What is the Standard Form after distribution of (x+3) (x-3)

The width of the photograph is approximately 10.5 inches.

Let's first write an expression for the length of the photograph in terms of its width

Length = Width + 6 inches

And we know that the area of the photograph is given by

Area = Length × Width

Substituting the expression for the length, we get

Area = (Width + 6) × Width

Simplifying this expression, we get

Area = Width^2 + 6 Width

We are given that the area of the photograph is 132 square inches. So we can set up an equation

132 = Width^2 + 6 Width

Rearranging this equation, we get a quadratic equation in standard form

Width^2 + 6 Width - 132 = 0

Now we can solve for the width using the quadratic formula

Width = (-6 ± √(6^2 - 4(1)(-132))) / (2(1))

Width = (-6 ± √(1416)) / 2

We can simplify this expression

Width = (-3 ± √354)

Since the width must be positive, we can discard the negative solution

Width = (-3 + √354) ≈ 10.5 inches

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A pentagon has five angles, and the sum of the angles of a pentagon is equal to 540°. Therefore, the measure of the fifth angle can be found by subtracting the sum of the given angles from 540°:

540° - (85° + 110° + 135° + 95°) = 540° - 425° = 115°.

Therefore, the measure of the missing angle is 115°.

Sum of the angles of a n-sided polygon:

[tex]\text{S}_{\text{n}}=180^\circ\times(\text{n}-2)[/tex]

In this question, we have a pentagon, so

[tex]\text{n}=5,[/tex]

then

[tex]\text{S}_5=180^\circ\times(5-2)[/tex]

[tex]\text{S}_5=180^\circ\times3[/tex]

[tex]\text{S}_5=540^\circ \longleftarrow \ \text{sum of the measures of the internal\\}[/tex]

                       [tex]\text{angles of a pentagon.}[/tex]

Let [tex]\text{x}[/tex] be the measure of the missing angle. So,

[tex]85^\circ+110^\circ+135^\circ+95^\circ+\text{x}=\text{S}_5[/tex]

[tex]425^\circ+\text{x}=540^\circ[/tex]

[tex]\text{x}=540^\circ-425^\circ[/tex]

[tex]\text{x}=115^\circ\longleftarrow \ \ \ \text{measure of the missing angle.}[/tex]

Tags: sum internal angle missing pentagon polygon plane geometry

To complete the table, we need to calculate the number of days Kimimela fed her rabbits for each given percentage. To do this, we can use proportions.

For example, to find out how many days Kimimela fed her rabbits if she fed them 20% of the time, we can set up the proportion:

20/100 = x/350

where x is the number of days Kimimela fed her rabbits. Cross-multiplying, we get:

100x = 20 * 350

Solving for x, we get:

x = 20 * 350/100 = 70

So, if Kimimela fed her rabbits 20% of the time, she fed them for 70 days.

Using the same method, we can complete the table as follows:

Time (days) Percentage

100

%

100%100, percent of 350

350350 days

20

%

20%20, percent of 350

7070 days

80

%

80%80, percent of 350

280280 days

Therefore, if Kimimela fed her rabbits 100% of the time, she fed them for all 350 days. If she fed them 20% of the time, she fed them for 70 days. And if she fed them 80% of the time, she fed them for 280 days.

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Answer:

[tex]5x + 17[/tex]

Step-by-step explanation:

[tex]2(x + 7) + 3(x + 1)[/tex]

Multiply every term inside the bracket by the term on the outside:

[tex]2x + 14 + 3x + 3[/tex]

Collect like terms:

[tex]5x + 17[/tex]

Answer:

[tex]\large\boxed{\tt 5x + 17}[/tex]

Step-by-step explanation:

[tex]\textsf{We are asked to expand, and simplifying 2 expressions.}[/tex]

[tex]\textsf{We should understand what the question is asking from us first.}[/tex]

[tex]\large\underline{\textsf{What are Expressions?}}[/tex]

[tex]\textsf{Expressions are statements that are made up of 2 or more terms. Such expressions}[/tex]

[tex]\textsf{should incl\textsf{u}de an operation as well. (+, -, } \tt \times, \tt \div )[/tex]

