What is the only way to increase power of a significance test when lowering the probability of committing a Type I error

Answer: 192

Step-by-step explanation:

2×8=16

2×6=12

17×12=192

So, the answer is 192.

Answer:

x+10

Step-by-step explanation:

x is any number and sum means addition, so x(number) +(sum) 10(10)

The correct option is (A). The solution of the equation is x=48.125.

Given an equation [tex]x+\frac{x}{7}+\frac{1}{11}\left(x+\frac{x}{7}\right)=60[/tex].

A statement that uses the equal sign to show that two expressions have the same value is called an equation. The equation is represented as ax+by=c.

Firstly, we will apply the distributive property to the given equation a×(b+c)=ab+ac, we get

[tex]x+\frac{x}{7}+\frac{x}{11}+\frac{x}{77}=60[/tex]

Now, we will simplify the right-hand side equation by taking LCM, we get

[tex]\begin{aligned}\frac{77x+11x+7x+x}{77}&=60\\ \frac{96x}{77}&=60\end[/tex]

Further, we will multiply both sides with 77, we get

77×(96x/77)=77×60

96x=4620

Then, we will divide both sides with 96, we get

96x/96=4620/96

x=48.125

Hence, the solution of the given equation [tex]x+\frac{x}{7}+\frac{x}{11}+\frac{x}{77}=60[/tex] is x=48.125.

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Answer:

The correct option is (A). The solution of the equation is x=48.125.

Given an equation .

A statement that uses the equal sign to show that two expressions have the same value is called an equation. The equation is represented as ax+by=c.

Firstly, we will apply the distributive property to the given equation a×(b+c)=ab+ac, we get

Now, we will simplify the right-hand side equation by taking LCM, we get

Further, we will multiply both sides with 77, we get

77×(96x/77)=77×60

96x=4620

Then, we will divide both sides with 96, we get

96x/96=4620/96

x=48.125

Hence, the solution of the given equation is x=48.125.

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Step-by-step explanation:

The perimeter of the entire figure is 62.8 cm.

Given, that the diameter of the inner semicircles is 9 cm and the width between the outer and inner semicircles is 2 cm.

The radius of the inner semicircle =4.5 cm and the radius of outer semicircle =5.5 cm (∵Diameter=9+2=11 cm)

We need to find the perimeter of the entire figure.

A perimeter is a closed path that encompasses, surrounds, or outlines either a two-dimensional shape or a one-dimensional length. The perimeter of a circle or an ellipse is called its circumference.

We know that, the circumference of a semicircle=πr and the circumference of two semicircles=2πr

Thus, the circumference of inner semicircles=2×3.14×4.5=28.26 cm

The circumference of outer semicircles=2×3.14×5.5=34.54 cm

The perimeter of the entire figure=28.26+34.54=62.8 cm

Therefore, the perimeter of the entire figure is 62.8 cm.

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The perimeter of whole path is 12.56 cm

Perimeter is the distance around the edge of a shape.

r1= 4.5cm, r2= 6.5cm

Perimeter of semicircle 1,

=πr1

= 3. 14 * 4.5

= 14.13 cm

Perimeter of semicircle 2

=πr2

= 3. 14 * 6.5

= 20.41 cm

Perimeter of path = 20.41 - 14.13 =6.28 cm

Perimeter of whole path = 2 * 6.28 =12.56

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Answer:

H (3x,3y)

Step-by-step explanation:

Dilation is a transformation when figure is enlarged or reduced by scale factor. During dilation, every point of preimage is dilated by scale factor, so it means each coordinate of that point must be multiplied by scale factor.

Hence, H(3x,3y) can be the result of dilation with scale factor of 3.

Answer:

78 m²

Step-by-step explanation:

V = bh

V = 13 m × 6 m

V = 78 m²

The area of the playground of dimensions 18 and 240is 432 [tex]m^{2}[/tex].

Given  If the dimensions of the playground in the layout are given by 180 cm x 240 cm.

The area of rectangle is length *breadth.

The given playground is in the shape of rectangle because there are two sides given and both the sides are different. In a rectangle length of opposite side found to be equal and each side form an angle of 90 degree.

Area=length*breadth

=180*240

=43200 [tex]cm^{2}[/tex]

We have to convert in meters so we have to divide by 100.

Area=43200 [tex]cm^{2}[/tex]

=43200/100

=432 [tex]m^{2}[/tex]

Hence the area of playground having dimensions 180 cm and 240 cm is 432[tex]m^{2}[/tex].

