Need some help with Algebra!

Today's my last day to get missing work in and I currently have an F in Math and I'm just trying to get it up so I don't fail. Help me out? Heres the picture:

To calculate the expected value of the insurance policy, we multiply the potential outcomes by their respective probabilities and sum them up. In this case, the insurance company sells a $250,000 policy to a 20-year-old female for $350.

The probability that she survives the year is given as 0.99654. The expected value represents the average value the insurance company can expect to receive from selling this policy.

The potential outcomes of the policy are two-fold: either the female survives the year, or she does not. If she survives, the insurance company does not have to pay out any benefits. However, if she does not survive, the company has to pay the policy amount of $250,000.

To calculate the expected value, we multiply each outcome by its corresponding probability and sum them up:

Expected value = (Probability of survival) * (Amount received) + (Probability of non-survival) * (Amount paid out)

Expected value = (0.99654) * (0) + (1 - 0.99654) * (250,000)

Simplifying the calculation, we find:

Expected value = 0 + (0.00346) * (250,000) = $865

Interpreting the expected value, on average, the insurance company can expect to receive $865 for selling this policy to a 20-year-old female.

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The time complexity of the binary search is c) O(NLogN)

An effective searching technique that uses sorted arrays is a binary search. The search space is continually divided in half until the target element is located or the search space is empty, using a divide-and-conquer strategy. It makes the necessary adjustments to the search bounds at each stage by comparing the target element to the centre element of the active search area.

The divide-and-conquer strategy is logarithmic, hence binary search has an O(log n) time complexity. This implies that number of operations needed to discover the target element rises at a logarithmic rate as input size increases. When compared to linear search algorithms with O(n) complexity, where each element must be evaluated sequentially, binary search is significantly more efficient for big arrays due to its logarithmic time complexity.

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The capitalized cost of the equipment is $29,900 (option a).

To determine the capitalized cost of the equipment, we need to sum up all the costs associated with its acquisition and installation.

The costs include:

- Purchase price: $27,000

- Shipping cost: $400

- Installation cost: $1,000

- Increased insurance cost: $1,500

To find the capitalized cost, we add up these costs:

Capitalized cost = Purchase price + Shipping cost + Installation cost + Increased insurance cost

              = $27,000 + $400 + $1,000 + $1,500

              = $29,900

Therefore, the capitalized cost of the equipment is $29,900 (option a).

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Answer:

Kristy makes 16 per hour.

Step-by-step explanation: Divded 48 by 3.

Answer:

581

Step-by-step explanation:

(a) The sample size needed is 1,498 if the researcher uses the prior proportion of 0.669.

n = 1.497.87 or 1,498

(b)  The researcher should use a sample size of 1,692 if the proportion is unknown.

n = 1,691.06 or 1,692

When a researcher is designing a sampling study to measure a population parameter, like the proportion, the sample size must be determined. The key to calculating sample size is the margin of error the researcher desires. The smaller the margin of error, the larger the sample size needed.

(a) The sample size needed is 1,498 if the researcher uses the prior proportion of 0.669.

To find sample size for the proportion use the equation:

[tex]n=\frac{Z^2\pi (1-\pi )}{e^2}[/tex]

Where:

Substituting values:

[tex]n=\frac{1.6449^2*0.50*0.331}{0.02^2}[/tex]

n = 1.497.87 or 1,498

(b) The researcher should use a sample size of 1,692 if the proportion is unknown.

Use the same equation as in part (a). However, to maximize [tex]\pi (1-\pi )[/tex] use 0.5 for [tex]\pi[/tex].

Substituting the values:

[tex]n=\frac{1.6449^2*0.50*0.500}{0.02^2}[/tex]

n = 1,691.06 or 1,692

Always round up because it is impossible to have a fraction of an observation.

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Answer:

Steps:

tan(50.1°) = x/5

1.19.. = x/5

x = 5 × 1.19 ..

x = 5.97.. ≈ 5.98

Answer:

Solution : Let Tn=0. Then,21+(n-1)×(-3)=0⇒n=8. So, 8th term is zero.