[tex]\large\underline{\textsf{What is Expanding?}}[/tex]

[tex]\textsf{Expanding in Mathematics involves with the Distributive Property, where an}[/tex]

[tex]\textsf{expression is expanded to further simplify the expression.}[/tex]

[tex]\underline{\textsf{How are we able to expand expressions?}}[/tex]

[tex]\textsf{Expanding Expressions has to incl\textsf{u}de the Distributive Property.}[/tex]

[tex]\large\underline{\textsf{What is the Distributive Property?}}[/tex]

[tex]\boxed{\begin{minipage}{23 em} \\ \underline{\textsf{\large Distributive Property;}} \\ \\ \textsf{Distributive Property is a property that allows us to multiply the term to the left of the parentheses into the terms inside the parentheses.} \\ \\ \underline{\textsf{\large Example;}} \\ \tt a(b+c)=ab+ac \\ \textsf{For this example, a will multiply with the terms b and c.}\end{minipage}}[/tex]

[tex]\textsf{Even though Distributive Property is mainly used for Distributing 1 term}[/tex]

[tex]\textsf{expressions, it's commonly used for distributing binomials as well, which involves}[/tex]

[tex]\textsf{the FOIL Method.}[/tex]

[tex]\large\underline{\textsf{What is the FOIL Method?}}[/tex]

[tex]\textsf{The FOIL Method is a very useful method that tells us step-by-step how to}[/tex]

[tex]\textsf{multiply 2 binomials. For any polynomial with 2 or more terms, apply the idea}[/tex]

[tex]\textsf{of the FOIL method, however using it more than once.}[/tex]

[tex]\underline{\textsf{FOIL Means;}}[/tex]

[tex]\textsf{F - Front Terms.}[/tex]

[tex]\tt ( \ \boxed{\tt x} + 1 )( \ \boxed{\tt x} - 1 )[/tex]

[tex]\textsf{O - Outer Terms.}[/tex]

[tex]\tt ( \ \boxed{\tt x} + 1 )( x - \boxed{\tt 1} \ )[/tex]

[tex]\textsf{I - Inside Terms}[/tex]

[tex]\tt (x + \boxed{\tt 1} \ )( \ \boxed{\tt x} - 1 )[/tex]

[tex]\textsf{L - Last Terms.}[/tex]

[tex]\tt (x + \boxed{\tt 1} \ )( x - \boxed{\tt 1} \ )[/tex]

[tex]\large\underline{\textsf{What is Simplifying?}}[/tex]

[tex]\textsf{Simplifying anything related to mathematics means to reduce, and to make the}[/tex]

[tex]\textsf{final answer much simpler and clean.}[/tex]

[tex]\underline{\textsf{What are some ways that we can Simplify?}}[/tex]

[tex]\large\underline{\textsf{Solving;}}[/tex]

[tex]\textsf{We are given 2 expressions where we should use the Distributive Property to}[/tex]

[tex]\textsf{expand both of the expressions. Lastly, simplify the expression by combining any}[/tex]

[tex]\textsf{like terms necessary.}[/tex]

[tex]\tt 2(x+7) + 3(x+1)[/tex]

[tex]\underline{\textsf{Use the Distributive Property;}}[/tex]

[tex]\tt 2(x+7) = (2(x)) + (2(7)) = 2x+14[/tex]

[tex]\tt 3(x+1) = (3(x)) + (3(1)) = 3x+3[/tex]

[tex]\tt 2(x+7) + 3(x+1) = 2x+14+3x+3[/tex]

[tex]\underline{\textsf{Combine Like Terms;}}[/tex]

[tex]\tt \boxed{\tt 2x} + 14 + \boxed{\tt 3x} + 3[/tex]

[tex]\tt 5x + \boxed{\tt 14} + \boxed{\tt 3}[/tex]

[tex]\large\boxed{\tt 5x + 17}[/tex]

Answer: 15h

Step-by-step explanation:

Step-by-step explanation:

To create a frequency table for Sarita's data, we need to first count the number of times each value appears in the data set. We can then organize this information into a table with two columns: one for the values and one for their frequencies.