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The attached graph represents the location of the pH values 0 and 1

The given parameters are:

pH value 0 = (1, 0)

pH value 1 = (0.1, 1)

This means that we plot a point at the coordinate (1, 0) and another point at the coordinate (0.1, 1).

See attachment for the graph

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Complete question

The pH value is 0 at (1, 0) and 1 at (0.1, 1).

Using desmos

A. Locate, plot, and label on your graph where the pH value is 0.1

B. Locate, plot, and label on your graph where the pH value is 1.

Answer:

115

Step-by-step explanation:

this app wont be with you forever

Answer: Option (1)

Step-by-step explanation:

The domain and range of all linear functions is the set of real numbers.

The domain and range of the function is

Domain = (-∞, ∞) and Range = (-∞, ∞)

The domain of a function is the set of values that we are allowed to plug into our function. This set is the x values in a function such as f(x). The range of a function is the set of values that the function assumes. This set is the values that the function shoots out after we plug an x value in.

The range is the set of outputs of a relation or function. In other words, it's the set of possible y values. Recall that ordered pairs are of the form (x,y) so the y coordinate is listed after the x. The output is listed after the input.

Given data ,

Let the function be represented as f ( x )

Now , the value of f ( x ) is

f ( x ) = ( 3/4 )x + 5

The domain of the function is the set of all possible input values for x, which in this case is all real numbers, since there are no restrictions on the value of x that can be plugged into the function. Therefore, the domain of the function is:

Domain = (-∞, ∞)

The range of the function is the set of all possible output values that the function can produce. Since the function is a linear function of the form y = mx + b, where m is the slope and b is the y-intercept, we know that the range of the function is all real numbers, since a line can take on any y-value.

Range = (-∞, ∞)

Hence , the domain and range are set of all real numbers

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The two inequalities are:

a) y ≤ -2x + 5

b) y < 3x - 14

On the first graph, we can see that the shaded region is below the line and that the line is solid, so the inequality is of the form:

y ≤ line.

Now, remember that if the line passes through (x₁, y₁) and (x₂, y₂).

Then the slope is:

[tex]a = \frac{y_2 - y_1}{x_2 -x_1}[/tex]

We can see that the line passes through (2, 1) and (-1, 3)

Then the slope is:

[tex]a = \frac{1 - 3}{2 - 1} = -2[/tex]

So the line is:

y = -2*x + b

To find the value of b, we use the point (2, 1), then we get:

1 = -2*2 + b

1 = -4 + b

1 + 4 = b = 5

Then the first inequality is:

y ≤ -2x + 5

Now let's go to the other line equation, again the shaded region is below the line and this time the line is dashed, then:

y < line.

Two points on the line are (5, 1) and (4, -2)

Then the slope is:

[tex]a = \frac{1 - (-2)}{5 - 4} = 3[/tex]

y = 3*x + b

To find the value of b, we use the point (5, 1)

Then:

1 = 3*5 + b

1 = 15 + b

1 - 15 = b = -14

Then the inequality is:

y < 3x - 14

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Answer:

9 students

Step-by-step explanation:

First find the amount of cross-country team students, 4% * 1125 = 45

Then 20% * 45 = 9, so 9 students in the team weigh between 60 and 70 kilos.

Answer: - 2.4x^2 - 1.5x - 11.4

Step-by-step explanation:

- 4.1x^2 + 0.9x - 9.81.7x^2 - 2.4x - 1.6 = - 2.4x^2 - 1.5x - 11.4

The solutions for the given quadratic equation are:

[tex]x = 8 \pm i[/tex]

Here we have the quadratic equation:

[tex]x^2 = 16x - 65[/tex]

We can rewrite it as:

[tex]x^2 - 16x + 65 = 0[/tex]

Using Bhaskara's formulas, we get:

[tex]x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4*65} }{2} \\\\x = \frac{16 \pm \sqrt{(-4)} }{2}[/tex]

Now, remember that:

[tex]\sqrt{-1} = i[/tex]

Then we can rewrite:

[tex]x = \frac{16 \pm \sqrt{-1} *\sqrt{4} }{2}\\\\x = 8 \pm i[/tex]

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[tex]y = 3x + 6 \\ switch \: y \: and \: x[/tex]

[tex]x = 3y + 6 \\ solve \: for \: y \\ x - 6 = 3y \\ y = \frac{x - 6}{3} [/tex]

Answer:

free tokens or something  

Step-by-step explanaon

um i do not know so um bye?

a. The equation of the ellipse is x²/900 + y²/400 = 1

b. the foci of the ellipse are at (±10√5,0)

c. Find the graph in the attachment

Since the ellipse is made from a block of wood and centered at the origin, and x- axis major axis.

