If your observed Z or t statistic falls in the "critical region," what can you conclude?

Answer:

2097150

Step-by-step explanation:

GIVEN :-

TO FIND :-

CONCEPT TO BE USED IN THIS QUESTION :-

Geometric Progression :-

[NOTE :- \(\frac{a(1-r^n)}{1-r}\) can also be the formula for "Sum of n terms of G.P." because if you put 'r' there (assuming r > 0) you'll get negative value in both the numerator &  denominator from which the negative sign will get cancelled from the numerator & denominator. YOU'LL BE GETTING THE SAME VALUE FROM BOTH THE FORMULAES.]

SOLUTION :-

Let the first term of the G.P. given in the question be 'a' and the common ratio between successive terms be 'r'.

⇒ a = 6

It's given that forth term is 384. So from "General form of G.P." , it can be stated that :-

\(=> ar^3 = 384\)

\(=> 6r^3 = 384\)

Divide both the sides by 6.

\(=> \frac{6r^3}{6} = \frac{384}{6}\)

\(=> r^3 = 64\)

\(=> r = \sqrt[3]{64} = 4\)

Sum of first 10 terms \(= \frac{6(4^{10}-1)}{4 - 1}\)

                                     \(= \frac{6(1048576 - 1)}{3}\)

                                     \(= 2 \times 1048575\)

                                     \(= 2097150\)

The square of a binomial, the value of the constant \(c\) should be equal to half the coefficient of the linear term squared, which in this case is \(c = \left(\frac{22}{2}\right)^2 = 121\). Therefore, the constant that needs to be filled in is 121.

To express the quadratic expression \(x^2 + 22x + c\) as the square of a binomial, we need to find a binomial of the form \((x + a)^2\) that expands to \(x^2 + 22x + c\). Expanding \((x + a)^2\) gives \(x^2 + 2ax + a^2\). Comparing the coefficients of the expanded binomial and the given quadratic expression, we can equate the linear terms to find \(2ax = 22x\), which gives \(a = 11\). Substituting this value of \(a\) back into the expanded binomial, we have \(x^2 + 22x + 121\). Therefore, the constant that needs to be filled in is 121.

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4/5 * 3 = 12/5

20 * 3 = 1 hour = 60 min

Answer:

Step-by-step explanation:

(5 - 1)/2 4/2= 2

(-3 + 7)/2= 4/2= 2

(2, 2) is the midpoint

Answer:

Step-by-step explanation:

Weight of one fruit fly = \(\frac{1.3*10^{2}}{1*10^{6}}\)

                                    \(=1.3*10^{2-6}\\\\=1.3*10^{-4}\)

4)

\(1.674*10^{-27}\)     > \(1.672*10^{-27}\)    >  9.109 * \(10^{-31}\)

Electron has the least negative exponent. So, it is the lightest

To show that each function is O(x^2) using the Definitional proof, we need to find a constant c and a positive number k such that |f(x)| ≤ c|x^2| for all values of x greater than some value k.

(a) For function f(x) = x:

We need to find a constant c and k such that |x| ≤ c|x^2| for all x > k.

Let's choose c = 1 and k = 1.

|f(x)| = |x| ≤ 1 * |x^2| for all x > 1.

Therefore, f(x) = x is O(x^2).

(b) For function f(x) = 9x + 5:

We need to find a constant c and k such that |9x + 5| ≤ c|x^2| for all x > k.

Let's choose c = 14 and k = 1.

|f(x)| = |9x + 5| ≤ 14 * |x^2| for all x > 1.

Therefore, f(x) = 9x + 5 is O(x^2).

(c) For function f(x) = 2x^2 + x + 5:

We need to find a constant c and k such that |2x^2 + x + 5| ≤ c|x^2| for all x > k.

Let's choose c = 8 and k = 1.

|f(x)| = |2x^2 + x + 5| ≤ 8 * |x^2| for all x > 1.

Therefore, f(x) = 2x^2 + x + 5 is O(x^2).

(d) For function f(x) = 10x^2 + log(x):

We need to find a constant c and k such that |10x^2 + log(x)| ≤ c|x^2| for all x > k.

