What value of t makes the statement true? (4x3 5)(2x3 3) = 8x6 tx3 15.

(B)

Step-by-step explanation:

Look at the triangle with the 65° angle. Note that the 65° angle is opposite the 24-ft side and p is the length of the triangle's hypotenuse so let's use the definition of sine 65°:

sin 65 = opposite/hypotenuse

= (24 ft)/p

Solving for p,

p = (24 ft)/(sin 65°)

= 26.48 ft

There is enough evidence to suggest p>0.5

What is hypothesis in statistics?

A type of statistical analysis called hypothesis testing involves testing your presumptions regarding a population parameter. It is used to estimate the relationship between 2 statistical variables.

Thus, a hypothesis is a set of possible explanations or resolutions applicable to a situation. In other words, the hypotheses pose different scenarios in which the aim is to explain the origin and eventual resolution of the problem.

Let's discuss few examples of statistical hypothesis from real-life -

We are testing :

H0: p <= 0.5

H1: p > 0.5

Test statistic:

Z = [(364/562) - 0.5] / √[0.5(1-0.5)] / 562

= 7.00

P-value = P(Z>7.00) = 0.0000

Hence, there is enough evidence to suggest p>0.5.

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Answer: The slope is -2

Step-by-step explanation:

To get the slope, you need to use the slope formula y2-y1/x2-x1

So what I did was plug it in and subtract:

The two given coordinates are (1,4) and (2,2)​

2-4= -2

2-1= 1

-2/1 = -2

So the slope is -2

I hope this helps!

Answer:

Subtracting polynomials is quite similar to adding polynomials, but there are those pesky "minus" signs to deal with. If the subtraction is being done horizontally, then the "minus" signs will need to be taken carefully through the parentheses

Step-by-step explanation:

Answer:

True

Step-by-step explanation:

This is true: the result of adding two polynomials will always be another polynomial. A polynomial is an algebraic expression made up of the sum of monomials, which are products of numbers (coefficients) and variables in positive integer exponents.

Answer:

X=20º

Step-by-step explanation:

This is a vertical angle meaning both sides are congruent to each other.

Meaning: (x+100)º= 6x

Solve:

(x+100)º= 6x

-x              -x

100º=5x

/5 .     /5

x=20º

Answer:

3

Step-by-step explanation:

Vol = length x width x height

Vol = 3 x 3 x3

Vol = 27

y = 220/ ((10/3) * √22)

It is given that area is 200 square feet. Cost is 3 dollar per foot. Fourth side costs 12 dollars per foot.

This f(x,y) needs to represent the cost of the fence so we can look at each side's price. We can choose the fourth side to be along the x axis and be represented by 13x. The other sides therefore must be represented by 5y, 5y, and 5x.

So, our cost, f(x,y) = 13x + 5x + 5y + 5y = 18x + 10y

Our constraint is xy = 220 as defined by the area of a rectangle.

We can then take our constraint to be put in terms of exclusively x, giving us y = 220/x

Plugging this into our cost, the thing we are minimizing, we get f(x) = 18x + 2200/x

In order to find the minimum we use the first derivative test, taking f'(x) and finding the critical points.

f'(x) = 18 - 2200/x2

Setting this equation to be equal to 0 we find that x = ±√2200/18 . But the negative answer doesn't make sense because distance cannot be negative so we throw it out.

x = (10/3) * √22

We must verify that this is a minimum by confirming the following:

If x < (10/3) * √22, f'(x) < 0. So, f(x) is decreasing when x < (10/3) * √22.

If x > (10/3) * √22, f'(x) > 0. So, f(x) is increasing when x > (10/3) * √22.

Thus, we have guaranteed that (10/3) * √22 is the x dimension. Now we plug in this value in our original equation to find y and that is our y dimension. So, y = 220/ ((10/3) * √22)

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Kallie will perform these tasks in the Analysis phase of the Systems Development Life Cycle (SDLC).

