3.) SOLVE FOR X.


4.) FIND NM.


ATTACHMENTS FOR BOTH BELOW. LMK IF YOU CAN SEE THEM. PLEASE HELP ASAP, I AM VEERRYY MUCH CONFUSED. TYSM TO WHOEVER HELPS!

The factors indicated, are classified according to whether it is associated with a movement along the aggregate demand curve or a shift of the aggregate demand curve.

Movement along the aggregate demand curve

Shift of the aggregate demand curve

An aggregate demand curve depicts the total domestic expenditure on goods and services at each price level.

The AD curve is trending higher due to rising economic prosperity. People's spending grows as their income rises, leading to an increase in AD and vice versa. As a result of the positive link between income and AD, the AD curve slopes higher.

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Full Question:

Classify each factor according to whether it is associated with a movement along the aggregate demand curve or a shift of the aggregate demand curve.

To find the probability P(X ≤ \(X_{max\)), we need to find the cumulative probability corresponding to the calculated z-score using a standard normal distribution table or a calculator.

Probability is a way to gauge how likely something is to happen. Many things are difficult to forecast with absolute confidence. Using it, we can only make predictions about the likelihood of an event happening, or how likely it is.

To find the probability that a randomly selected examinee will complete the exam on time, we need to calculate the z-score and then use the standard normal distribution.

Given:

Mean time per question (μ) = minutes

Standard deviation (σ) = minutes

Time allowed for the exam (X) = minutes

We want to find P(X ≤ \(X_{max\)), where \(X_{max\) is the maximum time allowed for the exam. Let's assume the maximum time allowed is \(T_{max\).

To calculate the z-score, we use the formula:

z = (X - μ) / σ

z = (\(T_{max\) - μ) / σ

The z-score tells us how many standard deviations an observation is from the mean.

To find the probability P(X ≤ \(X_{max\)), we can use a standard normal distribution table or a calculator to find the cumulative probability associated with the calculated z-score.

Now, let's calculate the z-score using the given values:

z = (\(T_{max\) - μ) / σ

z = (\(T_{max\) - minutes) / minutes

To find the probability P(X ≤ \(X_{max\)), we need to find the cumulative probability corresponding to the calculated z-score using a standard normal distribution table or a calculator.

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Other things being equal, the width of a confidence interval gets smaller as the sample size increases.

The width of a confidence interval gets smaller as:

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The acceleration vector of the car at time t=1 is -18i + 2e²j, where i and j are the unit vectors in the x and y directions, respectively. The position of the car at time t=2 is (5.52, 9.86) feet.

To find the acceleration vector of the car at time t=1, we need to find the derivative of the velocity vector v(t) with respect to time t:

a(t) = d/dt [3+6sin(3t)]i + d/dt [1+e²t]j

= 18cos(3t)i + 2e²j

Plugging in t=1, we get:

a(1) = 18cos(3)(1)i + 2e²j

= -18i + 2e²²j

To find the position of the car at time t=2, we need to integrate the velocity vector v(t) from t=0 to t=2:

r(t) = \(\int_{v(0)}^{v(t)} v(u) du\)

where v(u) = {3+6sin(3u),1+e²u}.

We can integrate each component of the velocity vector separately:

x(t) = \(\int_0^t (3+6sin(3u)) du\) = 3t - 2cos(3t) + 2

y(t) = \(\int_0^t (1+e^2u) du\) = t + (1/2)e²t

Plugging in t=2, we get:

x(2) = 3(2) - 2cos(3(2)) + 2 ≈ 5.52 feet

y(2) = 2 + (1/2)e⁴ ≈ 9.86 feet

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The pentagonal prism with volume 360 in³ and height of 3 inches have a base area of 120 in²

A pentagonal prism is a prism that has two pentagonal bases like top and bottom and five rectangular sides.