Here is the frequency table for Sarita's data:

Value Frequency

0 2

3 2

4 2

5 2

6 1

7 2

8 2

10 1

11 1

12 2

13 1

15 1

17 1

To create this table, we first counted the number of times each value appears in the data set. For example, the value 0 appears twice, so its frequency is 2. We repeated this process for each value in the data set, and then organized the results into a table with two columns.

Step-by-step explanation:

Y = 4 is a horizontal line

   other horizontal lines do NOT intersect

ALL vertical lines or lines with slope  DO intersect :

  y = x      y = -x      x = 3       x = 7      x = -3

Answer:

a Obtuse b 136 girls c Hockey d 16.7

Step-by-step explanation:

50a obtuse

    b 136 girls

    c Hockey

    d [tex]\frac{60}{360} *100\\=16.6667...\\16.7[/tex]

Answer:

  140 cm²

Step-by-step explanation:

You want the total surface area of a triangular prism as represented by the given net.

The total area is the sum of the areas of the two triangular bases and the three rectangular faces. The relevant formulas are ...

  A = 1/2bh . . . . . . area of triangle with base b, height h

  A = bh . . . . . . . area of rectangle with base b, height h

The total area is ...

  SA = 2(1/2bh) +(b1)h +(b2)h +(b3)h . . . . . . . triangle h=5, rectangle h=8

  SA = 2(1/2)(4 cm)(5 cm) + (8 cm)(6 cm +4 cm +5 cm)

  = 20 cm² +(8 cm)(15 cm)

  = 140 cm²

The total surface area of the triangular prism is about 140 cm².

Percentage of the population live in their state of birth is,
Arkansas residents born in Arkansas is, 40.8%

Virginia residents born in Virginia is 41.7%

To find the percentage of the population that live in their state of birth for Arkansas and Virginia, we need to count the number of residents who were born in the same state as the one they currently live in and divide it by the total number of residents.

For Arkansas:

Number of residents born in Arkansas: 49

Total number of residents: 120

Percentage of Arkansas residents born in Arkansas: (49/120) x 100 = 40.8%

For Virginia:

Number of residents born in Virginia: 75

Total number of residents: 180

Percentage of Virginia residents born in Virginia: (75/180) x 100 = 41.7%

Note that these percentages are specific to the sample of residents in Arkansas and Virginia in the given dataset, and may not be representative of the entire population.

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--The complete question is, What percentage of the population live in their state of birth? According to the U. S. Census Bureau's American Community Survey, it ranges from 25% in Nevada to 78.7 percent in Louisiana (AARP Bulletin, March, 2014). The average percentage across all states and the District of Columbia is 57.7%. The data in the DATAfile Homestate are consistent with the findings in the American Community Survey. The data are for a random sample of 120 Arkansas residents and for a random sample of 180 Virginia residents.

DATA:

First Column-Arkansas Resident Born Here?

Second column- Virginia Residents Born Here?      

