√175x²y³ simplify radical expression

To plot points along the portion of the firm's short-run supply curve that corresponds to prices where there is positive output, we need to identify the portion of the graph where the firm is producing output.

We can observe from the graph that the firm's short-run supply curve is the component of the marginal cost curve that is higher than the average variable cost curve. The company will shut down and create no production if prices fall below the minimum point of the average variable cost curve. However, as long as the price is above the marginal cost of production, the company will create output at prices above the minimum point of the average variable cost curve.

We may use the orange square symbols to represent the price and matching amount provided at each point where the company is generating output to plot points along this segment of the short-run supply curve. We may advance up the marginal cost curve from the last point of the average variable cost curve until we reach the maximum price at which the company is generating output. The firm's short-run supply curve may then be created by marking each price and quantity combination along this segment of the curve.

It is important to note that the firm's short-run supply curve is a reflection of its marginal cost curve above the average variable cost curve, and will shift as the firm's costs of production change or its technology improve.

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A) The length MN of the given wooden block is: 12

B) 80400 in²

1) The formula for volume of a cuboid is:

Volume = Length * Width * Height

Thus:

427 = (MN * 7 * 3) + (5 * 5 * 7)

427 = 21MN + 175

21MN = 252

MN = 252/21

MN = 12

2) Surface area of entire object is:

TSA = 2(12 * 3) + 2(12 * 7) - (5 * 7) + 2(7 * 3) + 3(5 * 7) + 2(5 * 5)

= 402 in²

For 200 blocks:

TSA = 200 * 402 = 80400 in²

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Answer:

i) The horizontal and vertical components of the softball's motion can be described by the following parametric equations:

x(t) = v0x * t

y(t) = v0y * t - 1/2 * g * t^2 + h0

where v0x and v0y are the initial horizontal and vertical velocities, respectively, g is the acceleration due to gravity (32.2 ft/s^2), h0 is the initial height (5 ft), and t is time.

We can find v0x and v0y by resolving the initial velocity vector into horizontal and vertical components:

v0x = v0 * cos(45°) = 49 ft/s * cos(45°) ≈ 34.65 ft/s

v0y = v0 * sin(45°) = 49 ft/s * sin(45°) ≈ 34.65 ft/s

Substituting these values into the parametric equations, we get:

x(t) = 34.65 * t

y(t) = 34.65 * t - 16.1 * t^2 + 5

ii) The softball will be in the air until it hits the ground, which occurs when y(t) = 0. We can solve for t using the quadratic formula:

16.1 * t^2 - 34.65 * t + 5 = 0

t ≈ 2.19 s (rounded to two decimal places)

Therefore, the softball was in the air for approximately 2.19 seconds.

iii) The horizontal distance traveled by the softball can be found by evaluating x(t) at the time when the softball hits the ground:

x(2.19) ≈ 75.8 ft (rounded to one decimal place)

Therefore, the softball traveled approximately 75.8 feet in the air.

iv) The softball reaches its maximum height when its vertical velocity is zero. We can find the time when this occurs by setting v0y - g * t = 0 and solving for t:

t = v0y / g = 34.65 / 32.2 ≈ 1.08 s (rounded to two decimal places)

Therefore, the softball reaches its maximum height after approximately 1.08 seconds.

v) The maximum height reached by the softball can be found by evaluating y(t) at the time when the softball reaches its maximum height:

y(1.08) ≈

Step-by-step explanation:

The value of domain and range for this situation are,

Domain = (- ∞, ∞)

Range = (- ∞, ∞)

We have to given that;

The total cost for funding a trip for the senior class to go to the fall fair, C(x), is a function of the number of students that will make the trip, x.

Now, We have;

⇒ C (x) = 350 + 7.5x

Clearly, the function is a polynomial.

Hence, The value of domain and range for this situation are,

Domain = (- ∞, ∞)

Range = (- ∞, ∞)

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31.25 g on the 5th day

no it will never by empty because even when it gets down to one singular piece of sugar you would technically just cut it in half, then that in half, yes it would get impossible, that's why you wouldn't actually do it, but if you typed it into a calculator it would just keep getting a smaller and smaller decimal.

The probability that the weight will be less than 4504 grams is 0.9713.

We need to standardize the value 4504 using the given mean and variance, and then use the standard normal distribution table to find the corresponding probability.

The standard deviation is the square root of the variance: √391876≈626.05

So, the z-score for a weight of 4504 grams is:

z=(4504−3315)/626.05≈1.8974

Using a standard normal distribution table, we find that the probability of a z-score being less than 1.8974 is approximately 0.9713.

