What is a population parameter? a population parameter is a numerical descriptive measure of a population. give three examples. (select all that apply. )

Answer:

-8

Step-by-step explanation:

Remember pemdas

7 - 3*(10:2):1 =

7-3*5:1=

7-15:1=

7- 15=

-8

Answer:

-8

Step-by-step explanation:

7-3(10:2):1

We always solve what's in parenthesies first.

7 - 3*5:1

Multiplication and division have equal priority, so we solve that part from left to right:

7 - 15:1 =

= 7 - 15 =

= -8

The mean, median, and mode all have the same values in a symmetrical distribution. The mean is frequently chosen as the primary indicator of central tendency in these situations.

The mean, median, and mode frequently occur at the same location in a symmetrical distribution, where the values of variables exist at regular frequencies. The graph's middle can be divided into two sides that mirror one another by drawing a line through it.

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Answer:

4/10

Step-by-step explanation:

Let the numerator be 2x and the denominator is 5x. The denominator is 6 more than the numerator so if we add 6 to the numerator both numbers should be equal.

Solve:

2x+6 = 5x

6 = 3x

x = 2

2x = 4

5x = 10

Step-by-step explanation:

Let the numerator be 2x and denominator be 5x,

A.T.Q,

2x+6 = 5x

2x-5x = -6

-3x = -6

3x = 6

x = 2

So, the fraction is 4/10 = 2/5

Answer:

Let , the constant factor be x so 3 X + 5 x + 7 x = 180

Step-by-step explanation:

Find out the value of x

7x+3x+5x=180

15x=180

x=180/15

x=12

now,

     x=3 multiplied by 12=36

     x=7 multiplied by 12=84

      x=5 multiplied by 12=60

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The equivalent expression of the radical expression ∛1080 is 6∛5

The radical expression is given as

∛1080

Express 1080 as 216 * 5

∛1080 = ∛(216 * 5)

Split the factors

∛1080 = ∛216 * ∛5

Evaluate the cube root of 216

∛1080 = 6 * ∛5

Evaluate the product

∛1080 = 6∛5

Hence, the equivalent expression of the radical expression ∛1080 is 6∛5

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Hello,

[tex]3x + 4 = 8[/tex]

[tex]3x + 4 - 4 = 8 - 4[/tex]

[tex]3x = 4[/tex]

[tex] \frac{3x}{3 } = \frac{4}{3} [/tex]

[tex]x = \frac{4}{3} [/tex]

Answer:

-456 I think but I could be wrong though

Answer:

336

Step-by-step explanation:

They can place 1 of 8 frags on the bottom.

Now they have 7 flags left.

They can place 1 of 7 flags in the middle.

Now they have 6 flags left.

The can place 1 of 6 flags on top.

8 × 7 × 6 = 336

Answer:  2.5 meters

This value is exact without any rounding.

======================================================

Explanation:

The first task is to convert the mixed number 20⅚ into an improper fraction

I'll write 20⅚ as 20 & 5/6

The rule to use is a & b/c = (a*c + b)/c

So,

a & b/c = (a*c + b)/c

20 & 5/6 = (20*6+5)/6

20 & 5/6 = 125/6

This means the volume is exactly (125/6)pi cubic meters.

Plug this into the sphere volume formula and isolate the radius r like so

V = (4/3)*pi*r^3

(125/6)pi = (4/3)*pi*r^3

125/6 = (4/3)*r^3

(4/3)*r^3 = 125/6

4r^3 = 3*(125/6)

4r^3 = 125/2

r^3 = (125/2)*(1/4)

r^3 = 125/8

r = (125/8)^(1/3)

r = 5/2

r = 2.5

The correct equation is (a) because π/1 = C2/r2, [tex]C_2= 2\pi r[/tex]

The complete question is added as an attachment

The given parameters are:

d1 = 1

d2 = 2r

The circumferences of the circles are calculated as:

C = πd

This gives

C1 = π * 1 = π

C2 = π * 2r = 2πr

So, we have:

C1 = π

C2 = 2πr

and

d1 = 1

d2 = 2r

Divide both equations

[tex]\frac{C_2}{d_1} = \frac{C_2}{d_2}[/tex]

[tex]\frac{\pi}{1} = \frac{2\pi r}{2r}[/tex]

This gives

[tex]\frac{\pi}{1} = \frac{C_2}{2r}[/tex]

Hence, the correct equation is (a) because π/1 = C2/r2, [tex]C_2= 2\pi r[/tex]

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Answer:

Its A

Step-by-step explanation:

Step-by-step explanation:

3 meters = 300 cm

anyway, the ratio tells us that the whole ribbon can be seen as the combination of 3 + 2 = 5 equally long parts.

one such part is then

3 m / 5 = 0.6 m or 60 cm

the shorter piece is 2 of these parts long :

2 × 0.6 = 1.2 m or 120 cm

The area of the region cleaned by the 7-in wiper blade with the angle subtended being 92⁰ as described in the task content is; 39.32in².

According to the task content, it follows that the wiper blade is 7-in long and subtends an angle of 92°.

Consequently, the area swept by the blade in discuss can be determined by the Area of sector formula and hence, is;

Area = (92/360) × (22/7) × 7².

Area = 39.32 in².

Ultimately, it follows from the solving steps above that the area of the region cleaned by the 7-in wiper blade with the angle subtended being, 92⁰ is; Area = 39.32 in².

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The y-intercept of the linear equation that models the data indicates that:

There was 1 inch of water on the tub when the water was turned on.

A linear function is modeled by:

y = mx + b

In which:

In this problem, when x changes by 1, y changes by 2, hence the slope is of 2. Hence, since when x = 1, y = 3, when x = 0, y = 3 - 2 = 1, which means that the y-intercept is of 1, and the correct interpretation is:

There was 1 inch of water on the tub when the water was turned on.

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The standard error of the sample mean to the nearest hundredth  is  mathematically given as

S.E=0.784

The amount by which the population means deviates from a sample mean is represented by the standard error of the mean, which is more often referred to as simply the standard error.

It informs you how much the sample means would change if you were to perform an experiment using fresh samples from the same population but this time uses different samples.

The standard error is obtained by calculating the standard deviation and then dividing that number by the square root of the sample size. It determines the accuracy of a sample mean by factoring in the variation in sample means that exists from one set of data to the next.

Generally, The population's mean height is 62.95 inches, so let's use that.

Taking into account that the population has a standard deviation of 5.65 inches

The sample mean is used to calculate the standard error of the mean.

[tex]SE=\frac{SD}{\sqrt{n}}[/tex]

Therefore

[tex]SE=\frac{5.65}{ \sqrt{52}}[/tex]

S.E=0.7835

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Answer:

9/4

Step-by-step explanation:

We follow bodmas

4/3 +( -1/6 + 13/12)

( lcm = 12)

( -2+ 13/12)

( 11/ 12)

4/3 + ( 11/12)

4/3 + 11/12

lcm = 12 also

and that will equal to

=16 + 11/ 12

= 27/ 12

divide by 3 to simplest form

= 9/4

The product of [tex](x - 8)^2[/tex] is not a polynomial equation because a polynomial equation means a coefficient of x that has a power of two or more powers.

The equations developed with variables, exponents, and coefficients exists named polynomial equations. It can include various exponents, where the higher one stands named the degree of the equation.

Closure property under multiplication notes that any two rational numbers' outcome will be a rational number, i.e. if a and b exist in any two rational numbers, ab will also be a rational number.

Example: (3/2) × (2/9) = 1/3.

The product of [tex]$(x - 8)^2[/tex]

Simplifying the equation as (x − 8)(x − 8)

Apply Perfect Square Formula, and we get

[tex]$(a-b)^{2}=a^{2}-2 a b+b^{2}$[/tex]

[tex]$&(x-8)^{2}=x^{2}-2 x \cdot 8+8^{2} \\[/tex]

[tex]$&=x^{2}-2 x \cdot 8+8^{2}[/tex]

Simplifying the above equation, we get

[tex]$x^{2}-2 x \cdot 8+8^{2}=x^{2}-16 x+64 \\[/tex]

[tex]$&=x^{2}-16 x+64[/tex]

This is not a polynomial equation because a polynomial equation means a coefficient of x that has a power of two or more powers.