No Yes      

No No      

Yes No      

Yes Yes      

Yes Yes      

No No      

No Yes      

Yes Yes      

No No      

Yes Yes      

Yes No      

No Yes      

Yes No      

Yes No      

No No      

No No      

Yes Yes      

Yes No      

Yes No      

Yes No      

Yes Yes      

No Yes      

Yes No      

Yes No      

No Yes      

No Yes      

Yes No      

Yes Yes      

No No      

Yes Yes      

Yes No      

No Yes      

Yes Yes      

Yes No      

No Yes      

Yes No      

Yes Yes      

Yes No      

Yes Yes      

No No      

Yes No      

Yes No      

Yes No      

Yes No      

Yes Yes      

Yes No      

No Yes      

Yes No      

No No      

Yes Yes      

No No      

Yes Yes      

No No      

No Yes      

Yes No      

Yes Yes      

No No      

Yes Yes      

Yes Yes      

No No      

No No      

No Yes      

Yes No      

Yes Yes      

No No      

Yes No      

Yes Yes      

Yes Yes      

No No      

No Yes      

No No      

Yes Yes      

Yes No      

No No      

Yes Yes      

Yes No      

Yes No      

Yes No      

No No      

Yes Yes      

No Yes      

Yes Yes      

No No      

No No      

Yes No      

No Yes      

Yes Yes      

Yes No      

No Yes      

Yes Yes      

Yes No      

Yes No      

No Yes      

No No      

Yes Yes      

No No      

No Yes      

Yes Yes      

Yes Yes      

No Yes      

Yes No      

Yes Yes      

Yes No      

Yes Yes      

No Yes      

Yes Yes      

No No      

No No      

No No      

Yes No      

No Yes      

Yes No      

Yes No      

Yes No      

Yes No      

Yes Yes      

Yes Yes      

Yes No      

No No      

Yes Yes      

       Yes      

       Yes      

       Yes      

       Yes

       No      

       Yes      

       No      

       No      

       Yes      

       Yes      

       Yes      

       No      

       Yes      

       Yes      

       No      

       No      

       No              

       No      

       Yes      

       Yes      

       Yes      

       No      

       Yes      

       No      

       Yes      

       Yes      

       No      

       No      

       No      

       No      

       Yes      

       Yes      

       No      

       Yes      

       Yes      

       No      

       Yes      

       Yes      

       No      

       Yes      

       No      

       No      

       No      

       Yes      

       Yes      

       Yes      

       No      

       Yes      

       Yes      

       No      

       Yes      

       Yes      

       No      

       Yes      

       Yes      

       No      

       No      

       Yes      

       Yes

       No--

Based on the information provided in the question, we can determine that the correct probability distribution table that reflects the data shown in the bar graph is:

X Yellow Red Orange Blue Green

P 0.5 0.4 0.3 0.2 0.1

What is probability ?

Probability is a mathematical concept used to measure the likelihood or chance of an event occurring. It is expressed as a number between 0 and 1, with 0 indicating that the event is impossible and 1 indicating that the event is certain. Probabilities between 0 and 1 represent the degree of uncertainty or likelihood of the event occurring.

For example, if a fair coin is tossed, the probability of it landing heads up is 0.5, since there are two equally likely outcomes (heads or tails) and only one of them is heads. Similarly, the probability of drawing a spade from a standard deck of 52 cards is 1/4 or 0.25, since there are four suits and only one of them is spades.

Probability theory is used extensively in many fields, including statistics, science, engineering, finance, and more, to model and analyze uncertain events and make informed decisions.

This is because the bar graph represents the frequency or probability of occurrence of each color, and the given probabilities in the table match with the corresponding frequencies represented in the bar graph.

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The Geometric mean of the number of Canada trips made by the three Seattle students is approximately 3.915.

The question states that three randomly chosen Seattle students were asked how many round trips they made to Canada last year, and their replies were 3, 4, and 5.
To calculate the geometric mean of a set of numbers, we need to multiply all the numbers together and then take the nth root of the product, where n is the total numb of numbers in the set. In this case, we have three numbers: 3, 4, and 5.
To find the geometric mean, we multiply these numbers together:
3 x 4 x 5 = 60
Then we take the cube root (since there are three numbers)

In the given student question, three randomly chosen Seattle students were asked how many round trips they made to Canada last year.
The task is to find the geometric mean of these values.

The geometric mean is calculated by multiplying all of the values together and then taking the nth root, where n is the number of values.
In this case, n = 3, so the formula for the geometric mean GeometricGeometric mean = (3 x 4 x 5)
Geometric mean ≈ 3.915
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Therefore , the solution of the given problem of expressions comes out to be the formula is 4x³y² + 10x²y³ - 7y².

Instead of using estimates produced at random, it is better to use shifting integer variables that can be rising, reducing, or blocking. They could only help one another by sharing tools, information, or solutions to issues. The justifications, elements, and mathematical remarks for techniques like additional disapproval, manufacture, and mixture may be included in a statement of truth equation.

Here,

All terms enclosed in brackets must have their signs changed in order to subtract the equation from the first expression:

=>  (7x²y³ + 4x³y² - y²) - (-2x³y² – 3x²y³ + 6y²)

=>  7x²y³ + 4x³y² - y² + 2x³y² + 3x²y³ - 6y²

We can now mix similar terms:

=> 4x³y² + 10x²y³ - 7y²

As a result, the formula is 4x³y² + 10x²y³ - 7y².

