Much of Ann’s investments are in Cilla Shipping. Ten years ago, Ann bought seven bonds issued by Cilla Shipping, each with a par value of $500. The bonds had a market rate of 95. 626. Ann also bought 125 shares of Cilla Shipping stock, which at the time sold for $28. 00 per share. Today, Cilla Shipping bonds have a market rate of 106. 384, and Cilla Shipping stock sells for $30. 65 per share. Which of Ann’s investments has increased in value more, and by how much? a. The value of Ann’s bonds has increased by $45. 28 more than the value of her stocks. B. The value of Ann’s bonds has increased by $22. 64 more than the value of her stocks. C. The value of Ann’s stocks has increased by $107. 81 more than the value of her bonds. D. The value of Ann’s stocks has increased by $8. 51 more than the value of her bonds.

The height of the scanner antenna is approximately 10.8 meters.

The distance from the point 24.0m away from the center of the house to the base of the antenna.

To do this, we can use the tangent function:
tan(18 degrees 10 minutes) = h / d
Where "d" is the distance from the point to the base of the antenna.
We can rearrange this equation to solve for "d":
d = h / tan(18 degrees 10 minutes)
Next, we need to find the distance from the point to the top of the antenna.

We can again use the tangent function:
tan(27 degrees 10 minutes) = (h + x) / d
Where "x" is the height of the bottom of the antenna above the ground.
We can rearrange this equation to solve for "x":
x = d * tan(27 degrees 10 minutes) - h
Now we can substitute the expression we found for "d" into the equation for "x":
x = (h / tan(18 degrees 10 minutes)) * tan(27 degrees 10 minutes) - h
We can simplify this equation:
x = h * (tan(27 degrees 10 minutes) / tan(18 degrees 10 minutes) - 1)
Finally, we know that the distance from the point to the top of the antenna is 24.0m, so:
24.0m = d + x
Substituting in the expressions we found for "d" and "x":
24.0m = h / tan(18 degrees 10 minutes) + h * (tan(27 degrees 10 minutes) / tan(18 degrees 10 minutes) - 1)
We can simplify this equation and solve for "h":
h = 24.0m / (tan(27 degrees 10 minutes) / tan(18 degrees 10 minutes) + 1)
Plugging this into a calculator or using trigonometric tables, we find that:
h ≈ 10.8 meters

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Question

A scanner antenna is on top of the center of a house. The angle of elevation from a point 24.0m from the center of the house to the top of the antenna is 27degrees and 10' and the angle of the elevation to the bottom of the antenna is 18degrees, and 10". Find the height of the antenna.

The area inside the ellipse is 8π. The area of the interior of the curve is 3π.

a) Using Green's theorem, we can compute the area inside the ellipse using the line integral around the boundary of the ellipse. Let C be the boundary of the ellipse. Then, by Green's theorem, the area inside the ellipse is given by A = (1/2) ∫(x dy - y dx) over C. Parameterizing the ellipse as x = 2 cos(t), y = 4 sin(t), where t varies from 0 to 2π, we have dx/dt = -2 sin(t) and dy/dt = 4 cos(t). Substituting these into the formula for the line integral and simplifying, we get A = 8π, so the area inside the ellipse is 8π.

b) To find a parametrization of the curve x^(2/3) + y^(2/3) = 4^(2/3), we can use x = 4 cos^3(t) and y = 4 sin^3(t), where t varies from 0 to 2π. Differentiating these expressions with respect to t, we get dx/dt = -12 sin^2(t) cos(t) and dy/dt = 12 sin(t) cos^2(t). Substituting these into the formula for the line integral, we get A = (3/2) ∫(sin^2(t) + cos^2(t)) dt = (3/2) ∫ dt = (3/2) * 2π = 3π, so the area of the interior of the curve is 3π.

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As the degrees of freedom for the numerator and denominator of a t-distribution get larger, the t-distribution approaches the standard normal distribution. This is known as the central limit theorem for the t-distribution.

In other words, as the sample size increases, the t-distribution becomes more and more similar to the standard normal distribution. This means that the distribution becomes more symmetric and bell-shaped, with less variability in the tails. The critical values of the t-distribution also become closer to those of the standard normal distribution as the sample size increases.

In practice, this means that for large sample sizes, we can use the standard normal distribution to make inferences about population means, even when the population standard deviation is unknown. This is because the t-distribution is a close approximation to the standard normal distribution when the sample size is large enough, and the properties of the two distributions are very similar.