Step-by-step explanation:

Answer:

[tex] x=\sqrt{10} [/tex]

[tex] y = \sqrt {14} [/tex]

[tex] z = \sqrt {35} [/tex]

Step-by-step explanation:

By geometric mean theorem:

[tex] x=\sqrt{2\times 5} [/tex]

[tex] x=\sqrt{10} [/tex]

By Pythagoras Theorem:

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

[tex] y^2 = (\sqrt{10}) ^2 + 2^2 [/tex]

[tex] y^2 = 10 + 4 [/tex]

[tex] y^2 = 14 [/tex]

[tex] y = \sqrt {14} [/tex]

Again by Pythagoras Theorem:

[tex] z^2 = x^2 + 5^2 [/tex]

[tex] z^2 = (\sqrt{10}) ^2 + 5^2 [/tex]

[tex] z^2 = 10 + 25 [/tex]

[tex] z^2 = 35 [/tex]

[tex] z = \sqrt {35} [/tex]

Answer:

1.2

Step-by-step explanation:

You have to subtract 7.2 - 6 = 1.2

East and south direction motion are perpendicular to each other. The hiker went approx 4 miles in south.

East or west are perpendicular (forming 90° ) to north or south directions.

North is 180° to south, and east is 180 degrees to west.

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]

where |AB| = length of line segment AB.

Considering the diagram attached below, we deduce that the triangle ABC is right angled triangles, and thus, its sides' lengths will follow Pythagoras theorem.

The movement the hiker did in the south direction is the length of the side BC, denoted by |BC| = x miles (assume)

Then, by applying Pythagoras theorem, we get:

[tex]|AC|^2 = |AB|^2 + |BC|^2\\(7.2)^2 = 6^2 + x^2\\51.84 = 36 + x^2\\\\\text{Subtracting 36 from both the sides and taking root, we get}\\\\x = \sqrt{15.86} \approx 3.98 \approx 4 \: \rm miles[/tex]

(took positive root since distance is a non-negative quantity)

Thus, the hiker went approx 4 miles in south.

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The system x = tx is Lyapunov stable.

To determine if the system x = tx is Lyapunov stable, we need to check if it satisfies the Lyapunov stability criterion. The criterion states that for every ε > 0, there exists a δ > 0 such that if ||x(0) - xe|| < δ, then ||x(t) - xe|| < ε for all t > 0.

Let's analyze the given system x = tx:

Taking the derivative of x with respect to t, we get:

dx/dt = x

This is a first-order linear ordinary differential equation.

The general solution to this equation is:

x(t) = C×[tex]e^{t}[/tex]

where C is the constant of integration.

Now, let's consider the initial condition x(0) = x0, where x0 is some constant.

Using this initial condition, we can solve for the constant C:

x(0) = C × e⁰

x0 = C

So, the specific solution to the initial value problem is:

x(t) = x₀ × [tex]e^{t}[/tex]

Now, we can check if the system is Lyapunov stable.

For any ε > 0, let's consider the case where δ = ε. Then, if ||x(0) - xe|| < δ, we have:

||x(0) - xe|| = ||x₀ - x₀ × [tex]e^{t}[/tex]|| = ||x₀(1 - [tex]e^{t}[/tex])||

Since e^t > 1 for t > 0, we can rewrite the expression as

||x₀(1 - [tex]e^{t}[/tex])|| = x₀ × ||[tex]e^{t}[/tex] - 1||

Now, we want to show that ||x(t) - xe|| < ε for all t > 0. We have

||x(t) - xe|| = ||x₀ × [tex]e^{t}[/tex] - x₀ × [tex]e^{t}[/tex]|| = 0

Since ||x(t) - xe|| = 0 for all t > 0, it satisfies the condition ||x(t) - xe|| < ε.

Therefore, the given system x = tx is Lyapunov stable.

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The question is incomplete the complete question is :

Determine whether the origin of the following system is Lyapunov stable: x= tx. Use Lyapunov's stability criterion .

Answer:

3150 ft³ or 89.1981 m³ or 116.6667 yd³

Step-by-step explanation:

The value of the integral ∫(3z + 1) / (z^2 + 2) dz, where C is the constant is ln|z^2 + 2| + 3/√2 arctan(z/√2) + C.

To evaluate the given integral, we first need to break it down using partial fraction decomposition. Therefore, we have:

(3z + 1) / (z^2 + 2) = (Az + B) / (z^2 + 2)

Multiplying both sides by (z^2 + 2), we get:

3z + 1 = (Az + B)

We now need to find the values of A and B. Setting z = 0, we get:

1 = B

Setting z = 1, we get:

3 + 1 = A + B
A = 2

Therefore, the integral becomes:

∫ (2z + 1) / (z^2 + 2) dz

We can now integrate using substitution, with u = z^2 + 2 and du/dz = 2z. This gives us:

∫ (2z + 1) / (z^2 + 2) dz = ∫ (1/u) du
= ln|u| + C
= ln|z^2 + 2| + C

Using trigonometric substitution, we can also evaluate the integral in terms of arctan:

Let z = √2 tanθ, dz = √2 sec^2θ dθ. Then the integral becomes:

∫ (3z + 1) / (z^2 + 2) dz = ∫ (3√2 tanθ + 1) / (2tan^2θ + 3) √2 sec^2θ dθ
= 3/√2 ∫ (tanθ) / (tan^2θ + (3/2)) d(tanθ) + 1/√2 ∫ (1) / (tan^2θ + (3/2)) dθ
= 3/√2 ln|tan^2θ + (3/2)| + 3/√2 arctan(tanθ/√2) + C
= 3/√2 ln|z^2 + 2| + 3/√2 arctan(z/√2) + C

Thus, we get the same result as before.