Therefore, the probability that a newborn baby boy born at the local hospital will weigh less than 4504 grams is approximately 0.9713.

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Using a calculator, the difference between the arithmetic average ROR and the weighted average ROR is Option B) 0.1%

arithmetic average ROR is given as

(2.1% + 4.4% - 7.8% + 10.5%) / 4 =  0.02300

The Weighted Average ROR

[(0.021 x 3200 ) + (0.044 x  1750) + (-0.078 x 1235) + (0.105 x   2300)] / (3200 + 1750 + 1235 + 2300) =  0.0341037124337065

Thus:

The Weighted Average ROR - arithmetic average ROR

= 0.0341037124337065 - 0.02300

= 0.01110371243 x 100

= 1.11037124337%

≈ 1.1%

So we are correct to state that Option B is the right answer.

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Answer:

Step-by-step explanation:

13. B

14. A

15. D

Answer:

B

Step-by-step explanation:

It is impossible for a possibility to be more than 1, or 100%.

Answer:

  A.  Bay Side: mean = 17.1, median = 16; Seaside: mean = 19.5, median = 18

  B.  Bay Side: σ = 8.96, IQR = 12, range = 37; Seaside: σ = 9.03, IQR = 16, range = 31

  C. Bay Side has lower center values and less variation.

Step-by-step explanation:

Given stem and leaf plots for 15 class sizes at each of two schools, you want to know (a) their measures of center, (b) their measures of variation, and (c) which would be preferred for lower class size.

The first attachment shows the statistics for Bay Side School. It tells you the measures of center for Bay Side are ...

The second attachment shows the statistics for Seaside School. The measures of center there are ...

We note the measures of center indicate smaller classes at Bay Side.

For Bay Side School, the measures of variation are ...

For Seaside School, the measures of variation are ...

The measures of variation are generally smaller for Bay Side.

The measures of center and the measures of variation both favor Bay Side School as the school of choice for smaller classes.

__

Additional comment

The arithmetic for these descriptive statistics can be tedious and error-prone. It is convenient to let a calculator do it. The lists of data points are given as L1 and L2 for the calculator screens attached. L1 is Bay Side data, and the result of the 1-Var Stats calculation is shown in the first attachment. Seaside data was put in L2, which was used for the calculations shown in the second attachment.

The Q1 and Q3 data values are the 4th lowest and 4th highest data values in each of the lists. The median is the 8th data value, counted from either end.

<95141404393>

The estimated sum of the given series using the sum of the first 10 terms is 302,500, the improved estimate for the sum of the given series is between 305,000 and 306,000, and the value of n is 8.

(a) Utilizing the equation for the entirety of the primary n terms of the arrangement, we have:

[tex]s10 = 1^3 + 2^3 + ... + 10^3[/tex]

= 1,000 + 8,000 + ... + 1,000,000

= 302,500

In this manner, the assessed whole of the given arrangement using the entirety of the primary 10 terms is 302,500.

(b) For n = 10, we have:

[tex]sn = 1^3 + 2^3 + ... + 10^3 ≈ 302,500[/tex]

Utilizing the disparities with[tex]f(x) = x^3[/tex], we have:

[tex]sn + ∫[10,∞] x^3 dx ≤ s ≤ sn + ∫[10,∞] x^3 dx + 10^3[/tex]

Utilizing calculus, ready to assess the integrand:

[tex]sn + ∫[10,∞] x^3 dx = sn + [1/4 x^4] [10,∞] = sn + 2500[/tex]

[tex]sn + ∫[10,∞] x^3 dx + 10^3 = sn + [1/4 x^4] [10,∞] + 10^3 = sn + 3500[/tex]

Substituting sn = 302,500, we get:

302,500 + 2500 ≤ s ≤ 302,500 + 3500

305,000 ≤ s ≤ 306,000

In this manner, the made strides assess for the sum of the given arrangement is between 305,000 and 306,000.

(c) The Leftover portion Gauge for the Necessarily Test states that the mistake E in approximating the whole s of an interminable arrangement by the nth halfway entirety sn is:

[tex]E ≤ ∫[n+1,∞] f(x) dx[/tex]

In this case, we need to discover mean of n such that E < 10 using the integral test

xss=removed xss=removed> [tex][(10^-5 x 4)^(1/4)] - 1[/tex]

n > 7.9378

Subsequently, we require n = 8 to guarantee that the blunder within the estimation s ≈ sn is less than[tex]10^-5.[/tex]

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The value of angle TUW is 32⁰.

The value of angle UTV is 25⁰.