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The correct option is d. Neither Type I nor Type II error.  The concepts of Type I and Type II errors, and to use appropriate methods and sample sizes to minimize the risk of making such errors.

To understand why, let's first define Type I and Type II errors. Type I error is rejecting a true null hypothesis, while Type II error is failing to reject a false null hypothesis.

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The given expression is 2s + 10 - 7s - 8 + 3s - 7. It has three different types of terms: 2s, 10, and -7s which are "like terms" because they have the same variable s with the same exponent 1.

This also goes with 3s.

There are also constant terms: -8 and -7.

Step-by-step explanation

To simplify this expression, we will combine the like terms and add the constant terms separately:

2s + 10 - 7s - 8 + 3s - 7

Collecting like terms:

2s - 7s + 3s + 10 - 8 - 7

Combine the like terms:

-2s - 5

Separating the constant terms:

2s - 7s + 3s - 2 - 5 = -2s - 7

Therefore, the simplified form of the given expression 2s + 10 - 7s - 8 + 3s - 7 is -2s - 7.

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The volume under the elliptic paraboloid [tex]z = 3x^2 + 6y^2[/tex] and over the rectangle R = [-4, 4] x [-1, 1] is 256/3 cubic units.

To calculate the volume under the elliptic paraboloid z = 3x^2 + 6y^2 and over the rectangle R = [-4, 4] x [-1, 1], we need to integrate the height of the paraboloid over the rectangle. That is, we need to evaluate the integral:

[tex]V =\int\limits\int\limitsR (3x^2 + 6y^2) dA[/tex]

where dA = dxdy is the area element.

We can evaluate this integral using iterated integrals as follows:

V = ∫[-1,1] ∫ [tex][-4,4] (3x^2 + 6y^2)[/tex] dxdy

= ∫[-1,1] [ [tex](x^3 + 2y^2x)[/tex] from x=-4 to x=4] dy

= ∫[-1,1] (128 + 16[tex]y^2[/tex]) dy

= [128y + (16/3)[tex]y^3[/tex]] from y=-1 to y=1

= 256/3

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The density function of the random variable |X| is f_Y(y) = 1 for 0 ≤ y ≤ 1.

a) Since X is uniformly distributed over (-1,1), the probability density function of X is f(x) = 1/2 for -1 < x < 1, and 0 otherwise. Therefore, the probability of the event {|X| > 1/2} can be computed as follows:

P{|X| > 1/2} = P{X < -1/2 or X > 1/2}

= P{X < -1/2} + P{X > 1/2}

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

= 1/4 + 1/4

= 1/2

Therefore, P{|X| > 1/2} = 1/2.

b) To find the density function of the random variable |X|, we can use the transformation method. Let Y = |X|. Then, for y > 0, we have:

F_Y(y) = P{Y ≤ y} = P{|X| ≤ y} = P{-y ≤ X ≤ y}

Since X is uniformly distributed over (-1,1), we have:

F_Y(y) = P{-y ≤ X ≤ y} = (1/2)(y - (-y)) = y

Therefore, the cumulative distribution function of Y is F_Y(y) = y for 0 ≤ y ≤ 1.

To find the density function of Y, we differentiate F_Y(y) with respect to y to obtain:

f_Y(y) = dF_Y(y)/dy = 1 for 0 ≤ y ≤ 1

Therefore, the density function of the random variable |X| is f_Y(y) = 1 for 0 ≤ y ≤ 1.

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Mrs. Falkener has written a company report every 3 months for the last 6 years, resulting in a total of 24 reports. Among these reports, 2/3 of them show the company earning more money than it spends. Therefore, 1/3 of the reports, or 8 reports, show the company spending more money than it earns.

In 6 years, there are 12 quarters since there are 4 quarters in a year. Mrs. Falkener has written a company report every 3 months, which means there are 12 * 3 = 36 periods in total. However, since each report covers a 3-month period, the total number of reports is 36 / 3 = 12.

Given that 2/3 of the reports show the company earning more money than it spends, we can calculate the number of reports showing the company spending more money than it earns. Since 2/3 of the reports represent the earnings being greater, the remaining 1/3 represents the expenses being greater. Therefore, 1/3 of 12 reports is 12 * (1/3) = 4 reports.