So, the equation of an ellipse with center at origin and x - axis major axis is

x²/a² + y²/b² = 1 where

Since the rectangular piece of wood  has a length of 60 inches and a width of 40 inches.

The length of the rectangle equals the length of the major axis

So, 2a = 60

a = 60/2

= 30

Also, the width of the rectangle equals the length of the minor axis

So, 2b = 40

b = 40/2

= 20

So, substituting a and b into the equation of the ellipse, we have

x²/a² + y²/b² = 1

x²/30² + y²/20² = 1

x²/900 + y²/400 = 1

So, the equation of the ellipse is x²/900 + y²/400 = 1

The foci of the ellipse are at (±c,0) where c² = a² - b²

= 30² - 20²

= 900 - 400

= 500

c = ±√500

c = ±10√5

So, the foci of the ellipse are at (±10√5,0)

c. Find the graph in the attachment

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The easiest way to solve a modulu problem is by

2^100/3

=3√2^100

=3√2^99×2

=2^33×3√2

The [tex]\sqrt{2}[/tex] is the reminder.

The easiest way to find the remainder that only work with specific numbers is that this can be done when one solve the division by the use of a calculator by:

First  if you answer is in decimal form, you have to subtract the whole number and later multiply the decimal value that is said to be left by the divisor of the original problem. The outcome is your remainder.

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Answer:

A 94

Step-by-step explanation:

If we take the mean of 91, 96, and 95, we find that it is equal to 94. Thus, if he scores a 94 on his last two quizzes, his average will remain a 94.

By using the area formulae for triangles and rectangles and the concept of surface area of the composite figure is equal to 1108 square inches.

The surface area is the area of all faces of a solid. In this case, we must sum the areas of seven rectangular and two triangular faces to determine the surface area:

A = 2 · (0.5) · (12 in) · (9 in) + 2 · (11 in) · (20 in) + 2 · (5 in) · (20 in) + 2 · (5 in) · (12 in) + (12 in) · (20 in)

A = 1108 in²

By using the area formulae for triangles and rectangles and the concept of surface area of the composite figure is equal to 1108 square inches.

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Step-by-step explanation:

[tex] = ( {x}^{3} ) { ({x}^{7} )}^{2} [/tex]

[tex] = {x}^{3} \times {x}^{7 \times 2} [/tex]

[tex] = {x}^{3} \times {x}^{14} [/tex]

[tex] = {x}^{3 + 14} [/tex]

[tex] = {x}^{17} [/tex]

The complement of y is 110 - 8x degrees.

Complementary angles are angles that add up to 90 degrees.

Therefore,

y = 8x - 20

The complement of y can be found as follows;

complement of y = 90 - (8x - 20)

complement of y = 90 - 8x + 20

complement of y = 110 - 8x

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Answer:

SOLUTION :-

There is theorem/proof that tells that in these types of figure the other angle is double of the previous one

Means , the required angle will be ( 36 x 2 ) i.e. 72°.

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The Prove that two non-zero vectors are collinear if and only if one vector is a scalar multiple of the other is given below.

1.  To know  collinear vectors:

∧ ⁻a ║ ⁻a

If  ⁻b = ∧ ⁻a

then |⁻b| = |∧ ⁻a|

So one can say that  line ⁻b and ⁻a are  collinear.

2.  If ⁻a and ⁻b  are collinear

Assuming |b| length  is  'μ' times of |⁻a |

Then | 'μ' ⁻a| = | 'μ' ⁻a|

So  ⁻b = 'μ' ⁻a

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The fourth proportional is found by applying the formula

The product of extremes = The product of means.

The fourth proportional is equal to bc/a and st/r

Let the fourth proportional be x then this can be written as

a:b::c:x

Applying the formula of product of extremes

ax= bc

or

x= bc/a

The fourth proportional can be obtained by multiplying the variables bc and dividing by a

Similarly

Let the fourth proportional be y then this can be written as

r:s::t:y

Applying the formula of product of extremes

ry= st

or

y= st/r

The fourth proportional can be obtained by multiplying the variables st and dividing by r

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A triangle undergoes a sequence of transformations.

First, the triangle is dilated by a scale factor of 1/3 about the origin. Then the triangle is reflected over the x-axis. Finally, the triangle is translated left 3 units and up 2 units.