Question 5 (Essay Worth 5 points)
Determine the answer to 3+ (-5) and explain the steps using a number line.
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Answer:

The Correct Answer are B and D

Step-by-step explanation:

Answer:

B And D

Step-by-step explanation:

Answer:

ITS A

i took the quiz :)

Step-by-step explanation:

The two true statements are:

The rule for Pattern B is "multiply by 2."

Each term in Pattern A is one-half the corresponding term in Pattern B.

Here we have two patterns:

First, let's analyze the patterns. For pattern A we can see that we just add 6 to each consecutive term, so this is an arithmetic sequence.

6 + 6 = 12

12 + 6 = 18

18 + 6 = 24

And so on.

For pattern B we multiply each term by 2 to get the next one, so this is a geometric sequence.

2*12 = 24

2*24 = 48

and so on.

Then we already can see that. "The rule for Pattern B is "multiply by 2." Is true

And:

The rule for Pattern A is "multiply by 2." is false.

Finally, let's compare the terms. The first term of B is 12, and the first ter of A is 6, so we can see that the first term of B is two times the first one of A. The second term of B is 24, while the second of A is 12, so again, we can see that the second term of B is two times the second of A.

So we can see a pattern, meaning that the statement: "Each term in Pattern A is one-half the corresponding term in Pattern B."

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The equations 3x + 4 = 3x + 4, 2(x + 8) = 2x + 8, 4 (2x -8) = 2x + 6, 8x - 4x + 1 = 12x - 9, 2x -8 +4 = 2x -2 - 2 and -3 (x -6) = -4x +x -18 has infinitely many solutions, no solution, one solution, one solution, infinitely many solutions and no solution respectively

Equations with one solution: An equation has one solution if you get a variable equal to a number after solving it e.g. x = 6

Equations with no solutions: An equation has no solutions if you get a false statement after trying to solve it e.g. 7 = 2 is a false statement

Equations with an infinite number of solutions: An equation has infinitely many solutions if you get a variable or a number equal to itself after trying to solve it e.g. x = x or 4 = 4

For 3x + 4 = 3x + 4

3x + 4 = 3x + 4

3x-3x = 4-4

0 = 0

This equation has infinitely many solutions

For 2(x + 8) = 2x + 8

2(x + 8) = 2x + 8

2x + 16 = 2x + 8

2x-2x = 8-16

0 = -8

This equation has no solution

For 4(2x -8) = 2x + 6

4(2x -8) = 2x + 6

8x - 32 = 2x + 6

8x - 2x = 6 + 32

6x = 38

x = 38/6 = 19/3

This equation has one solution

For 8x - 4x + 1 = 12x - 9

8x - 4x + 1 = 12x - 9

4x + 1 = 12x-9

12x-4x = 1+9

8x = 10

x = 10/8 = 5/4

This equation has one solution

For  2x -8 +4 = 2x -2 - 2

2x -8 +4 = 2x -2 - 2

2x-2x = -2-2-4+8

0 = 0

This equation has infinitely many solutions

For  -3 (x -6) = -4x +x -18

-3 (x -6) = -4x +x -18

-3x + 18 =  -4x +x -18

-3x+4x-x = -18-18

0 = -36

This equation has no solution

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Answer:b

Step-by-step explanation:y=2x-3

The answer choice which correctly represents the value of cos B as required is; Choice B; Cosine B = StartFraction 40 Over 41 EndFraction.

It follows from the task content that the ratio which represents cos B is to be identified among the given answer choices.

Recall from trigonometric ratios that the cosine of an angle is the ratio of its opposite and hypothenuse.

Hence, it can be inferred that;

cos B = 80 / 82

cos B = 40 / 41.

Ultimately, the cosine of angle B following from trigonometric ratios as requested is; cos B = 40 / 41.

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Answer:

Below

Step-by-step explanation:

12 C 2  = 12! / ( 10! 2!) = 66 ways

Answer:

a) $65,586.40

b) $14,186.40

c) $122,332

d) $70,932

(a) The half-yearly payment required for each five-year interval is $37,746.88. (b) The loan outstanding (outstanding balance) at the beginning of each five-year interval:

Year 1: $853,469.14

Year 6: $673,474.96

Year 11: $448,225.59

Year 16: $182,231.23

Year 21: $0.00. (c) The loan amortization table for the final 12 half-years is provided.

(a) The half-yearly payment required for each five-year interval is **$37,746.88**.

To calculate the half-yearly payment for each five-year interval, we need to consider the changing interest rates and the total number of periods.

For the first five years, the interest rate is 5.5% per annum, which is equivalent to 2.75% per half-year. The total number of periods for this interval is 10 (5 years multiplied by 2 half-years per year).

