to have the same variables to properly find the answer.

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An angle is an undefined term in plane geometry.

I've added as an attachment, the image of the complete question.

Vertex of \(\angle 4\)

The vertex of an angle is the point where the rays that form the angle meet.

From the diagram, rays BE and BA meet at point B to form \(\angle 1\).

Hence, B is the vertex

Sides of \(\angle 1\)

The sides of an angle are the rays that form the angle

\(\angle 1\)is formed by rays BC and BD

Hence, the sides are BC and BD

Another name for \(\angle 5\)

An angle is named by combining the two sides and a vertex.

This means that rays meet at point B

Hence, another name is \(\angle DBE\)

Classify the angles

Angles are classified based on their value  

Angle bisector

Ray BE is an angle bisector, because it divides \(\angle DBA\) into two equal halves.

Find \(\angle EBC\)

We have:

\(\angle EBD = 36\)

\(\angle DBC = 108\)

So:

\(\angle EBC = \angle EBD + \angle DBC\)

\(\angle EBC = 36 + 108\)

\(\angle EBC = 144\)

Find \(\angle ABE\)

We have:

\(\angle EBF = 117\)

\(\angle DBC = 108\)

So:

\(\angle EBF = \angle ABE + \angle EBD + \angle ABF\)

Where

\(\angle ABF = 90\)

\(\angle ABE = \angle EBD\)

So:

\(117 = \angle ABE + \angle ABF + 90\)

\(117 = 2\angle ABF + 90\)

Subtract 90 from both sides

\(27 = 2\angle ABF\)

Divide through by 2

\(\angle ABF = 13.5\)

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Answer:

60% test score

Step-by-step explanation:

When x is 6, y is 60

The value of the test statistic is approximately 2.54.

To find the value of the test statistic, we can use the formula for calculating the test statistic for a single-sample hypothesis test for the population mean. The formula is:

t = (\(\bar x\) - μ) / (s / √n)

Where:

\(\bar x\) is the sample mean,

μ is the population mean,

s is the sample standard deviation,

n is the sample size.

Given information:

Sample mean (\(\bar x\)) = 5.8 ppm

Population mean (μ) = 5.5 ppm

Sample variance (σ^2) = 0.36 ppm^2

Sample size (n) = 26

To calculate the sample standard deviation (s), we take the square root of the sample variance:

s = √(σ^2)

s = √(0.36)

s = 0.6

Now we can calculate the test statistic (t):

t = (5.8 - 5.5) / (0.6 / √26)

t = 0.3 / (0.6 / √26)

t = 0.3 / (0.6 / 5.099)

t = 0.3 / 0.118

t ≈ 2.54 (rounded to two decimal places)

Therefore, the value of the test statistic is approximately 2.54.

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A) The control limits for the x-bar chart are approximately:

Upper Control Limit (UCL) = 55.752 lb

Lower Control Limit (LCL) = 53.748 lb

B) The control limits for the 3-sigma R-chart are approximately:

Upper Control Limit (UCR) = 3.759 lb

Lower Control Limit (LCR) = 1.363 lb

To calculate the control limits for the x-bar chart and the 3-sigma R-chart, we need to use the given information about the sample size and the average range.

a) Control limits for the x-bar chart:

The control limits for the x-bar chart are typically calculated using the formula:

UCL = x(bar) + A₂ × R-bar

LCL = x(bar) - A₂ × R-bar

Where:

UCL = Upper Control Limit

LCL = Lower Control Limit

x(bar) = Overall mean

A₂ = Constant depending on the sample size (from statistical tables)

R-bar = Average range

In this case, the sample size is 7, so we need to find the value of A₂ from the statistical tables. For a sample size of 7, A₂ is approximately 0.577.

Using the given information:

x(bar) = 54.75 lb (Overall mean)

R-bar = 1.78 lb (Average range)

A₂ = 0.577

Substituting these values into the formula, we can calculate the control limits for the x-bar chart:

UCL = 54.75 + 0.577 × 1.78

UCL ≈ 55.752

LCL = 54.75 - 0.577 × 1.78

LCL ≈ 53.748

Therefore, the control limits for the x-bar chart are approximately:

Upper Control Limit (UCL) = 55.752 lb

Lower Control Limit (LCL) = 53.748 lb

b) Control limits for the 3-sigma R-chart:

The control limits for the R-chart can be calculated using the formula:

UCR = D₄ × R-bar

LCR = D₃ × R-bar

Where:

UCR = Upper Control Limit for R-chart

LCR = Lower Control Limit for R-chart

D₄, D₃ = Constants depending on the sample size (from statistical tables)

For a sample size of 7, the values of D₄ and D₃ are approximately 2.115 and 0.765, respectively.