In the Systems Development Life Cycle (SDLC), the Analysis phase is where Kallie will create use cases, data flow diagrams, and entity relationship diagrams. This phase is the second phase of the SDLC, following the Planning phase. During the Analysis phase, Kallie will gather detailed requirements and analyze the current system or business processes to identify areas for improvement.

Use cases are used to describe interactions between actors (users or systems) and the system being developed. They outline the specific steps and interactions necessary to achieve a particular goal. By creating use cases, Kallie can better understand the requirements and functionality needed for the system.

Data flow diagrams (DFDs) are graphical representations that illustrate the flow of data within a system. They show how data moves through different processes, stores, and external entities. These diagrams help Kallie visualize the system's data requirements and identify any potential bottlenecks or inefficiencies.

Entity relationship diagrams (ERDs) are used to model the relationships between different entities or objects within a system. They depict the structure of a database and show how entities are related to each other through relationships. ERDs allow Kallie to define the data structure and relationships required for the system.

By creating use cases, data flow diagrams, and entity relationship diagrams during the Analysis phase, Kallie can gain a deeper understanding of the system's requirements, data flow, and structure. These artifacts serve as important documentation for the subsequent phases of the SDLC, guiding the design, development, and implementation processes.

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The length of the side of the smaller square is 7 inches.

Let x be the length of the side of the smaller square.

The length of each side of a square is 3 inches more than the length of each side of a smaller square. Thus the length of the side of the larger square is

(x + 3) inches.

The area of the smaller square is

x^2 square inches.

The area of the larger square is

(x + 3)^2 square inches.

We can set up the equation:

x^2 + (x + 3)^2 = 149

Expanding and simplifying the equation gives:

x^2 + x^2 + 6x + 9 = 149

Combining like terms:

2x^2 + 6x - 140 = 0

Using the quadratic formula, we can solve for x:

x = (-b ± √(b^2 - 4ac)) / 2a

x = (-6 ± √(6^2 - 4 * 2 * (-140))) / 2 * 2

x = (-6 ± √(36 + 1120)) / 4

x = (-6 ± √(1156)) / 4

x = (-6 ± √(596)) / 4

x = (-6 ± 34) / 4

x = -10 or x = 7

The side length of the smaller square can't be negative, so x = 7 inches.

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Compute the Gaussian density for the following cases. (a) x=0, m=0, s-1. Give the name of question5a (b) x-2, m=0, s-1. The value of the account on January 1, 2021, would be $2,331.57.

To calculate the value of the account on January 1, 2021, we need to consider the compounding interest for each year.

First, we calculate the value of the initial deposit after three years (12 quarters) using the formula for compound interest:

Principal = $1,000

Rate of interest per period = 8% / 4 = 2% per quarter

Number of periods = 12 quarters

Value after three years = Principal * (1 + Rate of interest per period)^(Number of periods)

                           = $1,000 * (1 + 0.02)^12

                           ≈ $1,166.41

Next, we calculate the value of the additional $1,000 deposit made on January 1, 2019, after two years (8 quarters):

Principal = $1,000

Rate of interest per period = 2% per quarter

Number of periods = 8 quarters

Value after two years = Principal * (1 + Rate of interest per period)^(Number of periods)

                         = $1,000 * (1 + 0.02)^8

                         ≈ $1,165.16

Finally, we add the two values to find the total value of the account on January 1, 2021:

Total value = Value after three years + Value after two years

                ≈ $1,166.41 + $1,165.16

                ≈ $2,331.57

Therefore, the value of the account on January 1, 2021, is approximately $2,331.57.

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The total number of ways 'n' stairs can be climbed in are given by:\(a_{n} = a_{n-1} + a_{n-2}, n\geq 2\)

The principle states that if we have P number of ways of doing something and Q number of ways of doing another thing and we can not do both at the same time, then there are P+Q ways to do things by one of the methods.

Given:  \(a_{n}\) = Number of ways to climb 'n' stairs, if a person is taking one or two stairs at a time.