Using the formula for calculating the loan repayment amount for an annuity:

\(A = P * (r / (1 - (1 + r)^(-n)))\)

Where:

A is the periodic payment

P is the principal amount of the loan ($900,000)

r is the interest rate per period (2.75%)

n is the total number of periods

By substituting the values into the formula, we can calculate the half-yearly payment for this interval:

A = 900,000 * (0.0275 / (1 - (1 + 0.0275)^(-10)))

 ≈ 900,000 * 0.0275 / 0.383

 ≈ 37,746.88

Therefore, the half-yearly payment required for each five-year interval is $37,746.88.

(b) The loan outstanding (outstanding balance) at the beginning of each five-year interval is as follows:

Year 1: $853,469.14

Year 6: $673,474.96

Year 11: $448,225.59

Year 16: $182,231.23

Year 21: $0.00

At the beginning of the loan, the outstanding balance is the same as the principal amount, which is $900,000.

To calculate the outstanding balance at the beginning of each five-year interval, we need to consider the changing interest rates and the number of periods.

After the first five years, the interest rate increases by 12% every five years. We can calculate the new interest rate for each interval:

Year 6-10: 5.5% + (5 * 12%) = 11.5%

Year 11-15: 5.5% + (10 * 12%) = 17.5%

Year 16-20: 5.5% + (15 * 12%) = 23.5%

Year 21-25: 5.5% + (20 * 12%) = 29.5%

Using the formula for the outstanding balance of an annuity, we can calculate the outstanding balance at the beginning of each five-year interval, considering the new interest rate and the number of periods:

Year 1: $900,000

Year 6: $900,000 * (1 + 0.115)^10 - (37,746.88 * ((1 + 0.115)^10 - 1) / 0.115) ≈ $853,469.14

Year 11: $853,469.14 * (1 + 0.175)^10 - (37,746.88 * ((1 + 0.175)^10 - 1) / 0.175) ≈ $673,474.96

Year 16: $673,474.96 * (1 + 0.235)^10 - (37,746.88 * ((1 + 0.235)^10 - 1) / 0.235) ≈ $448,225.59

Year 21: $448,225.59 * (1 + 0.295)^10 - (37

,746.88 * ((1 + 0.295)^10 - 1) / 0.295) ≈ $182,231.23

Year 26: $0.00 (loan fully paid off)

Therefore, the outstanding balance at the beginning of each five-year interval is as mentioned above.

(c) The loan amortization table for the final 12 half-years of the loan term is as follows:

| Payment | Principal Payment | Interest Payment | Outstanding Balance |

|---------|------------------|-----------------|---------------------|

| 1       | $30,273.59       | $12,473.29      | $151,957.38         |

| 2       | $31,095.26       | $11,651.62      | $120,862.12         |

| 3       | $31,933.51       | $10,813.38      | $88,928.61          |

| 4       | $32,788.28       | $9,958.61       | $56,140.33          |

| 5       | $33,659.48       | $9,087.41       | $22,480.85          |

| 6       | $22,480.85       | $4,937.01       | $0.00               |

The loan amortization table shows the payment number, the principal payment, the interest payment, and the outstanding balance for each half-year during the final 12 half-years of the loan term. The principal payment represents the portion of the payment used to reduce the loan balance, while the interest payment represents the interest charged on the outstanding balance. As the payments progress, the outstanding balance decreases until the loan is fully paid off in the final payment.

Please note that the calculations in the table are based on the given information and assumptions made for the loan.

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The upper limit of the 95% confidence interval for the population mean is approximately 77.50.

The upper limit of a 95% confidence interval for the population mean can be calculated using the formula:

Upper Limit = Sample Mean + (Z * (Standard Deviation / √Sample Size))

In this case, the sample mean is 75, the standard deviation is 5, and the sample size is 15.

To find the Z value for a 95% confidence interval, we need to look it up in the Z-table. A 95% confidence interval corresponds to a Z value of approximately 1.96.

Plugging these values into the formula, we get:

Upper Limit = 75 + (1.96 * (5 / √15))

Calculating this expression, we find that the upper limit of the 95% confidence interval for the population mean is approximately 77.50.

Therefore, the correct answer is approximately 77.50.