Given that, the volume of a pentagonal prism is 360 in³, with a height of 3 inches,

We need to find the area of the base,

We know that, the volume of a pentagonal prism is =

V = 1/4 √(5(5+2√5)·a²h

Where a is the base edge and h is the height,

360 = 1/4·3 √(5(5+2√5)·a²

1/4·√(5(5+2√5)·a² = 120

Since, the base of a pentagonal prism is a pentagon, and the area of a pentagon = 1/4 √(5(5+2√5)·a²

And we have,

1/4 √(5(5+2√5)·a² = 120

Therefore, the base area is 120 in²

Hence, the pentagonal prism with volume 360 in³ and height of 3 inches have a base area of 120 in²

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Answer:

-11/9 is the slope

Step-by-step explanation:

Use the formula and u will find this is the answer, hope this helped!

Y2 - Y1 / X2 - X1

(-9 - 2) / (6 - (-3))

<!> Brainliest is appreciated!

Answer:

-11/9

Step-by-step explanation:

In order to find the slope from 2 points, use the following formula: \(m=\frac{y_{2} - y_{1} }{x_{2}-x_{1} }\)

Plug in each of the numbers into their corresponding areas. Basically, we are subtracting the y values together and dividing it with the difference of the x values:

\(\frac{-9-2}{6-(-3)}\)

The negatives cancel out and become postive, so the denominator will then read to be 6+3:

\(\frac{-9-2}{6+3}\)

\(-\frac{11}{9}\)

The volume of the cylinder is 3041.06 inches square

A cylinder has a height of 8 inches and a radius of 11 inches.

We have to determine the volume of the cylinder.

A cylinder has a height is 8 inches

The radius of the cylinder is 11inches.

The volume of the cylinder is given by

\(V=\pi r^2h\)

Insert the values

\(V=(3.14)(11)^2(8)\)

\(V=3041.06 in^2\)

Therefore the volume of the cylinder is 3041.06 inches square

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The solution for question a is -2.00. The solution for question b is  0.6687. The solution for question c is 10 students. We can show the working in the following manner.

a. What is the probability of completing the exam in one hour or less (to 4 decimals)?

To answer this question, we need to convert the time of one hour (60 minutes) to a z-score using the mean and standard deviation provided. We have:

z = (60 - 80) / 10 = -2.00

Using a standard normal distribution table or calculator, we can find that the probability of completing the exam in one hour or less is approximately 0.0228, rounded to 4 decimal places.

b. What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes (to 4 decimals)?

To answer this question, we need to find the probability of completing the exam in less than 75 minutes and subtract the probability of completing the exam in less than 60 minutes. We have:

z1 = (60 - 80) / 10 = -2.00

z2 = (75 - 80) / 10 = -0.50

Using a standard normal distribution table or calculator, we can find that the probability of completing the exam in less than 75 minutes is approximately 0.6915 and the probability of completing the exam in less than 60 minutes is approximately 0.0228, as calculated in part a.

So, the probability of completing the exam in more than 60 minutes but less than 75 minutes is approximately:

0.6915 - 0.0228 = 0.6687, rounded to 4 decimal places.

c. Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time (to the next whole number)?

To answer this question, we need to find the number of students whose exam time is greater than 90 minutes, which is the maximum time allowed. We can use the normal distribution with the given mean and standard deviation to calculate this.

First, we need to find the z-score corresponding to a time of 90 minutes:

z = (90 - 80) / 10 = 1.00

Using a standard normal distribution table or calculator, we can find that the probability of completing the exam in more than 90 minutes is approximately 0.1587.

Therefore, the expected number of students who will be unable to complete the exam in the allotted time is:

60 x 0.1587 = 9.52, which rounds up to 10 students (to the next whole number).

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The objective function is 4l + 4w + 38, where "l" and "w" are the dimensions of the rectangular garden. To find the dimensions that minimize the objective function, we need to take partial derivatives with respect to "l" and "w" and set them equal to zero.