Then, this can be done in two ways:

Thus, the total number of ways 'n' stairs can be climbed in are given by:

\(a_{n} = a_{n-1} + a_{n-2}, n\geq 2\)

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Answer:

X= 43/3

Step-by-step explanation:

-1/4 x 5 = 3/4x - 12

-5/4=3/4x -12

-5=3x - 48

-3x-5 = -48

-3x = -48 + 5

-3x= -43

-3 x= -43

X= 43/3

2-1. a) The shear stress τ is constant across the flow. b) The velocity is maximum at the center (r = 0) and decreases linearly as the radial distance increases. c)v_max = (P₁ - P₂) / (4μL) * \(R^{2}\) and v_avg = (1 / (π(\(R^{2} -a^{2}\)))) * ∫[a to R] v * 2πr dr 2-2.a) The shear stress is constant for parallel plates. b) The velocity profile shows that the velocity is maximum at the centerline and decreases parabolically .c)v_max = (P₁ - P₂) / (2μh) and v_avg = (1 / (2h)) * ∫[-h to h] v dr.

2-1. Flow in an annular region between concentric cylinders:

(a) Shear stress profile:

In an incompressible fluid flow between concentric cylinders, the shear stress τ varies with radial distance r. The shear stress profile can be obtained using the Navier-Stokes equation:

τ = μ(dv/dr)

where τ is the shear stress, μ is the dynamic viscosity, v is the velocity of the fluid, and r is the radial distance.

Since the flow is at steady state, the velocity profile is independent of time. Therefore, dv/dr = 0, and the shear stress τ is constant across the flow.

(b) Velocity profile:

To determine the velocity profile in the annular region, we can use the Hagen-Poiseuille equation for flow between concentric cylinders:

v = (P₁ - P₂) / (4μL) * (\(R^{2} -r^{2}\))

where v is the velocity of the fluid, P₁ and P₂ are the pressures at the outer and inner cylinders respectively, μ is the dynamic viscosity, L is the length of the cylinders, R is the outer radius, and r is the radial distance.

The velocity profile shows that the velocity is maximum at the center (r = 0) and decreases linearly as the radial distance increases, reaching zero at the outer cylinder (r = R).

(c) Maximum and average velocities:

The maximum velocity occurs at the center (r = 0) and is given by:

v_max = (P₁ - P₂) / (4μL) * \(R^{2}\)

The average velocity can be obtained by integrating the velocity profile and dividing by the cross-sectional area:

v_avg = (1 / (π(\(R^{2} -a^{2}\)))) * ∫[a to R] v * 2πr dr

where a is the inner radius of the annular region.

2-2. The flow between parallel plates:

(a) Shear stress profile:

For flow between very wide or broad parallel plates, the shear stress profile can be obtained using the Navier-Stokes equation as mentioned in problem 2-1. The shear stress τ is constant across the flow.

(b) Velocity profile:

The velocity profile for flow between parallel plates can be obtained using the Hagen-Poiseuille equation, modified for this geometry:

v = (P₁ - P₂) / (2μh) * (1 - (\(r^{2} /h^{2}\)))

where v is the velocity of the fluid, P₁ and P₂ are the pressures at the top and bottom plates respectively, μ is the dynamic viscosity, h is the distance between the plates, and r is the radial distance from the centerline.

The velocity profile shows that the velocity is maximum at the centerline (r = 0) and decreases parabolically as the radial distance increases, reaching zero at the plates (r = ±h).

(c) Maximum and average velocities:

The maximum velocity occurs at the centerline (r = 0) and is given by:

v_max = (P₁ - P₂) / (2μh)

The average velocity can be obtained by integrating the velocity profile and dividing by the distance between the plates:

v_avg = (1 / (2h)) * ∫[-h to h] v dr

These formulas can be used to calculate the shear stress profile, velocity profile, and maximum/average velocities for the given geometries.

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If A is a subset of B, it does not necessarily mean that A is a proper subset of B. The union of any set A and its complement is indeed the universal set.