In conclusion, among the 24 company reports written by Mrs. Falkener in the last 6 years, 2/3 of them, or 16 reports, show the company earning more money than it spends. The remaining 1/3, or 8 reports, show the company spending more money than it earns.

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We have shown that [tex]cov(x_1, x_1) = v(x_1) = \sigma^2_1(x 1 ,x 1 ).[/tex]

To show that [tex]cov(x_1, x_1) = v(x_1) = \sigma^2_1(x 1 ,x 1 )[/tex], we need to first understand what each of these terms means:

[tex]cov(x_1, x_1)[/tex] represents the covariance between the random variable x_1 and itself. In other words, it is the measure of how two instances of x_1 vary together.

v(x_1) represents the variance of x_1. This is a measure of how much x_1 varies on its own, regardless of any other random variable.

[tex]\sigma^2_1(x 1 ,x 1 )[/tex]represents the second moment of x_1. This is the expected value of the squared deviation of x_1 from its mean.

Now, let's show that [tex]cov(x_1, x_1) = v(x_1) = \sigma^2_1(x 1 ,x 1 ):[/tex]

We know that the covariance between any random variable and itself is simply the variance of that random variable. Mathematically, we can write:

[tex]cov(x_1, x_1) = E[(x_1 - E[x_1])^2] - E[x_1 - E[x_1]]^2\\ = E[(x_1 - E[x_1])^2]\\ = v(x_1)[/tex]

Therefore, [tex]cov(x_1, x_1) = v(x_1).[/tex]

Similarly, we know that the variance of a random variable can be expressed as the second moment of that random variable minus the square of its mean. Mathematically, we can write:

[tex]v(x_1) = E[(x_1 - E[x_1])^2]\\ = E[x_1^2 - 2\times x_1\times E[x_1] + E[x_1]^2]\\ = E[x_1^2] - 2\times E[x_1]\times E[x_1] + E[x_1]^2\\ = E[x_1^2] - E[x_1]^2\\ = \sigma^2_1(x 1 ,x 1 )[/tex]

Therefore, [tex]v(x_1) = \sigma^2_1(x 1 ,x 1 ).[/tex]

Thus, we have shown that [tex]cov(x_1, x_1) = v(x_1) = \sigma^2_1(x 1 ,x 1 ).[/tex]

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the answer is: f · dr = -30

To find f · dr for the line c from (3, 2, -2) to (6, 1, 7), we first need to parametrize the line in terms of a vector function r(t). We can do this as follows:

r(t) = <3, 2, -2> + t<3, -1, 9>

This gives us a vector function that describes all the points on the line c as t varies.

Next, we need to calculate f · dr for this line. We can use the formula:

f · dr = ∫c f · dr

where the integral is taken over the line c. We can evaluate this integral by substituting r(t) for dr and evaluating the dot product:

f · dr = ∫c f · dr = ∫[3,6] f(r(t)) · r'(t) dt

where [3,6] is the interval of values for t that correspond to the endpoints of the line c. We can evaluate the dot product f(r(t)) · r'(t) as follows:

f(r(t)) · r'(t) = <5y, 2, -1> · <3, -1, 9>

= 15y - 2 - 9

= 15y - 11

where we used the given expression for f and the derivative of r(t), which is r'(t) = <3, -1, 9>.

Plugging this dot product back into the integral, we get:

f · dr = ∫[3,6] f(r(t)) · r'(t) dt

= ∫[3,6] (15y - 11) dt

To evaluate this integral, we need to express y in terms of t. We can do this by using the equation for the y-component of r(t):

y = 2 - t/3

Substituting this into the integral, we get:

f · dr = ∫[3,6] (15(2 - t/3) - 11) dt

= ∫[3,6] (19 - 5t) dt

= [(19t - 5t^2/2)]|[3,6]

= (57/2 - 117/2)

= -30

Therefore, the answer is:

f · dr = -30

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One regular price ticket to the town carnival costs $12.75 using equation.

Let's assume the cost of one regular price ticket is represented by the variable 'x'.

With the coupon for $20 off, the total bill for your group to get into the carnival is $31. Since there are four people in your group, the equation representing the total bill is:

4x - $20 = $31

To solve for 'x', we'll isolate it on one side of the equation:

4x = $31 + $20

4x = $51

Now, divide both sides of the equation by 4 to solve for 'x':

x = $51 / 4

x = $12.75

Therefore, one regular price ticket costs $12.75.