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The terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].

In this question,

The function is f(x) = [tex]\frac{sin(x)}{1-x}[/tex]

The general form of Maclaurin series is

[tex]\sum \limits^\infty_{k:0} \frac{f^{k}(0) }{k!}(x-0)^{k} = f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} +\frac{f'''(0)}{3!}x^{3}+......[/tex]

To find the Maclaurin series, let us split the terms as

[tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] ------- (1)

Now, consider f(x) =  sin(x)

Then, the derivatives of f(x) with respect to x, we get

f'(x) = cos(x), f'(0) = 1

f''(x) = -sin(x), f'(0) = 0

f'''(x) = -cos(x), f'(0) = -1

[tex]f^{iv}(x)[/tex] = cos(x), f'(0) = 0

Maclaurin series for sin(x) becomes,

[tex]f(x) = 0 +\frac{1}{1!}x +0+(-\frac{1}{3!} )x^{3} +....[/tex]

⇒ [tex]f(x)=x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+.....[/tex]

Now, consider [tex]f(x) = (1-x)^{-1}[/tex]

Then, the derivatives of f(x) with respect to x, we get

[tex]f'(x) = (1-x)^{-2}, f'(0) = 1[/tex]

[tex]f''(x) = 2(1-x)^{-3}, f''(0) = 2[/tex]

[tex]f'''(x) = 6(1-x)^{-4}, f'''(0) = 6[/tex]

[tex]f^{iv} (x) = 24(1-x)^{-5}, f^{iv}(0) = 24[/tex]

Maclaurin series for (1-x)^-1 becomes,

[tex]f(x) = 1 +\frac{1}{1!}x +\frac{2}{2!}x^{2} +(\frac{6}{3!} )x^{3} +....[/tex]

⇒ [tex]f(x)=1+x+x^{2} +x^{3} +......[/tex]

Thus the Maclaurin series for [tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] is

⇒ [tex]f(x)=(x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+..... )(1+x+x^{2} +x^{3} +......)[/tex]

⇒ [tex]f(x)=x+x^{2} +x^{3} - \frac{x^{3} }{6} +x^{4}-\frac{x^{4} }{6} +.....[/tex]

⇒ [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex]

Hence we can conclude that the terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].

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Answer:

[tex]\dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{x^2}{\sqrt{25-x^2}}\:\:\text{d}x[/tex]

Rewrite 25 as 5²:

[tex]\implies \displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x[/tex]

Integration by substitution

[tex]\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}[/tex]

[tex]\textsf{Let }x=5 \sin \theta[/tex]

[tex]\begin{aligned}\implies \sqrt{5^2-x^2} & =\sqrt{5^2-(5 \sin \theta)^2}\\ & = \sqrt{25-25 \sin^2 \theta}\\ & = \sqrt{25(1-\sin^2 \theta)}\\ & = \sqrt{25 \cos^2 \theta}\\ & = 5 \cos \theta\end{aligned}[/tex]

Find the derivative of x and rewrite it so that dx is on its own:

[tex]\implies \dfrac{\text{d}x}{\text{d}\theta}=5 \cos \theta[/tex]

[tex]\implies \text{d}x=5 \cos \theta\:\:\text{d}\theta[/tex]

Substitute everything into the original integral:

[tex]\begin{aligned}\displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x & = \int \dfrac{25 \sin^2 \theta}{5 \cos \theta}\:\:5 \cos \theta\:\:\text{d}\theta \\\\ & = \int 25 \sin^2 \theta\end{aligned}[/tex]

Take out the constant:

[tex]\implies \displaystyle 25 \int \sin^2 \theta\:\:\text{d}\theta[/tex]