1. False: If A is a subset of B, it does not necessarily mean that A is a proper subset of B. A proper subset is a subset that contains some, but not all, elements of the superset. In other words, if A is a proper subset of B, it means that there exists at least one element in B that is not in A. However, if A is simply a subset of B, it is possible for A and B to have the same elements, making them equal sets.

For example, let's consider two sets:

A = {1, 2}

B = {1, 2, 3}

In this case, A is a subset of B because every element in A (1 and 2) is also in B. However, A is not a proper subset of B since there are no elements in B that are not in A. Therefore, statement 1 is false.

2. True: The union of any set A and its complement is indeed the universal set. The complement of a set A, denoted as A', consists of all elements that are not in A but are in the universal set U. When we take the union of A and its complement, we are effectively combining all elements from A and all elements not in A.

Let's consider a set A and its complement A' within a universal set U. The union of A and A' is denoted as A ∪ A' and can be represented as U. This is because every element in the universal set U is either in A or not in A (in A').

For example, let's consider the following sets:

U = {1, 2, 3, 4, 5}

A = {1, 2}

The complement of A, A', would be {3, 4, 5}. The union of A and A' (A ∪ A') is {1, 2} ∪ {3, 4, 5}, which is equal to U: {1, 2, 3, 4, 5}. Thus, the union of any set A and its complement is indeed the universal set. Therefore, statement 2 is true.

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Answer:

first 4 terms: 3,1,-1,-3

10th: -35

Step-by-step explanation:

Please mark as brainliest if answer right

Have a great day, be safe and healthy  

Thank u  

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Answer:

3, 1, - 1, - 3 and - 15

Step-by-step explanation:

To find the first 4 terms substitute n = 1, 2, 3, 4 into the rule, that is

a₁ = 5 - 2(1) = 5 - 2 = 3

a₂ = 5 - 2(2) = 5 - 4 = 1

a₃ = 5 - 2(3) = 5 - 6 = - 1

a₄ = 5 - 2(4) = 5 - 8 = - 3

The first 4 terms are 3, 1, - 1, - 3

For 10th term substitute n = 10 into the rule

a₁₀ = 5 - 2(10) = 5 - 20 = - 15

C

Step-by-step explanation:

hope this helps

"The mean will be higher than the median in any distribution that" ;

has a positively skewed distribution

A positively skewed distribution is one in which the majority of data values are clustered towards the lower end of the range, while the rest of the values are spread out towards the higher end of the range. As a result, the mean (the average of all the data values) will be higher than the median (the middle value of the data set).

This is because the mean is affected by the higher values in the distribution, while the median is not.

Positively skewed distributions are common in real-world data sets and are often seen in income distributions, stock market returns, and other economic data.

For example, a distribution of incomes may have a few very high earners pushing up the mean, while the majority of people make much lower amounts, resulting in a median that is lower than the mean.

Similarly, stock market returns may have a few very large returns that pull the mean up, while the majority of returns are much lower, resulting in a median that is lower than the mean.

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Answer:

17.5%

Step-by-step explanation:

We know that there are 35 out of 200 allergic to peanut

So we take

35 divided by 200, then times 100% = 17.5%

So, 17.5  of children are allergic to peanut

the answer is OD graph d.

The simplified form of the expression \((-7a^3b^{-3})\) / (21ab) is \((-a^2) / (3b^3),\) removing negative exponents and canceling out common factors.

To simplify the given expression, let's break it down step by step.

The expression is:

\((-7a^3b^{-3}) / (21ab)\)

First, we can simplify the numerator by applying the exponent rules.

The negative exponent in the numerator can be rewritten as a positive exponent in the denominator:

\((-7a^3) / (21ab^3)\)

Next, we can simplify the fraction by canceling out common factors in the numerator and denominator. In this case, we can cancel out the common factor of 7:

\((-a^3) / (3ab^3)\)

Now, we can simplify the remaining terms by canceling out the common factor of 'a':

\((-a^2) / (3b^3)\)

Finally, we have simplified the expression to \((-a^2) / (3b^3)\).