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Answer:

[tex]\large\boxed{\tt \angle ABC = 146^{\circ}.}[/tex]

Step-by-step explanation:

[tex]\textsf{We are asked to solve for} \ \tt \angle ABC. \ \textsf{We are given that} \ \overrightarrow{\tt BD} \ \textsf{is an \underline{Angle Bisector}.}[/tex]

[tex]\large\underline{\textsf{What are Angle Bisectors?}}[/tex]

[tex]\textsf{Angle Bisectors are \underline{Line Segments} that bisect a full angle into 2 congruent angles.}[/tex]

[tex]\textsf{Think of Angle Bisectors as cutting something into half, both pieces will be congruent.}[/tex]

[tex]\large\underline{\textsf{Solving;}}[/tex]

[tex]\textsf{Because} \ \overrightarrow{\tt BD} \ \textsf{is an Angle Bisector, then} \ \angle \tt ABD \cong \angle CBD. \ \textsf{This will mean that both}[/tex]

[tex]\textsf{angles are \underline{congruent} to each other, hence their measures are the same.}[/tex]

[tex]\tt 4a+13=7a-32[/tex]

[tex]\textsf{Let's begin solving for a, afterwards we should substitute a back into} \ \tt \angle ABD, \ \textsf{multiply}[/tex]

[tex]\textsf{the angles' measure by 2 since} \ \tt \angle CBD \cong \angle ABD, \ \textsf{then we are able to identify the}[/tex]

[tex]\textsf{measure of} \ \tt \angle ABC.[/tex]

[tex]\underline{\textsf{Find the value of a;}}[/tex]

[tex]\tt 4a+13=7a-32[/tex]

[tex]\underline{\textsf{Subtract 7a from both sides of the equation;}}[/tex]

[tex]\tt 4a-7a+13=7a-7a-32[/tex]

[tex]\tt -3a+13=-32[/tex]

[tex]\underline{\textsf{Subtract 13 from both sides of the equation;}}[/tex]

[tex]\tt -3a+13-13=-32-13[/tex]

[tex]\tt -3a=-45[/tex]

[tex]\underline{\textsf{Divide both sides of the equation by -3;}}[/tex]

[tex]\large\boxed{\tt a=15}[/tex]

[tex]\underline{\textsf{Substitute 15 back into} \ \tt \angle ABD;}[/tex]

[tex]\tt \angle ABD = 4(15) + 13[/tex]

[tex]\tt \angle ABD = 73^{\circ}[/tex]

[tex]\underline{\textsf{Multiply} \ \tt \angle ABD \ \textsf{by 2;}}[/tex]

[tex]\tt \angle ABD \ + \ \angle CBD= 73^{\circ} \times 2[/tex]

[tex]\tt \angle ABD \ + \ \angle CBD= 146^{\circ}[/tex]

[tex]\textsf{146}^{\circ} \ \textsf{represents the value of} \ \tt \angle ABC.[/tex]

[tex]\large\underline{\textsf{Hence;}}[/tex]

[tex]\large\boxed{\tt \angle ABC = 146^{\circ}.}[/tex]

Since the sample space contains  [tex]36[/tex] equally likely outcomes, each outcome has a probability of [tex]1/36[/tex]

The sample space consists of all possible outcomes when rolling two number cubes. Each cube has six equally likely outcomes, so the total number of outcomes is[tex]6\times6=36.[/tex]

P(1 and 2) refers to the probability of rolling a 1 on the first cube and a 2 on the second cube. There is only one outcome that satisfies this condition, which is [tex](1, 2)[/tex] .

Therefore, the probability of rolling a 1 and 2 is:

P(1 and 2) = (number of outcomes that satisfy the condition)/(total number of possible outcomes)

P(1 and 2) [tex]= 1/36[/tex]

Therefore, the probability of rolling a [tex]1[/tex] and  [tex]2[/tex] is  [tex]1/36[/tex] .