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Answer:

Step-by-step explanation:

92 -- 2*2*23

146 ---2*73

The calculated expectation of X is 0.1

From the question, we have the following parameters that can be used in our computation:

P{X = -1}=0.2, P{X = 0} = 0.5, P{X = 1}=0.3

The expectation of X is calculated as

E(x) = ∑xp(x)

So, we have

E(x) = -1 * 0.2 + 0 * 0.5 + 1 * 0.3

Evaluate

E(x) = 0.1

Hence, the expectation of X is 0.1

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ok

Equations : 4x + y = 2 x - y = 3

Equation 1 4x + y = 2 x - y = 3

x y x y

-2 10 -2 -5

-1 6 -1 -4

0 2 0 -3

1 -2 1 -2

2 -6 2 -1

The point where both lines cross is (1, -2) so it is the solution.

Answer:

5% is the smallest (so it comes first) 1/5 (comes next) and 0.5 is the largest.

Step-by-step explanation:

The smallest value is 5%. Why? Well, if you convert 5% into decimal form, it comes out to 0.05. 1/5 converted to decimal form is 0.2. As you can see, 5% out of all of these is the smallest. So in order, it is 5%, 1/5, 0.5.

Hope this helps! Have an amazing day! :D

Answer:

108

Step-by-step explanation:

b1+b2/2×h

so 12+15/2×8

12+15=27

27÷2=13.5

13.5×8=108

108 is the answer

The probability in each case scenario as described in the task content are as follows;

a) P(a) = 7/16

b) P(b) = 7/16

c) P(c) = 7/16

d) P(d) = 7/16

Given: There are 14 freshmen and 18 sophomore students and the students were selected with replacement.

a) The required probability that the 2nd student is also a freshman if the first student was a freshman since the selection is with replacement is simply;

P(a) = 14/32 = 7/16.

b) The probability that the 2nd student was a

freshman if the first student was a sophomore is;

P(b) = 14/32 = 7/16.

c). The required probability that the 2nd student is also a freshman if the first student was a freshman since the selection is with replacement is simply;

P(c) = 14/32 = 7/16.

d) The probability that the 2nd student was a

freshman if the first student was a sophomore is;

P(d) = 14/32 = 7/16.

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Answer:

5.352

Step-by-step explanation:

Common misconcepton, but the thousands and three hundred just mean the spot after the decimal point.

So, five is 5. since 'Five and'

Three hundred, 5.3

Fifty-two thousandths, 5.352

Hope this helped! Have a good day.

The solution for x, restricted to the interval [0, 6], is x = (1/3) arcsec(y/6).

To solve the equation y = 6 sec(3x) for x, where x is restricted to the interval [0, 6], you correctly followed these steps:

Start with the equation: y = 6 sec(3x).

Divide both sides of the equation by 6: y/6 = sec(3x).

Take the inverse secant (arcsec) of both sides: arcsec(y/6) = 3x.

Divide both sides by 3: (1/3) arcsec(y/6) = x.

By following these steps, you isolated the variable x and expressed it in terms of the given equation. The inverse secant function, also denoted as arcsec or sec^(-1), allows you to find the angle whose secant is equal to the value inside the parentheses.

The resulting solution x = (1/3) arcsec(y/6) satisfies the original equation y = 6 sec(3x) and is restricted to the interval [0, 6] as specified.

Well done on solving the equation and providing a clear explanation of the steps involved!

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A number multiplied by 1 is the number itself. So,1 is a factor of every number.

1 is the factor of every number as one divides every number exactly, without leaving any remainder behind, and gives the quotient as the number itself.

1 is the greatest factor of itself and the smallest factor of any other number.

There are some properties of factors one:

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By answering the above question, we may state that Shade the region to line the left of the open circle, since x is less than 4.

A line is a geometric object that is infinitely long and has no width, depth, or curvature. A line, which can also exist in two-, three-, and more-dimensional spaces, is therefore a one-dimensional object. A line is often used to refer to a line segment with two points as its endpoints. A line is a flat, one-dimensional object that may extend indefinitely in both directions and has no thickness. The terms "straight line" or "old correct line" are sometimes used to describe a line that has no "wobble" anywhere along its length.

plot the inequality 4x-22<-6 on a number line,

4x - 22 + 22 < -6 + 22

4x < 16

4x/4 < 16/4

x < 4

Plot an open circle at 4 on the number line to indicate that x is not included in the solution set.

Shade the region to the left of the open circle, since x is less than 4.

The resulting graph should look like this:

     ○----------------->

     |  |  |  |  |  |  |

   -∞  -5  -4  -3  -2  -1  0  1  2  3  4  5  6

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Answer:

Step-by-step explanation:

The value of p-hat (the sample proportion) is 0.24, or 24% if total number of skittles in a sample is 300 out of which 72 skittles are purple.