The value of angle TUW is calculated by applying the following formula.

angle TVW = angle TUW (vertical opposite angles are equal)

angle TVW = 32⁰

So, angle TUW = 32⁰

The value of angle UTV is calculated as;

angle UTV = VWU (vertical opposite angles are equal)

3x + 4 = 2x + 11

3x - 2x = 11 - 4

x = 7

angle UTV = 3x + 4

= 3(7) + 4

= 25⁰

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The amount of ice cream consumed in the United State would be 5,650 millions of pounds in 2028.

Based on the information provided about the mount of ice cream consumed in the United State, a quadratic equation that models the amount of ice cream consumed, in millions of pounds, since 2010 is given by;

y = 12(x - 6)² + 3922

Where:

By substituting the value of y, the number of years can be calculated as follows;

5,650 = 12(x - 6)² + 3922

5,650 - 3922 = 12(x - 6)²

1728 = 12(x - 6)²

144 = (x - 6)²

12 = x - 6

x = 12 + 6

x = 18 years.

Since 2010; 2010 + 18 = 2028.

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Using the Law of Sines, we can solve the given triangle with B = 52, C = 15, and b = 43. The three angles of the triangle are approximately A = 112.94°, B = 52°, and C = 15°, and the lengths of the sides opposite these angles are approximately a = 28.29, b = 43, and c = 8.11.

The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is the same for all sides and their opposite angles. Therefore, we can use the Law of Sines to solve the given triangle as follows

sin(A)/a = sin(B)/b = sin(C)/c

We are given B = 52, C = 15, and b = 43, so we can use the Law of Sines to find a

sin(A)/a = sin(B)/b

sin(A)/a = sin(52)/43

sin(A) = a sin(52)/43

a = 43 sin(A)/sin(52)

Similarly, we can use the Law of Sines to find c

sin(A)/a = sin(C)/c

sin(A)/a = sin(15)/c

sin(15)c = a sin(A)

c = a sin(A)/sin(15)

Now, we can substitute the expressions for a and c into the equation sin(A)/a = sin(B)/b and solve for sin(A)

sin(A)/(43 sin(A)/sin(52)) = sin(52)/43

sin(A) = (43 sin(52) sin(15))/c

Substituting the expression for c, we get

sin(A) = (43 sin(52) sin(15))/(a sin(A)/sin(15))

Simplifying, we get

sin²(A) = (43²sin²(52) sin²(15))/(a²sin²(A))

Multiplying both sides by a^2 sin^2(A), we get

a²sin⁴(A) = 43^2 sin²(52) sin²(15)

Taking the square root of both sides and solving for a, we get

a = √((43² sin₂(52) sin²(15))/(sin⁴A)))

Substituting the given values and solving for A, we get

a = 28.29 (approx)

A = 112.94° (approx)

c = 8.11 (approx)

Therefore, the three angles of the triangle are approximately A = 112.94°, B = 52°, and C = 15°, and the lengths of the sides opposite these angles are approximately a = 28.29, b = 43, and c = 8.11.

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The error made by Beth is that:

She didn't apply the scale factor to each dimension of the triangle before multiplying

We are given the parameters as:

Initial height = 3.5cm

Initial width = 4cm

Formula for area of triangle is:

Area = ¹/₂ * base * height

If the triangle was enlarged by a scale factor of 5, then it means each of the dimensions should first be multiplied by 5 to get:

New width = 4 * 5 = 20 cm

New height = 5 * 3.5 = 17.5 cm

Thus:

New area = ¹/₂ * 20 * 17.5 = 175 cm²

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Answer:

Point C is in the second quadrant. C is the correct answer.

Answer:

  0.66

Step-by-step explanation:

You want the probability that a randomly chosen member from the 800 members of a gym takes either an aerobics class, as 450 members do, or does weight training, as 200 members do. 125 members do both.

Adding the numbers given for members who do aerobics or weight training will count the number who do both twice. Then the number who do either is ...

  (# of aerobics) + (# of weights) - (# of both)

  = 450 +200 -125 = 525

The probability that one of the 800 members is in this group is ...

  P(either) = 525/800 ≈ 0.66

<95141404393>

The variance of the given distribution is 0.9.

Calculating the anticipated value or distribution mean is the first step in determining a probability distribution's variance. The anticipated value is a weighted average of all potential outcomes, with each outcome's probability serving as the weight. The expected value can be expressed mathematically as follows:

[tex]E(X) =[/tex] Σ[tex][xi[/tex] × [tex]P(xi)][/tex]

where μ is the mean of the data.

Then, calculate μ:

μ [tex]= (1+2+3+4+5)/5[/tex]

[tex]=(15/5)[/tex]

[tex]= 3[/tex]

and replace this value and the values of xn and P(xn) into the formula for the variance, just as follow:

Hence, the variance of the given distribution is 0.9.