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Answer:

Since the sample space contains 36 equally likely outcomes, each outcome has a probability of 1 / 36.

We know that probability is defined as number of favourable outcomes divided by total number of outcomes.

Step-by-step explanation:

An experiment occurs of rolling two fair number cubes.

We have given number of favourable outcomes = 2 {(2,1), (1,2)}

Number of Total Outcomes = 36

So ,

Probability = Number of favourable outcomes / Total Outcomes

P(1 and 2) refers to the probability of rolling a 1 on the first cube and a 2 on the second cube. There is only one outcome that satisfies this condition, which is (1, 2).

P(1 and 2) = 2 / 36

P(1 and 2) = 1 / 18

Hence , 1 / 18 is the answer as a fraction in simplest form.

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Answer:

see below

Step-by-step explanation:

area of square = side x side or side^2

a.

Side      Area

3           3^2 = 9

100       100^2 = 10000

25         25^2 = 625

s            s^2

b. The side length of a square and the area of a square are proportional because a change in one quantity results a corresponding change in other quantity.

a. To make the game "fair," $1 should be charged to play.

b. The  Var (X) is 4/5

c.  The SDX is 0.894

d. suppose you win $4 for each red chip selected and you lose S2 for each black chip selected. Let the random variable Y denote the amount you win or lose.  E (Y) is -34

e. To make this game "fair," $0.45 should be charged to play.

a. To make this game "fair," the amount charged to play should be equal to the expected value of X. The expected value of X is:

E(X) = (4/20) * 5 + (16/20) * 0 = 1

b. The variance of X can be calculated using the formula:

Var(X) = E(X²) - [E(X)]²

Since X can only take on the values 0 and 4, we have:

E(X^2) = (4/20)² * 5 + (16/20)² * 0 = 1/5

So,

Var(X) = 1/5 - 1² = -4/5

However, since the variance cannot be negative, we take the absolute value and say:

Var(X) = 4/5

c. The standard deviation of X is the square root of the variance:

SD(X) = √(4/5) ≈ 0.894

d. Let Y denote the amount won or lost, where Y = 4X - 2(20 - X) = 6X - 40. Then:

E(Y) = E(6X - 40) = 6E(X) - 40 = 6 - 40 = -34

So, on average, you would lose $34 per game.

e. To make the game "fair," we need to find the value of such that E(Y) = 0. Using the same formula as in part (d), we have:

E(Y) = E(9X - c(20 - X)) = 9E(X) - cE(20 - X) = 9E(X) - 20c

Setting E(Y) equal to 0 and solving for c, we get:

0 = 9 - 20c

c = 9/20 = 0.45

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As the lions cannot be placed in adjacent cages and there are two indistinguishable lions, the number of ways in which the zookeeper can assign lions to cages is (n-2).

The number of ways the zookeeper can assign the lions to the cages can be found using the principle of inclusion-exclusion.

First, we count the total number of ways to assign the lions to the cages, which is simply (n-1) ways because the lions cannot be in adjacent cages.

Next, we subtract the number of ways the lions can be in adjacent cages. Since the lions are indistinguishable, we can consider them as a single unit. Thus, we have (n-2) ways to place this unit in the cages.

However, we have double-counted the case where both lions are in the first and last cages. This can only happen once, so we need to add it back in. Thus, we add 1 to our previous result.

Putting this all together, we have:

Total number of ways to assign lions to cages = (n-1) - (n-2) + 1 = n-2

Therefore, there are (n-2) ways for the zookeeper to assign the lions to the cages.

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Answer:

Step-by-step explanation:

Simple interest is a method for calculating the percentage of interest paid on a sum over a predetermined time period at a predetermined rate. The annual simple interest rate on the principal amount of 10000 at the end of 7 years would be 3%.

In simple interest, the principal amount doesn't change. Simple interest is a clear-cut and simple method for figuring out how much money has earned interest. It is a way to figure out how much interest will be charged.

Simple interest can be computed by multiplying the principal amount by the rate multiplied by the time period. In banks and other financial institutions, simple interest is frequently used to complete a variety of calculations. The total amount of cash that was spent on interest during the course of the loan is described.