The sample proportion, denoted by p-hat, is the proportion of purple skittles in the sample. We can calculate p-hat by dividing the number of purple skittles by the total number of skittles in the sample

p-hat = number of purple skittles / total number of skittles in sample

In this case, we have

Number of purple skittles = 72

Total number of skittles in sample = 300

Therefore,

p-hat = 72/300 = 0.24

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Answer:

1530 x 2.6

Step-by-step explanation:

Just multiply, 90 divided by 1530, multiply 1530 x 2.6

The required Matlab code to find the greatest common divisor of a number using the Euclidean algorithm is shown.

To find the greatest common divisor (GCD) of two numbers using the Euclidean algorithm in MATLAB, you can use the following code:

% Replace '12345678' with your actual student number

studentNumber = '12345678';

% Extract the first 4 digits as the first number

firstNumber = str2double(studentNumber(1:4));

% Extract the last 3 digits as the second number

secondNumber = str2double(studentNumber(end-2:end));

% Find the GCD using the Euclidean algorithm

gcdValue = gcd(firstNumber, secondNumber);

% Display the result

disp(['The GCD of ' num2str(firstNumber) ' and ' num2str(secondNumber) ' is ' num2str(gcdValue) '.']);

Make sure to replace '12345678' with your actual student number. The code extracts the first 4 digits as the first number and the last 3 digits as the second number using string indexing. Then, the gcd function in MATLAB is used to calculate the GCD of the two numbers. Finally, the result is displayed using the disp function.

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Answer:

Step-by-step explanation:

To find the area of a square using the length of its diagonal, use the formula area = d^2 divided by 2

so

5 × 5 : 2 =

25 : 2 =

12.5 cm²

To solve this problem, we need to use the formula:

Number of revolutions = Distance traveled / Circumference of wheel

First, we need to find the circumference of the wheel using the formula: Circumference = diameter x pi

To calculate the number of wheel revolutions for Jackson's dirtbike, we'll need to use the diameter of the wheel and the distance he rides. Here's a step-by-step explanation:

1. Convert the diameter of the wheel (19 inches) to feet: 19 inches × (1 foot / 12 inches) = 1.583 feet

2. Calculate the circumference of the wheel using the diameter and the value of pi (3.14): Circumference = Diameter × pi = 1.583 feet × 3.14 = 4.97062 feet

3. Now, divide the total distance ridden (500 feet) by the circumference of the wheel to find the number of revolutions: Revolutions = Distance / Circumference = 500 feet / 4.97062 feet ≈ 100.59 revolutions

The wheel on Jackson's dirtbike will make approximately 100.59 revolutions when he rides for 500 feet.

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Answer: C. 358,800,800

Step-by-step explanation:

Total digitrs ( from 0 to 9 ) = 10

Total letter ( from a to z ) = 26

To make the code , we require 4 letters followed by 3 digits .

If letter can be used once ( i.e. repetition is not allowed) , then the number of ways to arrange this sequence of letters = \(^{26}P_4\)  [By using Permutation]

\(=\dfrac{26!}{(26-4)!}=\dfrac{26\times25\times24\times23\times22!}{22!}\)

\(=26\times25\times24\times23=358800\)

But repetition is allowed for digits , so number of ways to arrnge 3 digits = \(10\times10\times10=1000\)

Now , the total number of ways tio make the code = 358800  x 1000

=358,800 ,000

Hence, the correct option is C. 358,800,800.

Answer:

b=−5

Step-by-step explanation:

Subtract the numbers

b=2−7                                                            

b= −5

Answer:

b = -5

Step-by-step explanation:

b = 2 - 7

b = -5

Answer:

Im bleeding i cant yet the hospital said i need to do math so

Step-by-step explanation:

Answer:

8 min

4 min

6 min

The second gardener trimmed the fastest

Step-by-step explanation:

You have to divide the amount of time by the number of bushes for each gardener.

First gardener: 72min / 9 bushes = 8min/bush

Second gardener:  20min / 5 bushes = 4min/bush

Third gardener: 48 min / 8 bushes = 6min/bush

9514 1404 393

Answer:

  A.  {1, 7, 11}

Step-by-step explanation:

The constant in the equation is 1, so that is the value when x=0. No answer choice can be correct unless it includes the value 1. The only viable choice is A.

_____

If you want to go to the trouble to evaluate the expression for the different values of x, you can. It often works well to rewrite the expression to "Horner form":

  (2x -1)x +1

For x = -2, this is (2(-2) -1)(-2) +1 = (-5)(-2) +1 = 11

For x = 0, this is 1

For x = 2, this is (2(2) -1)(2) +1 = (3)(2) +1 = 7

So, the range for the given domain is {1, 7, 11} . . . . . matches choice A