Using the given information:

R-bar = 1.78 lb (Average range)

D₄ = 2.115

D₃ = 0.765

Substituting these values into the formula, we can calculate the control limits for the 3-sigma R-chart:

UCR = 2.115 × 1.78

UCR ≈ 3.759

LCR = 0.765 × 1.78

LCR ≈ 1.363

Therefore, the control limits for the 3-sigma R-chart are approximately:

Upper Control Limit (UCR) = 3.759 lb

Lower Control Limit (LCR) = 1.363 lb

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The complete question is :

35 sample of size 7 each were taken from a fertiliser filling machine at lawm ltd. the result were overall mean = 54.75lb average range = 1.78lb

a) For the given sample size, the control 3-sigma lmits for x bar

Upper Contol Litit (UCL-) = . (round your response to thee decimal places) Lower Control Limit (LCL)= . (round your response to three decimat places).

b) The control lirtits for the 3-sigma R-chart are:

Upper Control Limit (UCR)=  . (cound your rospanse to thee decimal places).

Lower Control Limit (LC,R)=  .(round your response fo three decimal places)

1. False. Autocorrelations in an ACF do not provide information about the normality of the data. 2. True. The R2 value does not necessarily indicate out-of-sample performance; Model 1 may still perform better. 3. False. Validation RMSE may not continually decrease with additional predictor variables and can vary depending on the data. 4. False. AUC (Area Under the Curve) closer to 0.5 does not indicate poor classifier performance; it depends on the context and the specific problem.5. False. A p-value of 0.02 does not imply that the probability of the null hypothesis being true is 0.02; it represents the likelihood of observing the data under the null hypothesis.

1. The presence or absence of significant autocorrelations in an ACF does not provide any information about the normality of the data. Autocorrelations indicate the presence of correlation between lagged values but do not imply normality. 2. It is possible for Model 1 to have a smaller R2 value than Model 2 but still perform better out-of-sample. R2 measures the proportion of variance explained by the model using the observed data, but it may not necessarily reflect the model's performance on unseen data. 3. Validation RMSE (Root Mean Squared Error) does not necessarily decrease with additional predictor variables in multiple regression. The relationship between predictor variables and RMSE depends on the data, model specification, and other factors. Adding more variables may or may not improve the model's performance. 4. The AUC value ranges from 0 to 1, with 0.5 indicating random performance. However, the interpretation of AUC depends on the problem and the context. AUC closer to 0.5 does not automatically imply poor classifier performance; it needs to be evaluated based on the specific problem and the desired performance level. 5. A p-value of 0.02 indicates that the observed data is unlikely to occur if the null hypothesis is true. However, it does not provide direct information about the probability of the null hypothesis being true. The interpretation of a p-value requires considering the specific test, hypothesis, and the significance level chosen.

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1   Y is distributed as N(aμ + b, a^2σ^2), as desired.

2  We have shown that under these conditions, E[XY] = E[X]E[Y].

To prove that Y is distributed as N(aμ + b, a^2σ^2), we need to show that the mean and variance of Y match those of a normal distribution with parameters aμ + b and a^2σ^2, respectively.

First, let's find the mean of Y:

E(Y) = E(aX + b) = aE(X) + b = aμ + b

Next, let's find the variance of Y:

Var(Y) = Var(aX + b) = a^2Var(X) = a^2σ^2

Therefore, Y is distributed as N(aμ + b, a^2σ^2), as desired.

We can use the definition of covariance to prove that E[XY] = E[X]E[Y]. By the properties of expected value, we know that:

E[XY] = ∫∫ xy f(x,y) dxdy

where f(x,y) is the joint probability density function of X and Y.

Then, we can use the fact that X and Y are independent to simplify the expression:

E[XY] = ∫∫ xy f(x) f(y) dxdy

= ∫ x f(x) dx ∫ y f(y) dy

= E[X]E[Y]

where f(x) and f(y) are the marginal probability density functions of X and Y, respectively.

Therefore, we have shown that under these conditions, E[XY] = E[X]E[Y].