Let's choose c = 11 and k = 1.

|f(x)| = |10x^2 + log(x)| ≤ 11 * |x^2| for all x > 1.

Therefore, f(x) = 10x^2 + log(x) is O(x^2).

In each case, we have found a constant c and a value k such that the inequality |f(x)| ≤ c|x^2| holds for all x greater than k. This satisfies the definition of f(x) being O(x^2).

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B. Stacy states that a plane has two dimensions because it is made up of an infinite number of lines is the correct answer

So, Amy used 30 stamps of 41-cent stamps and 17 stamps of 7-cent stamps.

This is a system of equations problem. We can call the number of 41-cent stamps x and the number of 7-cent stamps y.

The two equations are:

x + y = total number of stamps used

0.41x + 0.07y = 1.79 (the total cost of the postage in dollars)

We can use the first equation to solve for one of the variables in terms of the other. For example, we can solve for y in terms of x:

y = total number of stamps used - x

Then we can substitute this into the second equation:

0.41x + 0.07(total number of stamps used - x) = 1.79

Solving for x,

x = (1.79 - 0.07*total number of stamps used) / 0.41

We can substitute this into the first equation to solve for total number of stamps used.

x + y = total number of stamps used

x + (total number of stamps used - x) = total number of stamps used

2x = 1.79/0.41 + 0.07/0.41

x = (1.79/0.41 + 0.07/0.41)/2

x = (1.79 + 0.07)/(0.41*2)

So, x = 30 stamps of 41-cent stamps

and y = (total number of stamps used - x) = (total number of stamps used - 30) = (1.79/0.48 - 30) = 17 stamps of 7-cent stamps

So, Amy used 30 stamps of 41-cent stamps and 17 stamps of 7-cent stamps.

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0.092011 is an approximation of the definite integral of x^2/(1+x^4) from 0 to 1/2 up to 6 decimal places.

In summary, to approximate the definite integral of x^2/(1+x^4) from 0 to 1/2 using a power series, we can use the geometric series formula to expand the denominator, integrate the resulting power series term by term, and then evaluate the resulting series up to a certain number of terms to obtain an approximation of the integral to a desired degree of accuracy.

To begin, we can use the formula for a geometric series to write the denominator as:

1/(1+x^4) = 1 - x^4 + x^8 - x^12 + ...

We can then multiply the numerator x^2 by this series and integrate term by term to obtain:

∫x^2/(1+x^4) dx = ∫[x^2 - x^6 + x^10 - x^14 + ...] dx

= (1/3)x^3 - (1/7)x^7 + (1/11)x^11 - (1/15)x^15 + ...

To approximate the integral up to 6 decimal places, we can evaluate this series up to a certain number of terms, say the first 10 terms:

(1/3)(1/2)^3 - (1/7)(1/2)^7 + (1/11)(1/2)^11 - (1/15)(1/2)^15 + ...

= 0.092011

This is an approximation of the definite integral of x^2/(1+x^4) from 0 to 1/2 up to 6 decimal places.

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Answer:

5/6 is the simplest form

Hope that helps

Answer: b(-2, 2) c(4,2) d (-6,7)

Step-by-step explanation:

The formula for reflecting over x axis is (x,y) - > (x,-y)

the current points are

-2,-2

4, -2

-6, -7

which turn into -2, 2

4, 2

-6,7

In this case, the scientist is studying if students' beliefs about illegal drug use change as they go through college. The null hypothesis (H0) for McNemar's test in this context would state that there is no significant change in students' beliefs about illegal drug use between the time they enter college and four years later. In other words, the proportion of students who change their beliefs about illegal drug use is not significantly different from the proportion who do not change their beliefs.

Null hypothesis (H0): There is no significant difference in students' beliefs about illegal drug use before and after going through college.

Work out the difference (increase) between the two numbers you are comparing.

Increase = New Number - Original Number.

Then: divide the increase by the original number and multiply the answer by 100.

% increase = Increase ÷ Original Number × 100.

Answer:

(x + 6)(x + 7)

Step-by-step explanation:

To factor the trinomial x^2 + 13x + 42, we need to find two numbers that multiply to 42 and add up to 13.