[tex]\textsf{Use the trigonometric identity}: \quad \cos (2 \theta)=1 - 2 \sin^2 \theta[/tex]

[tex]\implies \displaystyle 25 \int \dfrac{1}{2}(1-\cos 2 \theta)\:\:\text{d}\theta[/tex]

[tex]\implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating $\cos kx$}\\\\$\displaystyle \int \cos kx\:\text{d}x=\dfrac{1}{k} \sin kx\:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\begin{aligned} \implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta & =\dfrac{25}{2}\left[\theta-\dfrac{1}{2} \sin 2\theta \right]\:+\text{C}\\\\ & = \dfrac{25}{2} \theta-\dfrac{25}{4}\sin 2\theta + \text{C}\end{aligned}[/tex]

[tex]\textsf{Use the trigonometric identity}: \quad \sin (2 \theta)= 2 \sin \theta \cos \theta[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{25}{4}(2 \sin \theta \cos \theta) + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{25}{2}\sin \theta \cos \theta + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\sin \theta \cdot 5 \cos \theta + \text{C}[/tex]

[tex]\textsf{Substitute back in } \sin \theta=\dfrac{x}{5} \textsf{ and }5 \cos \theta = \sqrt{25-x^2}:[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\cdot \dfrac{x}{5} \cdot \sqrt{25-x^2} + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}[/tex]

[tex]\textsf{Substitute back in } \theta=\arcsin \left(\dfrac{x}{5}\right) :[/tex]

[tex]\implies \dfrac{25}{2} \arcsin \left(\dfrac{x}{5}\right) -\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}[/tex]

Take out the common factor 1/2:

[tex]\implies \dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}[/tex]

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The type of model that best fits the given situation is; A linear equation  Model

Right inside the local reservoir we will have an initial amount of water A.

Now, for every hour that passes by, the amount of water in the reservoir decreases by 500 gals.

Thus, after t hours, the amount of water in the reservoir is expressed as:

W = A - 500gal * t

This is clearly a linear equation model and so we can conclude that the model that fits best in the given situation is a linear model.

The domain of this model is restricted because we can't have a negative amount of water in the reservoir,  and as such the maximum value of t accepted is: W = 0 = A - 500gal*t

t = A/500 hours

Therefore, the domain of this linear relation is: t ∈ {0h, A/500 }

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Answer:

25.12 rounded.

Step-by-step explanation:

C = 2[tex]\pi[/tex]r

C=[tex]\pi[/tex]d

C = [tex]\pi[/tex](8)

C = 25.12 rounded.

Answer:

25 and 1/7 mm

Step-by-step explanation:

Diameter = circunference / π

π = 3.1416    aprox.

Then:

8 = citcunference / 3.1416

8 * 3.1416 = circunference

25.1428 = circunference

1/7 = 0.1428

Then:

25.1428 = 25 + 1/7

= 25 1/7

Answer:

109°

Step-by-step explanation:

The sum of angles in all triangles is equal to 180. For this given triangle, we can solve for x.

48° + 23° + x° = 180°
180° - 71° = x°
x = 109°

Answer:

First and second angles going upward

Answer: 2, 4, 6

Step-by-step explanation:

The distance the bird has flown by the time Alice and Bob meet is 40 feet.

Given that the distance between Alice and Bob is 1000 feet and their running speed is 10 feet per second and the speed of bird is 20 feet per second.

Distance equals speed multiplied by time.

Distance between Alice and Bob=1000 feet.

Distance between the bird and Bob=1000 feet.

Speed of Alice and Bob=10 feet per second.

The combined speed of Alice and Bob=20 feet per second.

Since the two are running directly toward each other the distance each will cover at the meeting point is 50 feet (1000/20)

The time covered at the meeting point=20 second (1000/50)

Speed of the bird=20 feet per second.

The distance covered by the bird towards Bob at their meeting point is 40 feet(20 feet*20 seconds).

Hence the distance the bird has flown by the time Alice and Bob meet is 40 feet.