In this simplified form, the expression no longer contains negative exponents or common factors in the numerator and denominator.

To summarize, the simplified form of the expression \((-7a^3b^{-3}) / (21ab)\) is \((-a^2) / (3b^3).\)

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Answer:

Step-by-step explanation:

The derivative of the function f is f'(x) = x^3 - 8x^2, and the original function f can be obtained by integrating the derivative.

The given derivative, f'(x) = x^3 - 8x^2, represents the rate of change of the function f with respect to x. To find the original function f, we need to integrate the derivative.

Integrating the derivative f'(x), we obtain:

f(x) = ∫(x^3 - 8x^2) dx

To integrate x^3, we add 1 to the exponent and divide by the new exponent:

∫x^3 dx = (1/4)x^4 + C1, where C1 is the constant of integration.

To integrate -8x^2, we use the same process:

∫-8x^2 dx = (-8/3)x^3 + C2, where C2 is another constant of integration.

Combining the two results, we have:

f(x) = (1/4)x^4 - (8/3)x^3 + C, where C = C1 + C2 is the overall constant of integration.

Thus, the original function f, corresponding to the given derivative, is f(x) = (1/4)x^4 - (8/3)x^3 + C.



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Theorem 4.3.9 can be proved using either the 0 characterisation of continuity or the sequential characterisation of continuity. Both characterisations are important and useful in different situations.

Theorem 4.3.9: Suppose f: X → Y. TFAE:(i) f is continuous on X(ii) For every open subset V of Y, the inverse image f^-1 (V) is open in X(iii) For every convergent sequence x_n→x, we have f(x_n)→f(x).

A proof for Theorem 4.3.9 using the 0 characterisation of continuity is given below:

Suppose f: X → Y is continuous on X. Let V be open in Y. Let a∈f^-1 (V). Then f(a)∈V, so there exists an ɛ > 0 such that B(f(a), ɛ) ⊆ V.

Since f is continuous at a, there exists a δ > 0 such that x ∈ X and d(x, a) < δ implies d(f(x), f(a)) < ɛ. That is, f(B(a, δ)) ⊆ B(f(a), ɛ) ⊆ V.

This implies B(a, δ) ⊆ f^-1 (V). Thus f^-1 (V) is open in X.For the other direction, suppose for every open subset V of Y, the inverse image f^-1 (V) is open in X. Let a ∈ X. Let ɛ > 0. Set V = B(f(a), ɛ). Then V is open in Y, so f^-1 (V) is open in X.

Since a ∈ f^-1 (V), there exists a δ > 0 such that B(a, δ) ⊆ f^-1 (V). Then f(B(a, δ)) ⊆ V. That is, d(f(x), f(a)) < ɛ whenever d(x, a) < δ. Thus f is continuous at a.This gives us a proof of Theorem 4.3.9 using the 0 characterisation of continuity.

A proof for Theorem 4.3.9 using the sequential characterisation of continuity is given below:Suppose f: X → Y. Suppose (x_n) is a sequence in X converging to x. Then (x_n) is a net in X. By Theorem 4.3.2(iii), if f is continuous at x, then f(x_n) converges to f(x).Now suppose that for every convergent sequence (x_n) in X that converges to x, we have f(x_n) converges to f(x).

Let V be open in Y. Let a∈f^-1 (V). Suppose that f is not continuous at a. Then there exists an ɛ > 0 such that for every δ > 0, there exists an x ∈ B(a, δ) such that d(f(x), f(a)) ≥ ɛ. Let δ_n = 1/n. Then for each n, there exists x_n ∈ B(a, δ_n) such that d(f(x_n), f(a)) ≥ ɛ. Then (x_n) converges to a, but (f(x_n)) does not converge to f(a). This contradicts our assumption.

Thus f is continuous at a.Hence we have a proof for Theorem 4.3.9 using the sequential characterisation of continuity.