One way to do this is to list all the pairs of factors of 42 and see which pair adds up to 13:

1, 42 -> 1 + 42 = 43

2, 21 -> 2 + 21 = 23

3, 14 -> 3 + 14 = 17

6, 7 -> 6 + 7 = 13

So the pair of factors that we want is 6 and 7. We can use these numbers to rewrite the middle term of the trinomial:

x^2 + 13x + 42 = x^2 + 6x + 7x + 42

Next, we can group the first two terms and the last two terms:

x^2 + 6x + 7x + 42 = (x^2 + 6x) + (7x + 42)

Now, we can factor out the greatest common factor from each group:

x(x + 6) + 7(x + 6)

Notice that we have a common factor of (x + 6) in both terms. We can factor this out:

(x + 6)(x + 7)

Answer: Area: (100+75π/2)cm^2 Perimeter: (10+75π/2)cm

Step-by-step explanation:

Area of total shape

= square+3 semi-circles

=100cm^2 + 5*5πcm^2*3/2 = 100cm^2+75/2*πcm^2

= (100+75π/2)cm^2

Perimeter of shape

= 10cm + 25/2π*3cm

= (10+75π/2)cm

Step-by-step explanation:

in this figure you can observe 1 cube and 3 half cirles

the flield of the square would by using a formula a(lenght) * b(height)

in this case it would be 10m * 10m = 100m

and the formula for a cirle would be

A = pir^2

knowing that we can connect two of the halfs and make a cirle with a radius of 5

A= pi 5^2

A= pi 25

25pi approximatly equals to 78.53982m

knowing that information we can devide that value by 2 and get the value of the 3rd half cirlce

which is 39.26991m

than we add all the values we collected

39.26991m + 78.53982m + 100m = 217.80973 m

Answer:

This does not factor

Step-by-step explanation:

2x^2-17x-35=0

Since the x^2 has a coefficient of 2

(2x    ) ( x     )

Looking at the factors of -35

-1 35       1, -35

-5   7       5, -7

We need to  get to -17

You cannot factor this

-

The cost of goods sold for the period is $748,400.By accurately calculating the cost of goods sold, businesses can make informed decisions regarding pricing, inventory management, and overall financial performance.

To calculate the cost of goods sold (COGS), we add up the values provided in the data:

COGS = $246,800 + $242,800 + $258,800

= $748,400

The cost of goods sold for the period, based on the given data, is $748,400. The COGS represents the direct costs incurred by a company to produce the goods sold during a specific period. It includes the cost of raw materials, direct labor, and other direct production expenses.

Accurately calculating the COGS is crucial for financial analysis and decision-making. By subtracting the COGS from the total sales revenue, we obtain the gross profit, which is a key indicator of a company's profitability. The COGS also helps determine the company's gross margin, which indicates the percentage of revenue that remains after accounting for the cost of producing goods.

For effective inventory management, it is important for businesses to monitor and control their COGS. It allows them to assess the efficiency of their production processes, identify cost-saving opportunities, and evaluate pricing strategies. Additionally, tracking the COGS helps businesses determine the profitability of specific product lines or services.

The COGS provides valuable insights into a company's operational efficiency and profitability.

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To show that the wave function y=e^(b(x-vt)) is a solution of the linear wave equation (eq. 16.27), we need to plug the wave function into the equation and see if it satisfies the equation.

The linear wave equation (eq. 16.27) is given by:

∂^2y/∂x^2 = (1/v^2) ∂^2y/∂t^2

Plugging in the wave function y=e^(b(x-vt)) into the equation, we get:

∂^2(e^(b(x-vt)))/∂x^2 = (1/v^2) ∂^2(e^(b(x-vt)))/∂t^2

Taking the second derivative with respect to x, we get:

b^2 e^(b(x-vt)) = (1/v^2) ∂^2(e^(b(x-vt)))/∂t^2

Taking the second derivative with respect to t, we get:

b^2 e^(b(x-vt)) = (1/v^2) b^2 v^2 e^(b(x-vt))

Simplifying, we get:

b^2 e^(b(x-vt)) = b^2 e^(b(x-vt))

This shows that the wave function y=e^(b(x-vt)) satisfies the linear wave equation (eq. 16.27) and is therefore a solution of the equation.