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Answer: (x+4)² = -8 (y+3)

Step-by-step explanation: I got this correct on Edmentum.

Answer:

there are 8 packages left

Step-by-step explanation:

120÷12=10. So there are 10 packages of pencils to begin with. She loses 24 pencils which means that she lost two packages worth of pencils (12+12=24). So, 2 packages-10 packages=8 packages.

If Harriet Hanson Robinson (Larcom) had started her career in the mill as a machine worker instead of as a doffer, her initial impression of the mill may have been significantly different.

Here are a few ways her perspective might have been altered:

Physical Experience: As a machine worker, Larcom would have been directly involved in operating the machinery. She would have experienced the physical demands and potentially dangerous conditions associated with working with heavy machinery. This hands-on experience might have given her a greater appreciation for the risks and challenges faced by the workers. Social Interaction: Machine workers often worked in close proximity to each other, tending to the machines and coordinating their tasks. By working alongside her fellow machine workers, Larcom would have had more direct contact and interaction with her peers. This could have provided her with a deeper understanding of the camaraderie, unity, and challenges faced by the workers as a community.

Skill Development: Operating the machines required a certain level of technical skill and knowledge. If Larcom had started as a machine worker, she would have gained expertise in machine operation and maintenance. This technical knowledge could have given her a different perspective on the machinery, its intricacies, and its impact on the workers. Perspective on Management: As a machine worker, Larcom might have had more direct interactions with mill managers and overseers. This could have given her insights into the management practices, decision-making processes, and power dynamics within the mill. Such firsthand experiences might have influenced her initial impression of the mill's management and their treatment of workers.

Overall, starting as a machine worker would have provided Larcom with a different vantage point and firsthand experience of the mill's operations. This could have shaped her understanding of the work environment, the challenges faced by the workers, and the dynamics between management and labor.

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Answer:

in terms of degree it is linear equation

in terms of number of terms it is binomial

1. False: The sum of the probabilities can never exceed one.

2. False: The alpha value indicates the maximum level of error allowed in the investigation.

3. True: The critical value is obtained from the formulas applied to each test.

4. False: H0 is rejected when the test statistic value is greater than the critical value.

5. True: All probability is between zero and one.

6. False: The ANOVA test uses only a portion of the bell curve.

7. True: There are two types of hypotheses.

8. True: The null hypothesis may posit that there is no significant difference between the statistic and the parameter or between two parameters.