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Let's use x to represent the number of adoptions during the first week. In this problem  there were 10 more adoptions during the second week than during the first week. This means that the number of adoptions during the second week was x + 10.

During the third week, there were twice as many adoptions as during the first week. This means that the number of adoptions during the third week was 2x.

We are given that the total number of adoptions during the three weeks was at least 50. This means that the sum of the number of adoptions during the three weeks is greater than or equal to 50. We can write this as x + (x + 10) + 2x ≥ 50

Simplifying this inequality, we get:

4x + 10 ≥ 50

4x ≥ 40

x ≥ 10

Therefore, the possible values of x, the number of adoptions from the animal rescue group during the first week, are all numbers greater than or equal to 10. We can represent this as x ≥ 10

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The derivative  y′y′ at the point [tex]y'(-sqrt(e^(8-64))) = 7e^84/4097.[/tex]

To find the derivative of y with respect to x, we need to use the chain rule and the partial derivative of y with respect to x and y.

Let's begin by taking the partial derivative of y with respect to x:

[tex]∂y/∂x = 2x/(x^2 + y^2)[/tex]

Now, let's take the partial derivative of y with respect to y:

[tex]∂y/∂y = 2y/(x^2 + y^2)[/tex]Using the chain rule, the derivative of y with respect to x can be found as:

[tex]dy/dx = (dy/dt) / (dx/dt)[/tex], where t is a parameter such that x = f(t) and y = g(t).

Let's set[tex]t = x^2 + y^2[/tex], then we have:

[tex]dy/dt = 1/t * (∂y/∂x + ∂y/∂y)[/tex]

[tex]= 1/(x^2 + y^2) * (2x/(x^2 + y^2) + 2y/(x^2 + y^2))[/tex]

[tex]= 2(x+y)/(x^2 + y^2)^2[/tex]

dx/dt = 2x

Therefore, the derivative of y with respect to x is:

dy/dx = (dy/dt) / (dx/dt)

[tex]= (2(x+y)/(x^2 + y^2)^2) / 2x[/tex]

[tex]= (x+y)/(x^2 + y^2)^2[/tex]

Now, we can evaluate the derivative at the point [tex](-sqrt(e^(8-64)), 8)[/tex]:

[tex]x = -sqrt(e^(8-64)) = -sqrt(e^-56) = -1/e^28[/tex]

y = 8

Therefore, we have:

[tex]dy/dx = (x+y)/(x^2 + y^2)^2[/tex]

[tex]= (-1/e^28 + 8)/(1/e^56 + 64)^2[/tex]

[tex]= (-1/e^28 + 8)/(1/e^112 + 4096)[/tex]

We can simplify the denominator by using a common denominator:

[tex]1/e^112 + 4096 = 4096/e^112 + 1/e^112 = (4097/e^112)[/tex]

So, the derivative at the point (-sqrt(e^(8-64)), 8) is:

[tex]dy/dx = (-1/e^28 + 8)/(4097/e^112)[/tex]

[tex]= (-e^84 + 8e^84)/4097[/tex]

[tex]= (8e^84 - e^84)/4097[/tex]

[tex]= 7e^84/4097[/tex]

Therefore,the derivative  y′y′ at the point [tex]y'(-sqrt(e^(8-64))) = 7e^84/4097.[/tex]

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To determine the derivative y′ of y=ln(x2+y2) at the point (−√e8−64,8)(−e8−64,8), we first need to find the partial derivatives of y with respect to x and y. Using the chain rule, we get: ∂y/∂x = 2x/(x2+y2) ∂y/∂y = 2y/(x2+y2)
Then, we can find the derivative y′ using the formula: y′ = (∂y/∂x) * x' + (∂y/∂y) * y'

Therefore, the derivative y′ at the point (−√e8−64,8)(−e8−64,8) is (8-√e8−64)/(32-e8).
Given the function y = ln(x^2 + y^2), we want to find the derivative y′ at the point (-√(e^8 - 64), 8).
1. Differentiate the function with respect to x using the chain rule:
y′ = (1 / (x^2 + y^2)) * (2x + 2yy′)
2. Solve for y′:
y′(1 - y^2) = 2x
y′ = 2x / (1 - y^2)
3. Substitute the given point into the expression for y′:
y′(-√(e^8 - 64)) = 2(-√(e^8 - 64)) / (1 - 8^2)
4. Calculate the derivative:
y′(-√(e^8 - 64)) = -2√(e^8 - 64) / -63
Thus, the derivative y′ at the point (-√(e^8 - 64), 8) is y′(-√(e^8 - 64)) = 2√(e^8 - 64) / 63.