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Answer:

70 degrees total degree 540

Based on the drawing obtained from the description, the location of

Manu's farm is located relatively south to Dogbe's farm.

Reasons:

The bearing of the cottage to Dogbe's farm = 200°

The distance Dogbe walks from the cottage to his farm = 5 km

The bearing of the cottage from Manu's farm = 110°

The distance Manu walks from the cottage to his farm = 3 km

Required:

The distance and between the two farms.

Solution:

Please find attached a drawing showing the position of the cottage

By cosine rule, we have;

Where;

b = The distance Dogbe walks from the cottage to his farm = 5 km

c = The distance Manu walks to his farm from the cottage = 3 km

a = The distance between the two farms = d

A = The angle between the paths to the cottage from the farms = 50°

By plugging in the values, we have;

d² = 5² + 3² - 2 × 5 × 3 × cos(50°)

d = √(5² + 3² - 2 × 5 × 3 × cos(50°)) ≈ 3.8

Required:

The bearing of Manu's farm from Dogbe's

Solution:

By sine rule, we have;

Which gives;

\(\displaystyle sin(C) = \mathbf{\frac{3 \times sin(50^{\circ})}{\sqrt{5^2 + 3^2 - 2 \times 5 \times 3 \times cos(50^{\circ})} }}\)

\(\displaystyle \angle C = arcsine \left(\frac{3 \times sin(50^{\circ})}{\sqrt{5^2 + 3^2 - 2 \times 5 \times 3 \times cos(50^{\circ})} } \right) \approx 38 ^{\circ}\)

The bearing of Manu's farm from Dogbe's = 200° + ∠C

Therefore;

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Answer:

98%

Step-by-step explanation:

83%+91%+88%+?=90×4

262+?=360

?=360-262

?=98%

Answer:

98% i believe

Step-by-step explanation:

Graphing the equation 3x - 2y = -6 over the range x = -10 to x = 10:

To graph the equation 3x - 2y = -6, we need to rearrange it in the form y = mx + b, where m is the slope and b is the y-intercept.

3x - 2y = -6

-2y = -3x - 6

Divide both sides by -2:

y = (3/2)x + 3

Now we have the equation in slope-intercept form.

To graph the equation, we can plot a few points and draw a line through them. Let's choose some x-values from the range -10 to 10 and find the corresponding y-values.

For x = -10:

y = (3/2)(-10) + 3

y = -15 + 3

y = -12

For x = 0:

y = (3/2)(0) + 3

y = 0 + 3

y = 3

For x = 10:

y = (3/2)(10) + 3

y = 15 + 3

y = 18

Plotting these points (-10, -12), (0, 3), and (10, 18) on the graph and drawing a line through them, we get the graph of the equation 3x - 2y = -6.

Using the graphical method to solve the pair of equations:

The given equations are:

10x = 5y

-3x + y = 1

To solve these equations graphically, we need to plot their graphs on the same coordinate plane and find the point where they intersect, which represents the solution.

Rearranging the second equation in slope-intercept form:

y = 3x + 1

Now we have the equations in the form y = mx + b.

Plotting the graphs of the equations 10x = 5y and y = 3x + 1, we can find the point of intersection, which represents the solution to the system of equations.

The point of intersection is the solution to the system of equations.

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Answer:

oh.. maybe you can install photomath

Answer:

y=(x+4)^2

Step-by-step explanation:

A trick I use is if it is a plus sign then it goes left, if it is a minus sign then it goes right.

9514 1404 393

Answer:

  73.4°

Step-by-step explanation:

The relevant trig relation is ...

  Sin = Opposite/Hypotenuse

  sin(x°) = 23/24

  x° = arcsin(23/24) ≈ 73.4°

Answer:

Step-by-step explanation:

To simplify the expression (√5-3√2)^2, we can use the formula for squaring a binomial:

(a - b)^2 = a^2 - 2ab + b^2

In this case, we have:

a = √5

b = 3√2

So we can substitute these values into the formula:

(√5 - 3√2)^2 = (√5)^2 - 2(√5)(3√2) + (3√2)^2

Simplifying, we get:

(√5 - 3√2)^2 = 5 - 6√10 + 18

= 23 - 6√10

Therefore, (√5 - 3√2)^2 simplifies to 23 - 6√10.