Let's call the length of the garden (in meters) as "l" and the width of the garden (in meters) as "w". Then we can express the area of the garden as:

Area of garden = l x w = 30 m²

We can also express the total area of the garden and borders as:

Total area (a) = (l + 2) x (w + 2) + 2 x (l + 2) x 1

where (l + 2) and (w + 2) represent the dimensions of the garden including the border, and 2 x (l + 2) x 1 represents the two smaller borders.

To minimize the combined area of the garden and borders, we need to find the values of "l" and "w" that minimize the objective function, which is:

Objective function = a = (l + 2) x (w + 2) + 2 x (l + 2) x 1

Simplifying this expression, we get:

Objective function = a = l x w + 4l + 4w + 8

Substituting the expression for the area of the garden, we get:

Objective function = a = 30 + 4l + 4w + 8

Objective function = a = 4l + 4w + 38

Therefore, the objective function is:

a(l,w) = 4l + 4w + 38

To find the dimensions of the garden that minimize this objective function, we need to take the partial derivatives of "a" with respect to "l" and "w" and set them equal to zero. This will give us the values of "l" and "w" that minimize the objective function.

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the probability of experiencing neither side effect is 0.37 or 37%.let's denote the probability of experiencing nausea by P(n) and the probability of experiencing decreased sexual drive by P(d). We know that:

P(n) = 0.23
P(d) = 0.52
P(n ∩ d) = 0.12

We want to find the probability of experiencing neither side effect, which can be denoted by P(~n ∩ ~d), where ~n and ~d represent the complements of nausea and decreased sexual drive, respectively.

We can use the formula for the probability of the union of two events to find P(~n ∪ ~d):

P(~n ∪ ~d) = 1 - P(n ∪ d)

We know that P(n ∪ d) = P(n) + P(d) - P(n ∩ d), so we can substitute the given values to get:

P(n ∪ d) = 0.23 + 0.52 - 0.12 = 0.63

Therefore,

P(~n ∪ ~d) = 1 - 0.63 = 0.37

So the probability of experiencing neither side effect is 0.37 or 37%.

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We assume that with a linear relationship between two variables, for any fixed value of x, the observed residuals follows a normal distribution.

This assumption is based on the Central Limit Theorem, which states that when the sample size is large enough, the distribution of sample means will be approximately normal, regardless of the shape of the underlying population distribution.

In the case of a linear relationship between two variables, we can assume that the residuals (the difference between the observed y values and the predicted values based on the linear regression model) follow a normal distribution with mean 0 and constant variance. This assumption is important because it allows us to use statistical methods that rely on normality, such as hypothesis testing and confidence intervals.

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The value of x for which the equality (h⁰h⁻¹)(x) ≠ x does not hold is x = 7/3.

Therefore, Option (C) is the correct answer.

We have to find the value of x for which the equality (h⁰h⁻¹)(x) ≠ x

if h(x)=1/x+2.

Function h(x) is given as  h(x)=1/x+2  ...[1]

We have to find the inverse of the given function. To find the inverse of function h(x), we will interchange the variables x and y in the given function. After the interchange, we will get

,x = 1/y+2

Now, we will solve the above equation for y. Subtracting 2 from both sides, we getx - 2 = 1/y

Multiplying by y on both sides, we getyx - 2 = 1

Dividing both sides by x - 2, we get y = 1/(x - 2)

The inverse of h(x) is y = 1/(x - 2).

Now, we will find the value of x for which the equality (h⁰h⁻¹)(x) ≠ x does not hold.

h⁰h⁻¹(x) = xh⁻¹(x) = (h⁰)⁻¹(x)

We know that

h⁰(x) = x and h⁻¹(x)

= 1/(x - 2)h⁰h⁻¹(x)

= x or h⁰(h⁻¹(x))

= x ⇒ h(h⁻¹(x))

= x ⇒ h(1/(x - 2))

= xh(1/(x - 2)) = 1/(1/(x - 2)) + 2

= x⇒ x - 2 + 2(x - 2)

= 7/3.