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Tracy works at North College as a math teacher. She will be paid $900 for each credit hour she teaches. During the course of her first year of teaching, she would teach a total of 50 credit hours.

The college expects her to work a minimum of 170 days (and less and her salary would be reduced) and 8 hours each day. Her gross monthly income is $12,150.

The total number of hours Tracy works is given by;

Total number of hours Tracy works = Number of days she works in a year x Number of hours per day.

Number of days she works in a year = 170Number of hours per day = 8.

Total number of hours Tracy works = 170 × 8

= 1360.

Each credit hour Tracy teaches is paid for $900.

Therefore, for all the credit hours she teaches in a year, she will be paid for $900 × 50 = $45,000.In order to get Tracy's monthly gross income, we need to divide the total amount of money Tracy will be paid in a year by 12 months.$45,000 ÷ 12 = $3750.

Then, we can calculate the gross monthly income of Tracy by adding her salary per month and her total hourly work salary. The total hourly work salary is equal to the product of the total number of hours Tracy works and the amount she is paid per hour which is $900. Therefore, her monthly gross income will be:$3750 + ($900 × 1360) = $12,150. Answer: $12,150.

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The most probable number of heads becomes more and more likely as the number of tosses increases.

Let's denote the probability of observing tails as q (which is 1/2 for a fair coin). Then the probability of observing exactly n heads in 4n tosses is given by the binomial distribution:

p(n) = (4n choose n) * (1/2)^(4n)

where (4n choose n) is the number of ways to choose n heads out of 4n tosses. We can express this in terms of the most probable number of heads, which is 2n:

p(n) = (4n choose n) * (1/2)^(4n) * (2^(2n))/(2^(2n))

= (4n choose 2n) * (1/4)^n * 2^(2n)

where we used the identity (4n choose n) = (4n choose 2n) * (1/4)^n * 2^(2n). This identity follows from the fact that we can choose 2n heads out of 4n tosses by first choosing n heads out of the first 2n tosses, and then choosing the remaining n heads out of the last 2n tosses.

Now we can express the ratio p(n)/p(2n) as:

p(n)/p(2n) = [(4n choose 2n) * (1/4)^n * 2^(2n)] / [(4n choose 4n) * (1/4)^(2n) * 2^(4n)]

= [(4n)! / (2n)!^2 / 2^(2n)] / [(4n)! / (4n)! / 2^(4n)]

= [(2n)! / (n!)^2] / 2^(2n)

= (2n-1)!! / (n!)^2 / 2^n

where (2n-1)!! is the double factorial of 2n-1. Note that (2n-1)!! is the product of all odd integers from 1 to 2n-1, which is always less than or equal to the product of all integers from 1 to n, which is n!. Therefore,

p(n)/p(2n) = (2n-1)!! / (n!)^2 / 2^n <= n! / (n!)^2 / 2^n = 1/(n * 2^n)

As n increases, the denominator n * 2^n grows much faster than the numerator (2n-1)!!, so the ratio p(n)/p(2n) approaches zero. This means that the probability of observing n heads relative to the most probable number of heads becomes vanishingly small as n increases, which is consistent with the intuition that the most probable number of heads becomes more and more likely as the number of tosses increases.

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To calculate the total number of ways in which the committee of 3 women and 2 men can be formed from a pool of 11 women and 7 men, we can use the combination formula. The combination formula is C(n, r) = n! / (r! * (n-r)!) where n is the total number of items and r is the number of items to choose.

First, we'll calculate the number of ways to select 3 women from a pool of 11 women:
C(11, 3) = 11! / (3! * (11-3)!)
C(11, 3) = 11! / (3! * 8!)
C(11, 3) = 165

Next, we'll calculate the number of ways to select 2 men from a pool of 7 men:
C(7, 2) = 7! / (2! * (7-2)!)
C(7, 2) = 7! / (2! * 5!)
C(7, 2) = 21

Now, to find the total number of ways in which the committee can be formed, we'll multiply the number of ways to choose women and the number of ways to choose men:
Total number of ways = 165 (ways to choose women) * 21 (ways to choose men)
Total number of ways = 3,465

Therefore, the total number of ways in which the committee can be formed is 3,465 (Option A).