9. True: When α = 0.05, it means that the level of significance is 5%, not that you allow 95% error in the study.

10. True: The critical value is the point at which the rejection zone for H0 begins.

11. False: In binomial distribution, r cannot be greater than n.

12. False: The value of H1 depends on the alternative hypothesis.

13. False: In a normal distribution, the probabilities can never exceed one.

14. True: In incompatible events, the sum of probabilities must be one.

15. True: In classical probability, the sample space is always known.

16. False: When n = 30 in a binomial distribution, Pr can never be greater than one.

17. True: All probability values are between 0 and 1.

18. True: In binomial distribution, Pr(0) is one of the probabilities.

19. True: In Poisson distribution, N is known, and its average N is used.

20. True: Incompatible events are mutually exclusive.

21. False: The sum of Pr(Oc) and Pr(Not Oc) can never exceed one.

22. True: Incompatible events can be seen in classical probability.

23. True: The variable "eye color" is a qualitative variable, nominal.

24. True: The "scatter diagram" illustrates the relationship between two variables.

25. True: In the range of grouped data, you subtract the smallest value from the largest value.

26. True: The independent variable is manipulated by the researcher.

27. False: In Poisson distribution, Pr(X > 1) = 1 - Pr(0) - Pr(1).

28. False: In binomial distribution, Pr(r > 1) = 1 - Pr(0) - Pr(1).

29. True: The sum of Pr(Oc) and Pr(Not Oc) is always one.

30. True: In normal distribution, the sum of probabilities equals one.

31. False: In binomial distribution, r cannot be greater than n.

32. True: The bar chart is used to illustrate the relationship of nominal qualitative variables.

33. True: In binomial distribution, the sum of all Pr is 1.

34. False: In the Poisson distribution, Pr(X = 31) is always less than or equal to 1.

35. False: The null and alternative hypotheses are not mutually exclusive.

36. False: The given formula is only true for independent events.

37. True: The given formula is only true for dependent events.

38. True: The correlation coefficient can be negative.

39. True: The scatter plot illustrates the dispersion of the data.

40. True: The sum of all Pr(Oc) is always one.

41. True: When Z is negative, the probability is also negative.

42. False: In normal distribution, Pr(Z ≤ 3) = Pr(Z = 3) + Pr(Z < 3).

43. False: The arithmetic average is found by adding all the data and dividing by n.

44. True: The location of the median is found by looking for (n + 1) / 2.

45. True: Median of ungrouped data does not use outliers.

46. False: In normal distribution, Z is equal to the number of standard deviations away from the mean.

47. False: When Z = 1, the probability is 0.3413, not 0.5 - 0.3413.

48. True: Variance in grouped data is divided by n - 1.

49. False: If r > 1, then the variables are not directly proportional.

50. True: Every probability is a proper fraction.

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Answer:

(6, 6)

Step-by-step explanation:

ordered pair (6, 6) will be solution for the function y = 12 - x.

When x = 6

y = 12 - 6

y = 6

(x, y) = (6, 6)

Answer:

6,6

Step-by-step explanation:

Answer:

-6x + 56

Step-by-step explanation:

-2(3x) + 56

-6x + 56

The probability that a randomly selected card does not have a face value of 9 is approximately 0.9231 or 92.31%.

In a standard deck of 52 cards, there are 4 cards of each face value (Ace through 10) for each of the four suits (hearts, diamonds, clubs, and spades). Since the face value of 9 is one of the possible values, there are 4 cards with a face value of 9.

To find the probability that a randomly selected card does not have a face value of 9, we need to determine the number of cards that do not have this particular value. There are 52 cards in total, and we subtract the 4 cards with a face value of 9:

52 - 4 = 48

So, there are 48 cards that do not have a face value of 9. Therefore, the probability of selecting a card without a face value of 9 is:

P(not 9) = number of favorable outcomes / total number of outcomes
= 48 / 52
= 12 / 13
≈ 0.9231

The probability that a randomly selected card does not have a face value of 9 is approximately 0.9231 or 92.31%.

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Answer:

P = Parker

According to the question,

Equation for Jessica's height = 8 - 2P

Answer:

The Impressionist period was distinctive because composers were focused on creating an impression through building atmospheres, pictures, and sound worlds with music. Composers Debussy and Ravel were at the forefront of this ground-breaking musical style.

The function v = x + 1, the norm is v = sqrt(integral of (x+1)^2 dx from 0 to 1), which evaluates to sqrt(5/3). The angle between u and v is arccos(1/√5).

a) The norm f of a function f in the vector space C[0, 1] with the given inner product is defined as the square root of the inner product of the function with itself. In this case, f = sqrt((f, f)) = sqrt(integral of f(x) * f(x) dx over the interval [0, 1]). For the function u = x, the norm is u = sqrt(integral of x^2 dx from 0 to 1), which evaluates to sqrt(1/3). For the function v = x + 1, the norm is v = sqrt(integral of (x+1)^2 dx from 0 to 1), which evaluates to sqrt(5/3).

b) The inner product space C[0, 1] induces a norm on the set of functions, and we can define the distance between two functions as the norm of their difference. In this case, the norm of the difference between the functions u and v is |u - v| = sqrt((u - v, u - v)) = sqrt(integral of (x - (x+1))^2 dx from 0 to 1), which simplifies to sqrt(1/3). Therefore, |u - v| = sqrt(1/3).

c) The inner product (f, g) between the functions f and g is defined as the integral of their pointwise product over the interval [0, 1]. For the functions u = x and v = x + 1, (u, v) = integral of (x * (x + 1)) dx from 0 to 1, which evaluates to 7/6.

d) The angle between two functions u and v in an inner product space can be computed using the definition of the inner product and the norms of the functions. The angle theta between u and v is given by the equation cos(theta) = (u, v) / (u * v). In this case, cos(theta) = (7/6) / (sqrt(1/3) * sqrt(5/3)), which simplifies to 1/√5. Therefore, the angle between u and v is arccos(1/√5).

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Answer:

15 yards amd 3 feet.

Step-by-step explanation:

I wish to get extra three points :)

Answer:

This is 15 yards and 3 feets.

[tex]\log_7x=\dfrac{1}{2}\\\\x=7^{1/2}\\\\x=\sqrt{7}\approx2.646[/tex]