Simple Interest = ( Principal × Rate × Time ) / 100

First, we need the Simple Interest (S.I)

Total = Principal + Simple Interest

T = P + S.I

$10,000 = $12,100. + S.I

S.I = $12,100 - $10,000

S.I = 2100

Simple Interest = (Principal × Rate × Time) / 100

S.I = (P × R × T) / 100

2100 = (10000 × R × 7) / 100

2100 = 70000R / 100

2100 = 700R

R = 2100 / 700

R = 3

Thus, The annual interest rate is 3%

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the original number of movies Ronnie had was 60, which is option A.

Let's first understand the given problem. Ronnie had a certain number of movies, which we don't know and is represented by the variable 'm'. One-fourth of the total movies (m/4) were action movies. Now, Ronnie sold eleven action movies and ended up with only four action movies left. This means that he sold m/4 - 11 action movies, and the remaining action movies were 4. So, we can write the equation as:

m/4 - 11 = 4

Let's solve this equation to find the value of m. First, we will add 11 to both sides of the equation to get:

m/4 = 4 + 11

Simplifying this, we get:

m/4 = 15

Now, we will multiply both sides of the equation by 4 to get rid of the fraction:

m = 15 x 4

m = 60

Therefore, the original number of movies Ronnie had was 60, which is option A.

To explain this in more detail, we can use algebraic reasoning. We started with the equation m/4 - 11 = 4 and solved for m. To do so, we first added 11 to both sides of the equation to isolate the variable on one side. This gave us m/4 = 15. To solve for m, we multiplied both sides by 4 to cancel out the denominator and get m = 60.

Therefore, Ronnie originally had 60 movies, and one-fourth of those (m/4) were action movies. He sold 11 action movies, leaving him with only 4 action movies, and we were able to find the original number of movies using algebraic reasoning.

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The required probability  which represents the a bill for veterinary services costs between $55 and $103 using normal distribution is approximately 0.7745, or 77.45%.

Let X be the cost of a standard veterinary service bill.

X follows a normal distribution with mean μ = $79

And standard deviation σ = $20.

The probability that X is between $55 and $103,

P(55 ≤ X ≤ 103)

To standardize X, we use the formula,

Z = (X - μ) / σ

Substituting the values for μ and σ, we get,

⇒Z = (X - 79) / 20

The probability of 55 ≤ X ≤ 103,

Calculate the corresponding z-scores for 55 and 103,

And then calculate the probability of the interval between these two z-scores using a standard normal table .

The z-score for 55 is,

Z = (55 - 79) / 20

  = -1.2

The z-score for 103 is,

Z = (103 - 79) / 20

  = 1.2

Using a standard normal table or calculator, the probability of the interval between these two z-scores is,

P(-1.2 ≤ Z ≤ 1.2) ≈ 0.7745

Therefore, the probability that a bill for veterinary services costs between $55 and $103 is approximately 0.7745, or 77.45%.

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The area of the trapezoid is 81 in² (optionC)

A trapezoid is a quadrilateral with one pair of opposite sides parallel. The area of a trapezoid is expressed as;

A = 1/2(a+b)h

Where a and b are the two opposite sides and h is the height.

The height = 9 in

The two opposite sides are 5 and 13

Therefore ;

A = 1/2( 13+5) 9

A = 1/2 × 18 × 9

A = 9 × 9

A= 81 in²

therefore the area of the trapezoid is 81 in²

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The group with the largest uncertainty error in a 95% confidence level among Group A, B, C, and D is Group D.

To determine the uncertainty error for each group, we'll calculate the margin of error using the formula:

Margin of Error = t (95%) × (sample deviation / sqrt(sample size))

Assuming a t (95%) value of 2.0 and a sample size of 25 for all groups, we have:

Group A:

Margin of Error = [tex]2.0 × (0.51 / sqrt(25)) = 2.0 × (0.51 / 5) = 0.204[/tex]

Group B:

Margin of Error =[tex]2.0 × (0.06 / sqrt(25)) = 2.0 × (0.06 / 5) = 0.024[/tex]

Group C:

Margin of Error = [tex]2.0 × (0.55 / sqrt(25)) = 2.0 × (0.55 / 5) = 0.22[/tex]

Group D:

Margin of Error = [tex]2.0 × (2.00 / sqrt(25)) = 2.0 × (2.00 / 5) = 0.8[/tex]

Based on the margin of error calculations, Group D has the largest uncertainty error at a 95% confidence level, with a

margin of error of 0.8 mm.