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Type I error says that we suppose that the null hypothesis exists rejected when in reality the null hypothesis was actually true.

Type II error says that we suppose that the null hypothesis exists taken when in fact the null hypothesis stood actually false.

In statistics, a Type I error exists as a false positive conclusion, while a Type II error exists as a false negative conclusion.

Making a statistical conclusion still applies uncertainties, so the risks of creating these errors exist unavoidable in hypothesis testing.

The probability of creating a Type I error exists at the significance level, or alpha (α), while the probability of making a Type II error exists at beta (β). These risks can be minimized through careful planning in your analysis design.

Examples of Type I and Type II error

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Answer:

6x³ + 14x² + 6x - 12

Step-by-step explanation:

(3x - 2)(2x² + 6x + 6)

6x³ - 4x² + 18x² - 12x + 18x - 12

6x³ + 14x² + 6x - 12

Answer: 6x³+14x²+6x-12

Step-by-step explanation:

Here, we can use the distributive property to multiply one of the factors by each term in the other. Below, I multiplied [tex]2x^2 + 6x + 6[/tex] by both 3x and -2.

[tex](2x^2+ 6x + 6) (3x - 2)\\3x(2x^2+ 6x + 6)-2(2x^2+ 6x + 6)\\6x^3+18x^2+18x-4x^2-12x-12[/tex]

Now, we can rearrange so that like terms are next to each other. Then, we can combine them to simplify.

[tex]6x^3+18x^2+18x-4x^2-12x-12\\6x^3+18x^2-4x^2+18x-12x-12\\6x^3+14x^2+6x-12[/tex]

Option ( C ) is correct for this expression . 100 inches Over 1 minute times × 1 foot Over 12 inches.

What is a basic expression?

Given that the expression 100 inches.

We need to convert 100 inches per minute to feet per minute.

Since, we know that 1ft = 12 inch

Then,

1 in = 1/12 ft

Now, we shall convert 100 inches per minute to feet per minute.

To convert in/min to ft/min, let us multiply by 1/12

Thus, we have,

[tex]\frac{100 in}{min} * \frac{1 ft }{12 in}[/tex]

Therefore, option ( c ) is correct for this expression .

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The complete question is -

Which expression converts 100 inches per minute to feet per minute?

A) Start Fraction #1 100 inches Over 1 minute End Fraction × Start Fraction 60 minutes Over 1 hour End Fraction

B) Start Fraction #2 100 inches Over 1 minute End Fraction × Start Fraction 1 hour Over 60 minutes End Fraction

C) 100 inches Over 1 minute times × Start Fraction 1 foot Over 12 inches End Fraction

D) 100 inches Over 1 minute times × Start Fraction 12 inches Over 1 foot End Fraction

Answer:

The length of the diagonal of a rectangle is 10m

Step-by-step explanation:

There is a visual below,

The diagonal of a rectangle has formed 2 right triangles.

This diagonal is called hypotenuse side length of a triangle

To find the hypotenuse, we can use the Pythagorean Theorem

a² + b² = c²

we are given the length of a and b, where a = 6 and b = 8

(6)² + (8)² = c²

36 + 64 = c²

100 = c²

√100 = c

10 = c

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Answer:

K(6, 9)

Step-by-step explanation:

Let the K coordinates be (x, y)

Mid-point formula:

[tex]\sf (x_m, y_m) = (\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2})[/tex]

Applying formula:

[tex]\sf (-3,\:-1) = (\:\dfrac{-12+x}{2} ,\: \dfrac{-11+y}{2} )[/tex]

Comparing expression:

[tex]\sf \dfrac{-12+x}{2} = -3 \quad and \quad \dfrac{-11+y}{2} = -1[/tex]

[tex]\sf -12+x = 2(-3) \quad and \quad -11+y = 2(-1)[/tex]

[tex]\sf x = -6 + 12 \quad and \quad y = -2 + 11[/tex]

[tex]\sf x = 6 \quad and \quad y = 9[/tex]

So, coordinates of K is (6, 9)