We first showed the proof of Theorem 4.3.9 using the 0 characterisation of continuity.

We then showed the proof of Theorem 4.3.9 using the sequential characterisation of continuity.We used the 0 characterisation of continuity to show that f is continuous on X if and only if for every open subset V of Y, the inverse image f^-1 (V) is open in X. This result follows from the definitions of continuity and openness.

We used this characterisation to prove Theorem 4.3.9. We showed that f is continuous on X if and only if for every convergent sequence x_n→x, we have f(x_n)→f(x).

This characterisation is important because it allows us to prove continuity of a function using the open sets of the codomain. We do not need to use sequences or nets to prove continuity.

This is useful in some cases where we cannot use sequences or nets, for example, in metric spaces that are not first-countable.We then used the sequential characterisation of continuity to show that f is continuous on X if and only if for every convergent sequence x_n→x, we have f(x_n)→f(x).

This characterisation follows from the definition of continuity and the sequential characterisation of convergence. We used this characterisation to prove Theorem 4.3.9. We showed that f is continuous on X if and only if for every open subset V of Y, the inverse image f^-1 (V) is open in X.

This characterisation is important because it allows us to prove continuity of a function using sequences.

This is useful in many cases, for example, in metric spaces that are first-countable. It also allows us to prove that some spaces are not first-countable by finding a function that is not continuous using sequences.

Therefore, Theorem 4.3.9 can be proved using either the 0 characterisation of continuity or the sequential characterisation of continuity. Both characterisations are important and useful in different situations.

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The total price of the land would increase by $2020 if you measured the parcel with a steel tape measure on a day when the temperature was 17°C above normal.

The coefficient of thermal expansion for steel is 0.0000116 m/m°C. This means that for every 1°C increase in temperature, a steel tape measure will expand by 0.0000116 m. On a day when the temperature is 17°C above normal, the steel tape measure will expand by 0.0000116 * 17 = 0.0002072 m.

The land title says the dimensions of the parcel are 30 m x 40 m. If the steel tape measure expands by 0.0002072 m, then the actual dimensions of the parcel are 30.0002072 m x 40.0002072 m. This means that the actual area of the parcel is 30.0002072 * 40.0002072 = 12000.8288 square meters.

The land title says the price of the land is $60,000 per square meter. So, the actual price of the land is 12000.8288 * 60,000 = $7200492.8. This is $2020 more than the price listed on the land title.

The coefficient of thermal expansion is a measure of how much a material expands when its temperature increases. The coefficient of thermal expansion for steel is very small,

but it is still significant enough to cause a measurable change in the length of a steel tape measure when the temperature changes.

In this case, the temperature is 17°C above normal, which is a significant change in temperature. The steel tape measure will expand by 0.0002072 m, which is a small change, but it is still enough to cause a measurable change in the area of the parcel.

The actual area of the parcel is 0.0002072 m larger than the area listed on the land title. This means that the actual price of the land is $2020 more than the price listed on the land title.

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Answer:

\(4^{12}\)

Step-by-step explanation:

using the rule of exponents

\(\frac{a^{m} }{a^{n} }\) = \(a^{(m-n)}\) , then

\(\frac{4^{8} }{4^{-4} }\) = \(4^{(8-(-4))}\) = \(4^{(8+4)}\) = \(4^{12}\)

Answer:

\( \frac{ - 9}{21} \)

then simplify

Step-by-step explanation:

Slope =

\( \frac{y2 - y1}{x2 - x1} \)

Rise

(-2)-(7)=-9

Run

\(8 - ( - 13) = 21\)

Answer: 3/-7

Step-by-step explanation:

y2-y1 over x2-x1 is slope. your parenthesis go in the order of (x,y)

8=x1 -2=y1 -13=x2 7=y2

7 - -2 = 9

-13 - 8 = -21

9/-21

simplify to get 3/-7

Answer:

C 3.5t + 15, where t is the number

of days

Answer:

2?

Step-by-step explanation:

a and b are both terms