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A baker purchased 14 lb of wheat flour and 11 lb of rye flour for a total cost of 13.75 dollars. A second purchase, at the same prices, included 12 lb of wheat flour and 13 lb of rye flour.

The cost of the second purchase was 13.75 dollars. We need to find the cost per pound of wheat flour and of the rye flour. Let x and y be the cost per pound of wheat flour and rye flour, respectively. According to the given conditions, we have the following system of equations:14x + 11y = 13.75 (1)12x + 13y = 13.75 (2)Using elimination method, we can find the value of x and y as follows:

Multiplying equation (1) by 13 and equation (2) by 11, we get:182x + 143y = 178.75 (3)132x + 143y = 151.25 (4)Subtracting equation (4) from equation (3), we get:50x = - 27.5=> x = - 27.5/50= - 0.55 centsTherefore, the cost per pound of wheat flour is 55 cents.

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The coefficient of x²y¹⁵ in the expansion of (5x² + 2y³)⁶ is 192.

-To find the coefficient of x²y¹⁵ in the expansion of (5x² + 2y³)⁶, you can use the binomial theorem. The binomial theorem states that [tex](a + b)^n[/tex] = Σ [tex][C(n, k) a^{n-k} b^k][/tex], where k goes from 0 to n, and C(n, k) represents the number of combinations of n things taken k at a time.

-Here, a = 5x², b = 2y³, and n = 6. We want to find the term with x²y¹⁵, which means we need a^(n-k) to be x² and [tex]b^k[/tex] to be y¹⁵.

-First, let's find the appropriate value of k:
[tex](5x^{2}) ^({6-k}) =x^{2} \\ 6-k = 1 \\k=5[/tex]

-Now, let's find the term with x²y¹⁵:
[tex]C(6,5) (5x^{2} )^{6-5} (2y^{3})^{5}[/tex]
= C(6, 5) (5x²)¹ (2y³)⁵
= [tex]\frac{6!}{5! 1!}  (5x²)  (32y¹⁵)[/tex]
= (6)  (5x²)  (32y¹⁵)
= 192x²y¹⁵

So, the coefficient of x²y¹⁵ in the expansion of (5x² + 2y³)⁶ is 192.

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The possible numbers of lawns the company could have mowed are 12 and 80.

A landscaper earns $30 for each lawn her company mows, but she pays $210 per day in salary to her employees. If her company made more than $150 profit from mowing lawns in a 7-day week, we can use the inequality equation below to solve for the possible numbers of lawns the company could have mowed:7(30x) - 210(7) > 150where x is the number of lawns the company mowed. The left side of the inequality represents the total income the company earned from mowing lawns, while the right side represents the total cost, which is the weekly salary plus the $150 profit we want to exceed. Simplifying the inequality, we get:210x > 5402100 > x. Since the number of lawns has to be a whole number, the possible numbers of lawns the company could have mowed are 12 and 80.

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The statement ''the q test is a mathematically simpler but more limited test for outliers than is the grubbs test'' is correct becauae the Q test is a simpler but less powerful test for detecting outliers compared to the Grubbs test.

The Q test and Grubbs test are statistical tests used to detect outliers in a dataset. The Q test is a simpler method that involves calculating the range of the data and comparing the distance of the suspected outlier from the mean to the range.

If the distance is greater than a certain critical value (Qcrit), the data point is considered an outlier. The Grubbs test, on the other hand, is a more powerful method that involves calculating the Z-score of the suspected outlier and comparing it to a critical value (Gcrit) based on the size of the dataset.

If the Z-score is greater than Gcrit, the data point is considered an outlier. While the Q test is easier to calculate, it is less powerful and may miss some outliers that the Grubbs test would detect.

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The rectangular equation given is x + 7√(x² + y²) = x² + y², which can be converted to the polar equation r = 7 - cos(θ).