.The value of x for which the equality (h⁰h⁻¹)(x) ≠ x does not hold is x = 7/3.

Therefore, Option (C) is the correct answer.

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a. The joint probability function for Y1 and Y2 can be expressed as follows:

P(Y1 = y1, Y2 = y2) = (2! / y1! y2! (2 - y1 - y2)!) * (1/3)^y1 * (1/3)^y2 * (1/3)^(2 - y1 - y2), where y1, y2 = 0, 1, 2.

b. The probability that firm A is assigned exactly one contract and firm B is assigned zero contracts is 1/3

The joint probability function for Y1 and Y2 is determined by considering the number of ways in which the two contracts can be assigned to the three firms. Since each firm can receive 0, 1, or 2 contracts, there are (3^2) = 9 possible outcomes for the pair (Y1, Y2). However, not all outcomes are equally likely. To determine the probability of each outcome, we use the following reasoning:

There are (2!) / (y1! y2! (2 - y1 - y2)!) ways to assign the two contracts to firms A, B, and C, where y1 and y2 represent the number of contracts assigned to firms A and B, respectively. The factor of 2! accounts for the fact that there are two contracts to be assigned, while the denominator takes into account the number of ways in which these contracts can be assigned to the three firms.

Since the contracts are randomly assigned, the probability of firm A receiving a particular contract is 1/3, and the same holds for firms B and C. Therefore, the probability of assigning y1 contracts to firm A and y2 contracts to firm B is (1/3)^y1 * (1/3)^y2.

The remaining contract must be assigned to firm C. Therefore, the probability of the remaining contract being assigned to firm C is (1/3)^(2 - y1 - y2).

Putting these three factors together yields the joint probability function given above.

To find F(1,0), we need to sum the joint probabilities for all outcomes where Y1 = 1 and Y2 = 0, as well as for all outcomes where Y1 < 1 and Y2 = 0. Therefore,

F(1,0) = P(Y1 = 1, Y2 = 0) + P(Y1 = 0, Y2 = 0)

= [(2! / 1! 0! (2 - 1 - 0)!) * (1/3)^1 * (1/3)^0 * (1/3)^(2 - 1 - 0)] + [(2! / 0! 0! (2 - 0 - 0)!) * (1/3)^0 * (1/3)^0 * (1/3)^(2 - 0 - 0)]

= (2/9) + (1/9)

= 1/3

Therefore, the probability that firm A is assigned exactly one contract and firm B is assigned zero contracts is 1/3

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Answer:

Step-by-step explanation:

The disCount will be the differenceb

The number of adult tickets that were sold would be = 435 tickets.

A ticket is an official document that gives an individual access to an event.

The cost of adult tickets = $8

The cost for student tickets = $4

The number of students tickets sold = X

The number of adults tickets sold = X +30

The told number of tickets sold = 840

To find X;

X + X + 30 = 840

2x + 30 = 840

2x = 840-30

2x = 810

X = 810/2

X = 405

Therefore, the number of tickets sold for adults = 405 +30 = 435 tickets.

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Answer:

15

Step-by-step explanation:

The length of the rectangle is,

⇒ (x - 5)

A rectangle is a two dimension figure with 4 sides, 4 corners and 4 right angles. The opposite sides of the rectangle are equal and parallel to each other.

Given that;

The area of a rectangle is, (x³ - 5x² + 3x - 15), and the width of the rectangle is (x² + 3)

Now, We know that;

Area = Length x Width

⇒ (x³ - 5x² + 3x - 15) = l × (x² + 3)

⇒ l = (x³ - 5x² + 3x - 15) / (x² + 3)

⇒ l = (x² (x - 5) + 3 (x - 5)) / (x² + 3)

⇒ l = (x² + 3) (x - 5) / (x² + 3)

⇒ l = (x - 5)

Thus, The length of the rectangle is,

⇒ (x - 5)

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Answer:

m = -20

Step-by-step explanation:

first distribute:

-4(m + 18) = 8

-4m - 72 = 8

Now add 72 to both sides to get m by itself

-4m = 80

and divide by -4 on both sides

m = -20

Hope this helped! :)

Answer: m = -20

Step-by-step explanation:

−4(m + 18) = 8

Step 1: Simplify both sides of the equation.