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Statistics and computer systems have a mutually beneficial relationship that has contributed to significant advancements in both fields. The use of statistics has played a crucial role in the development of computer systems, while computer systems have greatly impacted the field of statistics.

Statistics has been instrumental in the development of computer systems by providing a framework for data analysis and interpretation. Without statistics, computers would not be able to process and analyze large amounts of data efficiently. For example, statistical models and algorithms are used in machine learning and artificial intelligence to enable computers to learn and make decisions based on data. Additionally, statistics is used to test and validate the effectiveness of computer systems, ensuring that they are reliable and accurate.

On the other hand, computer systems have revolutionized the field of statistics by making data analysis faster and more accurate. With the availability of powerful computers, statisticians can analyze and interpret large datasets more quickly and accurately, leading to new insights and discoveries. Computer systems have also enabled the development of sophisticated statistical software and tools, making statistical analysis more accessible to a wider audience.

In conclusion, the relationship between statistics and computer systems has been symbiotic, with each field contributing to the growth and advancement of the other. The use of statistics has led to the development of sophisticated computer systems, while computer systems have greatly impacted the field of statistics, making data analysis faster and more accurate.
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It would take around 8 hours to bring each lake's pollution level down to 50% of its starting point, and around 22 hours to bring it down to 5%.

Assuming each lake is a separate tank with only clean water flowing in, we can use the exponential decay model [tex]$A = A_0e^{-kt}$[/tex], where $A$ is the amount of pollutant at time t, A₀ is the initial amount of pollutant, and k is the decay constant.

To find the time it would take to reduce the pollution level in each lake to 50% of its original level, we need to solve the equation [tex]$0.5A_0 = A_0e^{-kt}$[/tex] for t:

[tex]0.5A_0 &= A_0e^{-kt} \\frac{0.5A_0}{A_0} &= e^{-kt} \\ln\left(\frac{0.5A_0}{A_0}\right) &= -kt \\ln(0.5) &= -kt \t &= \frac{\ln(0.5)}{-k}\end{align*}[/tex]

To find the time it would take to reduce the pollution level in each lake to 5% of its original level, we need to solve the equation[tex]$0.05A_0 = A_0e^{-kt}$[/tex] for t:

[tex]0.05A_0 &= A_0e^{-kt} \\frac{0.05A_0}{A_0} &= e^{-kt} \\ln\left(\frac{0.05A_0}{A_0}\right) &= -kt \\ln(0.05) &= -kt \t &= \frac{\ln(0.05)}{-k}\end{align*}[/tex]

The decay constant $k$ can be found by using the given information that each lake is replaced by clean water every 8 hours. This means that the half-life of the pollutant is 8 hours, which gives us:

[tex]0.5A_0 &= A_0e^{-k(8)} \\ln(0.5) &= -8k \k &= -\frac{\ln(0.5)}{8} \approx 0.08664\end{align*}[/tex]

Substituting this value of k into the equations we derived earlier, we get:

[tex]t_{50} = \frac{\ln(0.5)}{-k} \approx 8.006 \text{ hours}[/tex]

[tex]t_{5} &= \frac{\ln(0.05)}{-k} \approx 22.133 \text{ hours}[/tex]

Therefore, it would take approximately 8 hours to reduce the pollution level in each lake to 50% of its original level, and approximately 22 hours to reduce it to 5% of its original level.

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Answer:turn right 45 degrees, then turn right another 45 degrees. flip the figure x-axis wise/horizontally

Step-by-step explanation:

We apply the Gram-Schmidt process to these vectors to find an orthonormal basis:

v1 = x1 = [3, -4, 1

To find a basis of the null space N(A), we need to find all vectors x such that Ax = 0, where 0 is the zero vector.

To do this, we set up the augmented matrix [A | 0] and row reduce:

[ 1 2 1 3 2 | 0 ]

[ 4 1 0 6 1 | 0 ]

[ 1 1 2 4 5 | 0 ]

R2 - 4R1 -> R2:

[ 1 2 1 3 2 | 0 ]

[ 0 -7 -4 6 -7 | 0 ]

[ 1 1 2 4 5 | 0 ]

R3 - R1 -> R3:

[ 1 2 1 3 2 | 0 ]

[ 0 -7 -4 6 -7 | 0 ]

[ 0 -1 1 1 3 | 0 ]

R2 / -7 -> R2:

[ 1 2 1 3 2 | 0 ]

[ 0 1 4/7 -6/7 1 | 0 ]