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Answer x=6 and y=10

Explanation:

Replace each variable in turn with 0 and solve

5x+3y=30  Put zero in place of x and solve for y

5(0)+3y=30

\(\frac{3y}{3} =\frac{30}{3}\)

y=10 The y- intercept is 10

5x+3y=30 Put zero in place of y and solve for x

5x+3(0)=30

\(\frac{5x}{5} =\frac{30}{5}\)

x=6  The x-intercept is 6

Answer:

Maybe

Step-by-step explanation:

I did but I forgot the answer

I am so sorry

Answer:

Point Slope Equation: Y+2=-3(x-6). Slope Intercept Equation: Y=-3x+16.

Step-by-step explanation:

If you are looking for an equation? Here you go.

Let me know if this is what you wanted!

Answer:

graph 1 perpendicular graph 2 neither graph3 neither graph 4 neither

graph 1 is perpendicular because of how the lines perfectly intersect

graph 2 is parrallel because those lines will never intersect to matter have far they are extended

graph 3 is neither

graph 4 is also neither

The equation "Y^2/9 - x^2/4=1" will produce the given graph. Therefore option B would be the correct answer.

Mathematically, an equation can be defined as a statement that supports the equality of two expressions, which are connected by the equals sign “=”. For example, 2x – 5 = 13.

An equation combines two expressions connected by an equal sign (“=”). These two expressions on either side of the equals sign are called the “left-hand side” and “right-hand side” of the equation. We generally assume the right-hand side of an equation is zero. This will not reduce the generality since we can balance this by subtracting the right-hand side expression from both sides’ expressions.

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1. The number of students choosing walking as their preferred method of transportation is 9. Therefore, the correct option is C.

2. The total students participated in the survey are 20. Therefore, the correct option is B.

3. The percentage of the total students choosing skateboarding is 10%. Therefore, the correct option is A.

4. The percentage of the boys choosing walking is 30%. Therefore, the correct option is A.

5. The percentage of the students who chose biking were girls is 37.5%. Therefore, the correct option is B.

1. The number of students who chose walking as their preferred method of transportation based on the information provided is 9. The total number of students who chose walking is given as 9 (3 boys + 6 girls). Hence the correct answer is option C.

2. The total students who participated in the survey based on the information provided are 20. The total number of students is given as 20 (10 boys + 10 girls). Hence the correct answer is option B.

3. The percentage of the total students who chose skateboarding based on the information provided is 10%. 3 students chose skateboarding out of 20 total students, so the percentage is (3/20) * 100% = 10%. Hence the correct answer is option A.

4. The percentage of the boys who chose walking based on the information provided is 30%. 3 boys chose walking out of 10 total boys, so the percentage is (3/10) * 100% = 30%. Hence the correct answer is option A.

5. The percentage of the students who chose biking were girls based on the information provided is 37.5%. 3 girls chose biking out of 8 total students who chose biking, so the percentage is (3/8) * 100% = 37.5%. Hence the correct answer is option B.

Note: The question is incomplete. The complete question probably is: Given the following data:

Activity Boys Girls Total

Walk 3 6 9

Bike 5 3 8

Skateboard 2 1 3

Total 10 10 20

1. How many students chose walking as their preferred method of transportation? A. 6 B. 3 C. 9 D. none of these. 2. How many total students participated in the survey? A. 10 B. 20 C. 15 D. none of these 3. What percentage of the total students chose skateboarding? A. 10% B. 20% C. 30% D. none of these. 4. What percentage of the boys chose walking? A. 30% B. 45% C. 25% D. none of these. 5. What percentage of the students who chose biking were girls? A. 30% B. 37. 5% C. 45% D. none of these.

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Answer:

2 long sleeve and then 7 short sleeve

Rob bought 2 tee shirts and 6 long sleeve shirts.

A straight line on the coordinate plane is represented by a linear function. As an illustration, the equation y = 3x - 2 depicts a linear function because it is a straight line in the coordinate plane.

Any function whose graph is not a straight line is said to be nonlinear. Any curve other than a straight line can be a graph of it.

An example of a non-linear function is a quadratic function.

Let, The number of tee shirts be 'x' and the number of long sleeve shirts be 'y'.

Therefore,

x + y = 8...(i) and 4x + 5y = 38...(ii)

Multiplying equation (i) by 4 we ave 4x + 4y = 32...(iii).

subtracting eqn(iii) from eqn(i) we have,

y = 6 and hence x = 2.