−4(m + 18) = 8

(−4)(m) + (−4)(18) = 8 (Distribute)

−4m + −72 = 8

−4m − 72 = 8

Step 2: Add 72 to both sides.

−4m − 72 + 72 = 8 + 72

−4m = 80

Step 3: Divide both sides by -4.

−4m / − 4 = 80 / −4

m = −20

Answer:

Ghe661.25

Step-by-step explanation:

First to get the interest gotten for the first year, you'll multiply 15/100 by the principal amount which is 500. The value gotten will be 75. This means that the principal amount for the second year will be 500 + 75 which will amount to 575. Again, 15% of 575 will be added to 575 and this will give you the total amount at the end of the two years

Answer:

C

and 10%

Step-by-step explanation:

To examine whether spherical refraction is different between the left and right eyes of 17 people, the appropriate technique to use would be a paired samples t-test.

The reason for this is that we are comparing the differences in refraction between the left and right eyes within the same individuals. A paired samples t-test is used to compare the means of two related groups (in this case, the left and right eyes) when the data is not normally distributed or when the variances are unequal. It also assumes that the differences between the pairs are normally distributed.

A t-test for two population means with independent samples would be appropriate if we were comparing the means of two separate groups (e.g., comparing the average refraction for a group of people with left-eye dominance to a group with right-eye dominance). However, since we are measuring both eyes within the same individuals, we cannot treat these measurements as independent samples.

A z-test for two population means assumes that the population variances are known, which is typically not the case in practice. Additionally, a z-test is typically only used for large sample sizes (typically greater than 30).

A test for two population proportions would be inappropriate since we are not dealing with proportions in this scenario.

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The accumulated amount after 25 years is , $70,702.80.

Now, We can use the formula for compound interest to find the accumulated amount after 25 years:

A = P(1 + r/k)^(kt)

Where A is the accumulated amount, P is the principal , r is the interest rate, n is the number of times the interest is compounded per year, and t is the time period.

In this case, we have:

P = $25,300

r = 0.045 (

k = 12 (monthly compounding)

t = 25

Substituting these values into the formula, we get:

A = $25,300(1 + 0.045/12)^(12 x 25)

A ≈ $70,702.80

Therefore, the accumulated amount after 25 years is ,

$70,702.80.

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Answer:

49 cupboards

Step-by-step explanation:

See the steps below, it is self-explanatory:

As 14*14/4= 49, the answer is 49 cupboards

The solution for r is given by \(r=\frac{\sqrt[3]{\frac{3}{2} x} }{\pi }\).

To solve the equation

\(x=(\frac{2}{3} )\pi r^3\) for r,

we need to isolate the variable r.

Let's follow the steps:

Multiply both sides of the equation by \((\frac{3}{2} )\) to cancel out the coefficient \((\frac{2}{3} )\) on the right side:

\((\frac{3}{2} )x=\pi r^3\).

Divide both sides of the equation by π to get rid of it on the right side:

\(\frac{\frac{3}{2} x}{\pi } = r^3\).

Take the cube root of both sides to eliminate the exponent:

\(\sqrt[3]{\frac{3}{2} x} = r\).

Therefore, the solution for r is given by \(r = \frac{\sqrt[3]{\frac{3}{2} x} }{\pi }\).

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Using strong induction, we can prove that the square root of -2 is irrational by showing that it cannot be expressed as a fraction of coprime odd integers.

To prove that √-2 is irrational using strong induction, we need to show that for any natural number n, if the square root of -2 can be expressed as a fraction a/b, where a and b are coprime integers, then a and b must be odd.