[ 0 -1 1 1 3 | 0 ]

R1 - 2R2 - R3 -> R1:

[ 0 0 0 0 0 | 0 ]

[ 0 1 4/7 -6/7 1 | 0 ]

[ 0 0 11/7 -1/7 1 | 0 ]

We can write the system of equations corresponding to this row echelon form as:

x2 + (4/7)x3 - (6/7)x4 + x5 = 0

(11/7)x3 - (1/7)x4 + x5 = 0

Solving for the variables in terms of the free variables x3, x4, and x5, we get:

x1 = -[(4/7)x3 - (6/7)x4 - x5]/2

x2 = -(4/7)x3 + (6/7)x4 - x5

x3 = x3 (free variable)

x4 = x4 (free variable)

x5 = x5 (free variable)

So the null space N(A) is the set of all vectors of the form:

x = [ -[(4/7)x3 - (6/7)x4 - x5]/2, -(4/7)x3 + (6/7)x4 - x5, x3, x4, x5 ]

To find an orthogonal basis for N(A), we can use the Gram-Schmidt process. Let's call the columns of A a1, a2, a3, a4, and a5.

First, we need to find a basis for N(A) by setting the free variables to 1 and the others to 0:

x1 = [3, -4, 1, 0, 0]

x2 = [-2, 3, 0, 1, 0]

x3 = [-2, 1, 0, 0, 1]

Next, we apply the Gram-Schmidt process to these vectors to find an orthonormal basis:

v1 = x1 = [3, -4, 1

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Answer:

15 Pi m

Step-by-step explanation:

arc ADC = 360 Degrees - 60 Degrees divided by 360 Degrees Multiplied by 2 Pi Multiplied by 9

= 5/6 Times 18 Pi

 =15 Pi m

To find the probability we do the following:

The probability that there really is a wolf when the boy cries wolf is 0.54.

Let A be the event that the boy cries wolf and B be the event that there is a wolf. We are given the following probabilities:

P(A|B) = 0.8 (the probability that the boy cries wolf when there is a wolf)

P(A|B') = 0.4 (the probability that the boy cries wolf when there is no wolf)

P(B) = 0.25 (the probability that there is a wolf)

We want to find P(B|A), the probability that there really is a wolf given that the boy cries wolf.

We can use Bayes' theorem to calculate this:

P(B|A) = P(A|B) * P(B) / [P(A|B) * P(B) + P(A|B') * P(B')]

Substituting the given values, we get:

P(B|A) = 0.8 * 0.25 / [0.8 * 0.25 + 0.4 * 0.75] = 0.54

Therefore, the probability that there really is a wolf when the boy cries wolf is 0.54.

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The value of cos S to the nearest hundredth is 0.54

Trigonometric Ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin(tetha) = opp/hyp

cos(tetha) = adj/hyp

tan(tetha) = opp/adj

Here in this triangle, the hypotenuse is 28

and the opposite to angle S is line TU

The adjascent is 15

therefore cos S = adj/hyp

= 15/28

= 0.54 ( nearest hundredth)

therefore the value of cos S is 0.54

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The housing supply function is given by h(I,R) = a log1 + bR + cR log1. The housing supply responds to changes in I with a rate of a, and to changes in R with a rate of b + c log1.

Based on the given equation, the housing supply (h) is determined by two variables: I and R. The equation shows that h is a function of R, with a log-linear relationship. The variable I only appears as a constant (a) in the equation, so changes in I do not directly affect the supply of houses.
On the other hand, changes in R (or OR, which is the same variable) do affect the supply of houses. Specifically, an increase in R leads to an increase in the supply of houses. The magnitude of this increase depends on the values of b and c in the equation.
To see this, we can take the partial derivative of h with respect to R:
dh/dR = b + cR/(ln(10))
This equation tells us how much the housing supply changes in response to a change in R. The derivative is positive (i.e. the supply increases) as long as c is positive. The larger c is, the greater the increase in supply for a given increase in R.
Therefore, the correct answer is:
b + cR/(ln(10))

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Answer:

To solve this problem, we can use a proportion.

Let x be the change of water in 12 weeks.

We know that in 7 weeks, the water decreased by 9 5/8 pints.

So, we can set up the proportion:

9 5/8 pints / 7 weeks = x / 12 weeks

To solve for x, we can cross-multiply and simplify:

9 5/8 pints * 12 weeks = 7 weeks * x

115 1/2 pints = 7 weeks * x

x = 115 1/2 pints / 7 weeks

x = 16 1/2 pints per 12 weeks

Therefore, the change of water in 12 weeks is 16 1/2 pints.

Step-by-step explanation: