Find the volume of this square pyramid

Answer:

1. and 2. I belive are true sorry if wrong

Step-by-step explanation:

Kohl's rectangular gold garden has an area of 54 square feet. To determine the dimensions of the garden, we need to find two numbers whose product is 54.

To find the dimensions of the garden, we can factorize the area of 54 square feet. The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. Since the garden is rectangular, we are looking for two numbers whose product is 54.

By examining the factors, we can see that the dimensions of the garden could be 6 feet by 9 feet, as their product is indeed 54. Alternatively, the garden could have dimensions of 3 feet by 18 feet, as their product is also 54. Both sets of dimensions result in an area of 54 square feet.

Therefore, the possible dimensions of Kohl's rectangular gold garden could be either 6 feet by 9 feet or 3 feet by 18 feet.

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5. Assuming that t0 = 200 find the value of the marginal tax rate that will yield the same level of equilibrium GDP as the one obtained

(1). Solution: Given, T = t0 + t1Y and T = 300

Substituting the given values, we get300 = 200 + t1YGDP, Y = C + I + G + X - M

where, Y = GDP; C = consumption; I = private investment; G = government spending; X = exports; M = imports

We know, C = c0 + c1 (Y − T) Disposable income, YD = Y − T

So, C = c0 + c1 (Y − T) = c0 + c1YD

From the question, S = 0.5Y − 500

We know that, private saving, S = Y − C − T

So, Y − C − T = 0.5Y − 500 ⇒ 0.5Y = C + T + 500

Putting the values,

0.5Y = (c0 + c1YD) + T + 500 ⇒ 0.5Y = (c0 + c1(Y - T)) + T + 500 ⇒ 0.5Y = c0 + c1Y - c1T + T + 500

Solving the above expression, we get

0.5Y - c1Y = c0 - 0.5T + 500 ⇒ 0.5(1-c1)Y = c0 - 0.5T + 500

Hence, Y = (c0 - 0.5T + 500) / (0.5 - c1)

Again, from the question, Y = C + I + G + X - M

Substituting the values we get,

(c0 + c1(Y − T)) + 400 = I + 400 + Y - 500 + X - X0.5Y − 500 + 400 = I + 300 + X − G ⇒ 0.5Y + I = 1200 + G + X

Assuming equilibrium GDP Y = Y*, private investment I = I*, government spending G = G* and net exports X = X*, so0.5Y* + I* = 1200 + G* + X*

Now, from the given information of S, we have S = Y* − C* − T.

Substituting for C* from the equation above, we get S = Y* − (c0 + c1(Y* − T)) − T ⇒ S = Y* − c0 − c1Y* + c1T − T

Substituting for Y* from above, we have S = ((c0 - 0.5T + 500) / (0.5 - c1)) - c0 - c1[((c0 - 0.5T + 500) / (0.5 - c1))] + c1T - T

Now, we need to find the value of t1 when t0 = 200. For this, we need to substitute the value of t0 and Y* in T = t0 + t1YSo, 300 = 200 + t1Y* ⇒ t1 = (300 - 200) / Y* ⇒ t1 = 0.1

Therefore, the value of the marginal tax rate t1 is 0.1.

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Answer:

450m

Step-by-step explanation:

Answer:

A. 5+x-12=2x-7

-7+x+2x-7

x=2x

x-2x=0

-x=0-> x=0

B.

when X=-0.5

5+(-0.5)-12=2 (-0.5)-7

-7.5=-8 (answer does not work)

when X=0

5+0-12=2(0)-7

-7=-7 (answer works)

when x=1

5+1-12=2(1)-7

18=-5 (answer does not work)

Step-by-step explanation:

Answer:

[tex]x=0[/tex]

Step-by-step explanation:

A) From this case you would need to narrow down the whole equation...

Like this: [tex]5+x-12=2x-7[/tex]  →  [tex]x-7=2x-7[/tex]  →  [tex]x=0[/tex]

B) To prove that one of these numbers solve the equation, we would have to check it ourselves.

Like this: [tex]5+(-0.5)-12=2(-0.5)-7[/tex]  →  [tex]-7.5=-8[/tex] (SO THIS WON'T WORK)

[tex]5+0-12=2*0-7[/tex]  →  [tex]0=0[/tex] (THIS WORKS!)

[tex]5+1-12=2(1)-7[/tex]  →  [tex]-6=-5[/tex] (NOR DOES THIS WORK)

Therefore: [tex]x=0[/tex] will work to solve the equation correctly

Answer:

A

Step-by-step explanation:

Because That's The Main Idea

The approximate value of y(1) using Euler's method with a step size of h = 0.1 is 1.

Using Euler's method and a step size of h = 0.1, we can use the given initial value problem (IVP) and iterate through the interval [0, 1] in steps of size h to approximate the value of y(1). Let's begin by determining the number of steps required: n = (1-0) / 0.1 = 10.

Utilizing the accompanying recipe, we can repeat through the span beginning with the underlying condition y(0) = 1.

y(i+1) is equivalent to y(i) + h * (t(i) * y(i) + t(i)2), where I is among 0 and n-1, t(i) is equivalent to I * h, and y(i) is the surmised worth of y at t(i).

At each step, we can inexact the upsides of y utilizing the recipe gave:

The approximate value of y(1) is 1, assuming Euler's method and a step size of h = 0.1. y(0) = 1 y(1)  y(0) + 0.1 * (0 * y(0) + 02) = 1 + 0.1 * (0 + 0) = 1.

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Answer:

Yes it will be appropriate to model the distribution of a sample mean with a normal model

Step-by-step explanation:

Given that the population is not normal, and the sample is sufficiently large, according to the Central Limit theorem, the distribution of the mean pf the sampling distribution will be approximately normal not withstanding the population from which the sample is obtained. Therefore, the mean, [tex]\overline x[/tex], and the standard deviation, [tex]\dfrac{\sigma}{\sqrt{n} }[/tex], of the sample will be equal to the mean, μ, and standard deviation, σ, of the of the population

Therefore, it will be appropriate to model the distribution of a sample mean with a normal model

Answer:

yes

Step-by-step explanation:

i got it right on plato

The particle never moves to the right of the origin. Hence, the farthest left of the origin that the particle moves is at x=-∞.

Given that, the position of the particle is [tex]x*d(t)=(1-2)^2*(1-6).[/tex]

Initially, the position of the particle is[tex]x*d(t)=(-1)^2*(-5)=5.[/tex]

The displacement of the particle can be given as the difference between the final position and the initial position. Therefore, the displacement of the particle is [tex]Δx=0-5=-5.[/tex]

(a) Since the value of the displacement is negative, the particle is moving to the left.

(b) The particle will be at rest when the velocity of the particle is zero. Here,[tex]v=d/dt (x*d(t))=x*d'(t)[/tex]

When the velocity of the particle is zero, then x*d'(t)=0.So, either x=0 or d'(t)=0.

When x=0, the particle will be at rest at the origin.

To find out the value of t, substitute x=0 in the given equation,

we get[tex]0=(-1)^2*(-5).[/tex] This is not possible. Therefore, the particle will never be at rest.

(c) When the direction of the velocity changes, the particle changes direction. The velocity of the particle is given as v=d/dt (x*d(t))=x*d'(t).

From the above equation, we see that the velocity of the particle changes direction when x changes direction. The direction of x changes at x=0.

(d) The farthest left of the origin that the particle moves is at x=-∞ when t=0.

Thus, the particle never moves to the right of the origin. Hence, the farthest left of the origin that the particle moves is at x=-∞.

Note: Here, we have considered the absolute values of -1 and -5.

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Answer:

If two secant segments intersect outside a circle, then the product of the secant segment with its external portion equals the product of the other secant segment with its external portion.

45(45) = (2x + 75)(27)

2,025 = (2x + 75)(27)

2x + 75 = 75, so x = 0 and GH = 48

Answer:

i don't know what goes up with three leg and goes down with 4

Answer:

u

Step-by-step explanation:

Since our test statistic is smaller than this critical value, we can conclude that the sample mean of 6.42 is not statistically significant and therefore cannot be used to reject the null hypothesis that the population mean is 6.0.

The first step here is to calculate the test statistic, which is also called a t-statistic or a z-score. This is done by subtracting the population mean from the sample mean, and dividing this difference by the standard deviation of the sample (dividing by the standard error if the population standard deviation is known). In this case, we have:

Test Statistic = (Sample Mean - Population Mean) / (Sample Standard Deviation / √n)

Test Statistic = (6.42 - 6.0) / (7.44 / √29)

Test Statistic = 0.42 / (7.44 / 5.385)

Test Statistic = 0.42 / 1.377

Test Statistic = 0.305

Now that we have the test statistic, we can compare it to the t-table or z-table (depending on if the population standard deviation is known) to determine whether or not the calculated value is statistically significant. For example, if we assume a 95% confidence level, the critical value of z (the p-value) is 1.96.

Since our test statistic is smaller than this critical value, we can conclude that the sample mean of 6.42 is not statistically significant and therefore cannot be used to reject the null hypothesis that the population mean is 6.0.

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Answer:

21 and 6

Step-by-step explanation:

a will be the larger, b the smaller:

a - b = 15

2b + 3a = 75

First, we'll solve the first equation for a in terms of b:

a = b + 15

Then substitute that in for a in the second equation to get a numerical value for b:

2b + 3(b + 15) = 75

2b + 3b + 45 = 75

5b = 30

b = 6

Next, we'll get a numerical value for a:

a - b = 15

a - 6 = 15

a = 21

Check the math:

2(6) + 3(21) = 12 + 63 = 75

Please lmk if you have questions.

Answer:

#1=8

#2=3

Step-by-step explanation:

Answer: 10/3 units

Step-by-step explanation: sqrt 100/9= 10/3

Answer: The correct answer will be 4

Step-by-step explanation:

Answer:

b(-1,-8)

Step-by-step explanation:

assuming x2 and y2 are the coordinates of b.

and x1 and y1 are the coordinates of A

midpoint are (xm, ym)

formula xm= (x1+x2)/2

ym=(y1+y2)/2

Answer:

3,410,000

Step-by-step explanation:

3.41 * 10^6 = 3.41 * 1,000,000 = 3,410,000

Answer:

3,410,000

plz mark me as brainliest

There is exactly one solution if v3 lies on this line, and no solution otherwise.

Given: vectors v1, v2, v3 in R2

We know that the vectors v1 and v2 are parallel, and we are asked to find the number of solutions of the system xv1 + yv2 = v3. We will argue geometrically.

Let us say that v1 and v2 are not equal to zero and are parallel to the x-axis. We can then write:

v1 = (a, 0)
v2 = (b, 0)

where a and b are nonzero constants. Since v1 and v2 are parallel, their cross-product is zero:

v1 × v2 = a*0 - 0*b = 0

This means that v1 and v2 are linearly dependent. Thus, we can express v2 as a scalar multiple of v1:

v2 = k*v1

where k is a nonzero constant. We can then substitute these expressions into the system and solve for x and y:

xv1 + yv2 = v3
xv1 + y(k*v1) = v3
(x + ky)v1 = v3

Since v1 is nonzero, the equation has a unique solution if and only if (x + ky) is nonzero. But (x + ky) is zero if and only if x = -ky, which is the equation of a line passing through the origin and perpendicular to v1 and v2. Thus, there is exactly one solution if v3 lies on this line, and no solution otherwise.

To see this geometrically, we can sketch the vectors v1, v2, and v3, and the line passing through the origin and perpendicular to v1 and v2. If v3 lies on this line, then there is exactly one solution, which corresponds to the intersection of the line and the vector v3. If v3 does not lie on this line, then there is no solution, since the line does not pass through v3.

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Answer:

C) 1/2

Step-by-step explanation:

3 of the 6 numbers on a cube are even. 3/6 = 1/2

First use the distributive property by multiplying the 3 with each term inside the parentheses:

-8 + 3c + 3 + c

Now combine like terms:

-8 + 3 = -5

3c + c = 4c

Combine for final answer:

-5 + 4c which can also be written as 4c-5

Answer:

first let's multiply 3 by the bracket

-8+3*c+3*1+c=-8+3c+3+c

then add the numbers that have the same variables -8+3+3c+c

=-8+3+4c then add the numbers

=-5+4c is the answer

you can also write it like this

4c-5

In this hypothesis testing scenario, the null hypothesis (H) states that the mean (μ) weight of a product is equal to 30 grams.

The one-sample t-test is computed using the formula t = (x - μ) / (s / √n), where x is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size. In this case, the calculated t-value is -0.3.

Based on the decision rule, since the calculated t-value of -0.3 falls within the non-critical region (-1.96 to 1.96), we do not reject the null hypothesis.

Therefore, the conclusion is that there is not enough evidence to support the claim that the mean weight of the product is 30 grams. The data does not provide sufficient statistical evidence to conclude that the mean weight significantly differs from 30 grams.

It's important to note that in hypothesis testing, the result does not prove that the null hypothesis is true; rather, it suggests that there is not enough evidence to reject it based on the given data and significance level.

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Answer:

I would need to see the graph to answer this question.

Given:

The expressions are:

[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}[/tex]

[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}[/tex]

To find:

The simplified form of the given expressions.

Solution:

We have,

[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}[/tex]

It can be written as:

[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}=\dfrac{12a^{2+1}bc^3}{40b^4c^2}[/tex]

[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}=\dfrac{3a^{3}c^{3-2}}{10b^{4-1}}[/tex]

[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}=\dfrac{3a^{3}c^{1}}{10b^3}[/tex]

Therefore, the value of the given expression is [tex]\dfrac{3a^{3}c}{10b^3}[/tex].

We have,

[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}[/tex]

It can be written as:

[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}=\dfrac{(x+y)(x-y)}{4(x+y)}\cdot \dfrac{x+y}{x-y}[/tex]

[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}=\dfrac{x+y}{4}[/tex]

Therefore, the value of the given expression is [tex]\dfrac{x+y}{4}[/tex].