Using the trigonometric identity cos(θ) = x/r, we can write:

r = 7 - x/r

Multiplying both sides by r, we get:

r² = 7r - x

Using the polar to rectangular conversion formulae x = r cos(θ) and y = r sin(θ), we can express r in terms of x and y:

r² = x² + y²

Substituting r² = x² + y² into the previous equation, we get:

x² + y² = 7r - x

Substituting cos(θ) = x/r, we can write:

x = r cos(θ)

Substituting this into the previous equation, we get:

x² + y² = 7r - r cos(θ)

Simplifying, we get:

x² + y² = 7√(x² + y²) - x

Rearranging, we get:

x + 7√(x² + y²) = x² + y²

This is the rectangular form of the polar equation r = 7 - cos(θ).

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The MAD is approximately 15.4. The MAD tells us that on average, the forecast errors are about 15.4 units away from the mean forecast error.

The Mean Absolute Deviation (MAD) is a measure of the variability of a set of data. It represents the average distance of the data points from the mean of the data set.

To calculate the MAD, we need to first find the mean of the forecast errors. The mean is the sum of the forecast errors divided by the number of errors:

Mean = (-22 - 10 + 15)/3 = -4/3

Next, we find the absolute deviation of each error by subtracting the mean from each error and taking the absolute value:

|-22 - (-4/3)| = 64/3

|-10 - (-4/3)| = 26/3

|15 - (-4/3)| = 49/3

Then, we find the average of these absolute deviations to get the MAD:

MAD = (64/3 + 26/3 + 49/3)/3 = 139/9

Therefore, the MAD is approximately 15.4. The MAD tells us that on average, the forecast errors are about 15.4 units away from the mean forecast error.

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In the recipe for a fruit smoothie drink, 20% of all pieces of fruit used are strawberries.

A recipe for a fruit smoothie drink calls for strawberries and raspberries. The ratio of strawberries to raspberries in the drink is 5:20.

The ratio of strawberries to raspberries in the drink is 5:20, i.e., the total parts are 5 + 20 = 25.

The fraction representing strawberries is: 5/25 = 1/5.

Now we have to convert this fraction to percent form.

This can be done using the following formula:

Percent = (Fraction × 100)%

Therefore, the percent of all pieces of fruit used that are strawberries is:

1/5 × 100% = 20%

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What is the name of a regular polygon with 45 sides?

A regular polygon with 45 sides is called a "45-gon."

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a. 1 multiplied by 0 with a bar over it is also equal to 0. b. the final value of the expression is 0. c.  0 with a bar over it multiplied by 0 is also equal to 0. d. we cannot give a definite value for this expression without additional context.

a) The value of the expression 1⋅0¯ is 0.

When we multiply any number by 0, the result is always 0. Therefore, 1 multiplied by 0 with a bar over it (representing a repeating decimal) is also equal to 0.

b) The value of the expression 1 1¯ is 0.

When a number has a bar over it, it represents a repeating decimal. Therefore, 1.111... is the same as the fraction 10/9. Subtracting 1 from 10/9 gives us 1/9, which is equal to 0.111... (or 0¯). Therefore, the value of 1 1¯ is 1 + 1/9, which simplifies to 10/9, or 1.111.... Subtracting 1 from this gives us 1/9, which is equal to 0.111... (or 0¯), so the final value of the expression is 0.

c) The value of the expression 0¯⋅0 is 0.

When we multiply any number by 0, the result is always 0. Therefore, 0 with a bar over it (representing a repeating decimal) multiplied by 0 is also equal to 0.

d) The value of the expression (1 0¯¯¯¯¯¯¯¯) is undefined.

The notation (1 0¯¯¯¯¯¯¯¯) is ambiguous and could be interpreted in different ways. One possible interpretation is that it represents the repeating decimal 10.999..., which is equivalent to the fraction 109/99. However, another possible interpretation is that it represents the mixed number 10 9/10, which is equivalent to the improper fraction 109/10. Depending on the intended interpretation, the value of the expression could be different. Therefore, we cannot give a definite value for this expression without additional context.

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Converting to Cartesian coordinates is one way to find alternate descriptions for (r,0) (-1,π) in polar coordinates.

Here,

When looking for alternate descriptions for the polar coordinates (r,0) (-1,π), converting them to Cartesian coordinates is one way to do it.

However, a better method is to remember the steps to graph a point in polar coordinates.

If r is greater than zero, start along the positive z-axis, and if r is less than zero, start along the negative z-axis.

Then, rotate counterclockwise if θ is greater than zero, and rotate clockwise if θ is less than zero.