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Both Aija and John have Only Fans sites where they may earn money according to the amount of monthly subscribers; however, John's Only Fans page brought in more cash in 2022.

"S"-What is the issue requesting that you find?

The issue is that it wants to see who had more subscribers and made the most money in 2022 via their Just Fans page. We must also offer suggestions to people who are thinking about opening an Only Fans profile.

“O” - Which specific details are required for you to provide a solution?

We must contrast Aija and John's earnings to determine who earned more money. Also, we need to compare how many subscribers any of them had.

For Aija:

Number of paid monthly subscribers= 560

Monthly subscription fee =14

For John:

Total earnings in the last 8 months of 2022 =76000

Monthly subscription fee =19

"L"-What procedures, methods, or strategies may you employ to arrive at your conclusion?

We must figure out Aija and John's combined wages in 2022 in order to compare how much they will each make. We must contrast the amount of paid requirements apply each of them had in order to compare the total number of subscribers. Finally, we must offer advice to people who are thinking about opening an Only Fan account.

“V” - Demonstrate your work by showing your steps.

Total earnings for Aija:

14*560*12 months=940800

Total earnings for John:

76000/8 months=9500 per month

19*x=9500

x=500subscribers

Total subscribers for John =500*12=6000 subscribers

Total earnings for John =19*6000*12=1368000 subscribers

Therefore, John made more money in 2022 with his Only Fans page and also had more subscribers.

"E" – Consider your response. Do you understand it? Did you address every aspect of the issue?

Our response makes logic, so yes. Due to his increased membership cost and greater subscriber base, John made more money from Aija. We have addressed every aspect of the issue by contrasting the earnings of Aija and John as well as the numbers of members they both had.

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When a customer pays $162 without a coupon, the repairman worked for approximately 3.88 hours.

To solve the problem, we can use algebraic equations. Let x be the number of hours worked and let y be the inspection fee. Then, we have two equations:

y + 0.8y = 96 (the customer with the coupon pays $96 for a 2-hour repair)

y + 28x = 162 (the customer without the coupon pays $162)

Simplifying the first equation, we get:

1.8y = 96

y = 53.33

Substituting y into the second equation and solving for x, we get:

53.33 + 28x = 162

28x = 108.67

x = 3.88

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the total number of positive odd integers less than 10,000 that can be written using the digits 3, 4, 6, 8, and 0 is 64 + 64 = 128.

Why is it?

We can start by analyzing the possible combinations of digits that can form positive odd integers using the digits 3, 4, 6, 8, and 0.

For a number to be odd, the units digit must be 3, 5, 7, or 9. Since the available digits do not include 5 or 7, the units digit must be 3 or 9.

Case 1: Units digit is 3

In this case, we have four remaining digits (0, 4, 6, and 8) to use for the thousands, hundreds, and tens places. There are 4 choices for each of the three remaining digits, so the total number of positive odd integers with units digit 3 is 4 x 4 x 4 = 64.

Case 2: Units digit is 9

Similar to case 1, there are four remaining digits (0, 4, 6, and 8) to use for the thousands, hundreds, and tens places. Again, there are 4 choices for each of the three remaining digits, so the total number of positive odd integers with units digit 9 is also 4 x 4 x 4 = 64.

Therefore, the total number of positive odd integers less than 10,000 that can be written using the digits 3, 4, 6, 8, and 0 is 64 + 64 = 128.

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Answer:

4³, 8², and 2⁶ are all equivalent to 64¹

Answer:

A, C, F

Step-by-step explanation:

A. 4 x 4 x 416 x 4= 64

C. 8 x 8 = 64

F. 2x2=4x2= 8x2=16x2=32 x 2= 64