We can start by using the base case, n = 1. Assume that √-2 can be expressed as a fraction a/b where a and b are coprime integers. Then, we have

√-2 = a/b

Squaring both sides gives

-2 = a^2/b^2

Multiplying both sides by b^2 gives

-2b^2 = a^2

This implies that a^2 is even, and therefore a is also even. We can express a as 2k for some integer k, which means

-2b^2 = (2k)^2

Simplifying, we get

-2b^2 = 4k^2

Dividing both sides by -2 gives

b^2 = -2k^2

This implies that b^2 is even, which means that b is also even. However, this contradicts our assumption that a and b are coprime integers. Therefore, √-2 cannot be expressed as a fraction a/b where a and b are coprime integers.

Now, let's assume that for all n ≤ k, the square root of -2 cannot be expressed as a fraction a/b where a and b are coprime integers with a and b odd. We want to prove that this also holds for n = k+1.

Assume that √-2 can be expressed as a fraction a/b where a and b are coprime integers with a and b odd. Then, we have

√-2 = a/b

Squaring both sides gives

-2 = a^2/b^2

Multiplying both sides by b^2 gives

-2b^2 = a^2

This implies that a^2 is even, and therefore a is also even. We can express a as 2k for some integer k, which means

-2b^2 = (2k)^2

Simplifying, we get

-2b^2 = 4k^2

Dividing both sides by -2 gives

b^2 = -2k^2

This implies that b^2 is even, which means that b is also even. However, this contradicts our assumption that a and b are coprime integers with a and b odd. Therefore, √-2 cannot be expressed as a fraction a/b where a and b are coprime integers with a and b odd.

By strong induction, we have proven that for any natural number n, the square root of -2 cannot be expressed as a fraction a/b where a and b are coprime integers with a and b odd. Therefore, √-2 is irrational.

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The radius of the cone is 783.84/√h

A cone is the surface traced by a moving straight line (the generatrix) that always passes through a fixed point (the vertex).

Volume is defined as the space occupied within the boundaries of an object in three-dimensional space.

The volume of a cone is expressed as;

V = 1/3πr²h

643072 × 3 = 3.14 × r²h

r²h = 614400

r² = 614400/h

r = 783.84/√h

therefore the radius of the cone is 783.84/√h

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When the  expression \(log_b{32} - log_b{7}\), b > 1, is written in the form \(log_b{a}\), the value of a is 25

We know that a logarithm of a number is nothing but the power to which a number must be raised to get some other values.

We know that the rules for the addition or subtraction (basic algebraic operations)

When the base od logarithms are same, then only we can add or subtract logarithms.

Consider logartithemic numbers : log₂7 , log₃8 and log₂5

Her we can observe that the base of  logartithemic numbers  log₂7 , log₃8  is different.

Here, the expression \(log_b{32} - log_b{7}\)

The base of logartithemic numbers  \(log_b{32} , log_b{7}\) is equal.

so, we can perform algebraic operation.

\(log_b{32} - log_b{7}\\\\=log_b{32-7}\\\\=log_b{25}\)

Comparing above expression with \(log_b{a}\) we have,

a = 25

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=========================================================

Explanation:

Draw a horizontal line to cut the figure into a rectangle on top and a triangle down below.

The rectangle has length and width of 6 and 4 (horizontal and vertical components respectively). The area of this rectangle is 6*4 = 24 square units.

The triangle has a base of 6 and a height of 4. The base and height of any triangle are always perpendicular.

The area of the triangle is base*height/2 = 6*4/2 = 24/2 = 12 square units. If you were to cut out half of the triangle and rearrange things, you'll find that a rectangle can be formed. This rectangle is half in area that of the first rectangle we found. This is why we divide by 2 when finding the area of the triangle.

Once you know the two sub-areas, we add them up to get the overall area: 24+12 = 36 square units.