Answer: 23 - 6√10

Step-by-step explanation:

(√5 - 3√2)²   ⇒   5 - 6√10 + 18  ⇒ 5 - 6√10 + 18  ⇒  23 - 6√10

To minimize the cost of the package:

Radius = 3.86 cm.

Height = 9.62 cm.

Minimum cost = 14.01 cents.

In Mathematics and Geometry, the volume of a cylinder can be calculated by using the following formula:

Volume of a cylinder, V = πr²h

Where:

By substituting the given parameters into the formula for the volume of a cylinder, we have the following;

450 = πr²h

h = 450/πr²

For the surface area of this cylinder, we have:

SA = 2πrh + 2πr²

By substituting the height (h) in terms of r, we have:

SA = 2πr(450/πr²) + 2πr²

SA = 900/r + 2πr²

Assuming the radius of the paper top is denoted by r, and the height of this cylinder is either greater than or equal to the radius of the paper top, so that the top can be completely covered with paper. Therefore, the cost of the container is given by:

Cost, C = 0.02(2πrh + πr²) + 0.08(πr²)

Cost, C = 0.04πr(900/πr²) + 0.02πr² + 0.08πr²

Cost, C = 36/r + 0.02πr² + 0.08πr²

Cost, C = 36/r + 0.1πr²

In order to minimize the cost, we have to find the values of r that minimizes the cost function, C, by taking the partial derivatives of C with respect to r, and then equate to zero:

dC/dr = -36/r² + 0.2πr = 0

36/r² = 0.2πr

r³ = 36/0.2π

r = ∛(36/0.628)

r = ∛(57.3248)

r = 3.86 cm.

h = 450/πr²

h = 450/(3.14 × 3.86²)

Height, h = 9.62 cm.

Minimum cost, C = 36/3.86 + 0.1π(3.86)²

Minimum cost, C = 9.33 + 4.68

Minimum cost, C = 14.01 cents.

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The alpha level for each hypothesis test made on the same set of data is called B. experimentwise alpha

When numerous suppositions are examined concurrently, the likelihood of committing at least one type I mistake grows.

In order to manage the probability of erroneously rejecting the null hypothesis in all tests, scientists usually modify the alpha level for each test, with the purpose of maintaining an experimentwise alpha that reflects the probability of making a type I error in the entire set of tests.

The Bonferroni procedure is a technique utilized to regulate the experimentwise error rate by adjusting the alpha level for each hypothesis test.

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Answer:

The value of x is dependant on the value of y.

Step-by-step explanation:

y = (number)x + (another number)

I need a PNG or  picture to give you a factual answer to the value of X.

The circulation integral of F around the given circle is 2π. To compute the circulation integral using the tangential vector form of Green's Theorem, we first need to parameterize the circle C.

The given circle has the equation x^2 + y^2 = 1, which can be parameterized as follows:

x = cos(t)

y = sin(t)

where t is the parameter ranging from 0 to 2π.

Next, we compute the tangential vector for the parameterization:

r(t) = cos(t)i + sin(t)j

Taking the derivative of r(t) with respect to t, we get:

r'(t) = -sin(t)i + cos(t)j

Now, we can compute the circulation integral using the formula:

∮C F · dr = ∫(F · T) ds

where F is the given vector field, T is the tangential vector, and ds is the differential arc length.

Plugging in the values, we have:

F · T = (-1 yi + 1 xj) · (-sin(t)i + cos(t)j) = -sin(t)y + cos(t)x

ds = ||r'(t)|| dt = dt

Now, we integrate over the parameter t from 0 to 2π:

∫[0 to 2π] (-sin(t)y + cos(t)x) dt

= ∫[0 to 2π] (-sin(t)sin(t) + cos(t)cos(t)) dt

= ∫[0 to 2π] (-sin^2(t) + cos^2(t)) dt

= ∫[0 to 2π] (1) dt

= [t] from 0 to 2π

= 2π

Therefore, the circulation integral of F around the given circle is 2π.

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