By following these steps, alternate descriptions for (r,0) (-1,π) in polar coordinates can be determined without having to convert them to Cartesian coordinates.

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Adding these amounts, we get : $33.90 + $25.96 = $59.86 Since this amount is greater than $50, we see that the inequality holds for this example.

To represent the given scenario as an inequality, we need to use the following expression: Total amount spent on peppers + Total amount spent on corn < $50We are given that Peppers cost $16.95 per pound, and the quantity of peppers is 'p' pounds.  

So the total amount spent on peppers is given by:16.95 × p

For corn, we are given that it costs $6.49 per pound, and the quantity of corn is 'c' pounds, so the total amount spent on corn is given by:6.49 × c .

Using these values, we can write the inequality as follows:16.95p + 6.49c < 50This is the required inequality. Let's verify this inequality using an example .

Suppose Rebecca buys 2 pounds of peppers and 4 pounds of corn. Then, the total amount spent on peppers is:16.95 × 2 = $33.90and the total amount spent on corn is:6.49 × 4 = $25.96.

Adding these amounts, we get:$33.90 + $25.96 = $59.86 Since this amount is greater than $50, we see that the inequality holds for this example.

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The number is 7800.

Counting is the process of expressing the number of elements or objects that are given.

Counting numbers include natural numbers which can be counted and which are always positive.

Counting is essential in day-to-day life because we need to count the number of hours, the days, money, and so on.

Numbers can be counted and written in words like one, two, three, four, and so on. They can be counted in order and backward too. Sometimes, we use skip counting, reverse counting, counting by 2s, counting by 5s, and many more.

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Two news websites have opened their memberships to the public, and their growth rates between Day 10 and Day 20 are compared to determine which website will have more members after 50 days.

To calculate the average rate of change for each website, we need to determine the difference in the number of members between Day 10 and Day 20 and divide it by the number of days in that period. Let's say Website A had 200 members on Day 10 and 500 members on Day 20, while Website B had 300 members on Day 10 and 600 members on Day 20.

For Website A, the rate of change is (500 - 200) / 10 = 30 members per day.

For Website B, the rate of change is (600 - 300) / 10 = 30 members per day.

Both websites have the same average rate of change, indicating that they are growing at the same pace during this period. To predict the number of members after 50 days, we can assume that the average rate of change will remain constant. Thus, after 50 days, Website A would have an estimated 200 + (30 * 50) = 1,700 members, and Website B would have an estimated 300 + (30 * 50) = 1,800 members.

Based on this calculation, Website B is projected to have more members after 50 days. However, it's important to note that this analysis assumes a constant growth rate, which might not necessarily hold true in the long run. Other factors such as website popularity, marketing efforts, and user retention can also influence the final number of members.

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Answer:

[tex]\frac{dy}{dx}[/tex] = ([tex]\frac{dy}{dt}[/tex]) / ([tex]\frac{dx}{dt}[/tex]) = (36[tex]e^{4}[/tex]) / (-5) = -7.2[tex]e^{4}[/tex]

Step-by-step explanation:

To find the slope of the tangent line, we need to find [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex], and then evaluate them at t=1 and compute [tex]\frac{dy}{dx}[/tex].

We have:

x(t) = 2[tex]t^{3}[/tex]  - 8[tex]t^{2}[/tex] + 5t + 3

Taking the derivative with respect to t, we get:

[tex]\frac{dx}{dt}[/tex] = 6[tex]t^{2}[/tex] - 16t + 5

Similarly,

y(t) = 9[tex]e^{4t-4}[/tex]

Taking the derivative with respect to t, we get:

[tex]\frac{dy}{dt}[/tex] = 36[tex]e^{4t-4}[/tex]

Now, we evaluate [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex] at t=1:

[tex]\frac{dx}{dt}[/tex]= [tex]6(1)^{2}[/tex] - 16(1) + 5 = -5

[tex]\frac{dy}{dt}[/tex] = 36[tex]e^{4}[/tex](4(1)) = 36[tex]e^{4}[/tex]

So the slope of the tangent line at t=1 is:

[tex]\frac{dy}{dx}[/tex]= ([tex]\frac{dy}{dt}[/tex]) / ([tex]\frac{dx}{dt}[/tex]) = (36[tex]e^{4}[/tex] / (-5) = -7.2[tex]e^{4}[/tex]

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