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Answer:

94.5 ft

Step-by-step explanation:

4x13 = 52

5x5 = 25

2.5x7 = 17.5

52+25+17.5 = 94.5ft

- Gage Millar, Algebra 2 tutor

Answer:

C

Step-by-step explanation:

The total probability rule is used when an an analyst is interested in a)all potential outcomes. So, correct option is a.

Probability is the branch of the mathematics concerning numerical descriptions of how likely an event is going to occur, or how likely it is that a proposition is true. The probability of an event is the number between 0 and 1, where, roughly speaking, 0 indicates that impossibility of the event and 1 indicates certainty. If  probability of an event is higher then, the more likely it is that the event will occur.

Probability can be defined as the proportion  of the number of favorable outcomes to the total number of outcomes of an event..

Hence, option a is correct.

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The order of the matrix is last 4 x 1

The order of the above matrix is (1×3), since the number of rows (m) = 1 and the number of columns (n) = 3. The order of the above matrix is (1×2) since the number of rows (m) = 1 and the number of columns (n) = 2.

In order for matrix multiplication to be defined, the number of columns in the first matrix must be equal to the number of rows in the second matrix. To find A B AB AB , we take the dot product of a row in A and a column in B.

To add the matrices, they must be in the same order.

Matrix addition entails joining related elements.

The matrix has four rows and one column.

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The probability that Marshall wins a prize on his second spin =  0.1875

Consider an event X = the number of spins until Marshall wins a prize.

Given that a prize wheel with 4 segments of equal size, one of which is labeled winner.

So, the sample space n = 4

For given event x, the possible outcomes = 1

Using the formula of probability,

p = x/n

p = 1/4

p = 0.25

So, the probability of success p = 0.25

q = 1 - p

q = 1 - 0.25

q = 0.75

To find the probability that Marshall wins a prize on his 2nd spin.

Using formula, \(P(X = k) = (1 - p)^{k-1}p\)

For  k = 2,

\(P(X = 2) = (1 - 0.25)^{2-1}\times 0.25\)

P(X = 2) = 0.75 × 0.25

P(X = 2) = 0.1875

Thus, the required probability is  0.1875

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The percent discount on the jackets is 33.33%. To find the percent of the discount, we need to calculate the difference between the original price and the discounted price, and then express it as a percentage of the original price.

The difference between the original price and the discounted price is $90 - $60 = $30.

To express this as a percentage of the original price, we can use the formula:

percent discount = (discount / original price) x 100

Plugging in the values we get:

percent discount = ($30 / $90) x 100 = 33.33%

Therefore, the percent discount on the jackets is 33.33%.

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Answer:

Step-by-step explanation:

-10 + -10 = -20

so

10 - _ = -20

10 - 30 = -20

Answer:

21:1

Step-by-step explanation:

840/40 = 21

Every teacher has 21 atusents

For every 21 stutent theare is  1 teacher

40 teachers .Every teacher 21 students  40 x 21 = 840

Answer:

27.5

Step-by-step explanation:

Well for me i would divide 110 by 4 because there are 4 sides and that will be your answer

The minimum number of hard disks that can be used for implementing a RAID 10 solution is four (option c).  

The minimum number of hard disks that can be used for implementing a RAID 10 solution on the video editor's desktop workstation is four (option c). RAID 10, also known as RAID 1+0, combines elements of RAID 1 (mirroring) and RAID 0 (striping) to provide both data redundancy and improved performance.

In RAID 10, data is mirrored across multiple drives to ensure redundancy. To implement RAID 10, we need at least two sets of mirrored drives, where each set contains a minimum of two hard disks.

The data is striped across these mirrored sets to enhance performance.

With four hard disks, you can create two mirrored sets, each consisting of two drives.

The data is then striped across these mirrored sets, providing both redundancy and improved read/write speeds.

If one drive fails in each mirrored set, the system can still function without data loss due to the mirroring.

This setup provides a balance between data protection and performance.

While more hard disks can be used in RAID 10 for larger capacities and better performance, the minimum requirement is four hard disks.

Using fewer drives would not allow for the necessary mirrored sets and striping required for RAID 10.

By implementing a RAID 10 solution with four hard disks on the video editor's workstation, you can help safeguard against data loss in case of drive failure, ensuring the integrity and availability of important video editing data.  

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