i need help with this ;n;​

a) 1. The backpack weight sample mean is 6.375 pounds.

a) 2. The backpack weight sample mean is 6.375 pounds.

a) 3. The backpack weight sample mean is 6.625 pounds.

b) The range of the sample means is between 6.375 and 6.625, which approaches 0.250 pounds, plus or minus.

c) Since the students are using the sample means to estimate the mean backpack weight of their schoolmates, the true statements are A. and D.

The farther the range of the sample means is from O, the less confident they can be in their estimate.

A sample mean is a statistic computed from a sample of data or variables.

The sample mean represents the average value of the sample data.

The range describes the difference between the largest number and the smallest number.

Thus, based on the range of the sample means, the farther it is from 0, the less confident the students can be in their estimate.

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The equation that represents the function g(x) is g(x) = log₅(x - 3)

The equation of the function f(x) is given as:

f(x) = log₅(x)

From the graph, we can see that the function g(x) is 3 units to the right of the function f(x)

This means that

g(x) = f(x - 3)

The function f(x - 3) is calculated as follows

f(x - 3) = log₅(x - 3)

Substitute f(x - 3) = log₅(x - 3) in g(x) = f(x - 3)

g(x) = log₅(x - 3)

Hence, the equation that represents the function g(x) is g(x) = log₅(x - 3)

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The shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet

The given parameter is

Perimeter, P = 520 feet

Represent the shorter side with x and the longer side with y

One side of the garden is bordered by a river:

So the perimeter is:

P = 2x + y

Substitute P = 520

2x + y = 520

Make y the subject

y = 520 - 2x

The area is

A = xy

Substitute y = 520 - 2x in A = xy

A = x(520 - 2x)

Expand

A = 520x - 2x^2

Differentiate

A' = 520 - 4x

Set to 0

520 - 4x = 0

Rewrite as:

4x= 520

Divide by 4

x= 130

Substitute x= 130 in y = 520 - 2x

y = 520 - 2 *130

Evaluate

y = 260

The area is then calculated as:

A = xy

This gives

A = 130 * 260

Evaluate

A = 33800

Hence, the shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet

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Answer:

23 children and 9 adults

p <= 70/9

Step-by-step explanation:

Let x = # of adults and y = # of children. The number of children is 5 more than twice the number of adults, so x = 2y + 5. The total cost of entrance is $6525, so 225x + 150y = 6525.

x = 2y + 5

225x + 150y = 6525

x - 2y = 5

225x + 150y = 6525

75x - 150y = 375

225x + 150y = 6525

300x = 6900

x = 23

x = 2y + 5

23 = 2y + 5

18 = 2y

y = 9

225p + 3750 <= 5500

225p <= 1750

p <= 70/9

Using the combination formula, it is found that Julia can take 15 combinations.

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

For this problem, 4 students are taken from a set of 6, hence the number of combinations is given as follows:

[tex]C_{6,4} = \frac{6!}{4!2!} = 15[/tex]

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Answer:

Step-by-step explanation:

answer is 1/2

yes it will be a rational number.

square having area 400 sq units will have a length of 20 on each side by increasing the length of each side by 1 we get 21 then the area will be

21 ^ 2=441

Then we will have a rational number.

Answer:

○ [tex]F = \frac{S + 24}{3}[/tex]

Step-by-step explanation:

We know that:

[tex]S = 3F -24[/tex],

where F is the length of the foot of a person, and S is their shoe size.

To solve for the length of a person's foot, we have to rearrange the given equation to make F the subject:

[tex]S = 3F -24[/tex]

⇒ [tex]S + 24 = 3F[/tex]     [adding 24 to both sides]

⇒ [tex]\frac{S + 24}{3} = F[/tex]          [dividing both sides by3]

⇒ [tex]F = \frac{S + 24}{3}[/tex]          [swapping the sides]

Answer:

1) 9.69

2) 87.36

3) 171.42

4) 18.75

5) 16.29

Step-by-step explanation:

Treat each value's full price as x. then, all of the values are (2/3)x. To get (2/3)x back to x, we must multiply (2/3)x*(3/2)=x. Multiply each of the questions given by 3/2 or 1.5

Answer:

Below in bold.

Step-by-step explanation:

This figure is a trapezoid.

We can find the missing side  length by applying the Pythagoras Theorem to the right triangle on the left of the figure.

The dotted line has length 12 cm and the base of the triangle = 25 - 18 = 7 cm

So x^2  = 12^2 + 7^2     (where x = missing length)

            =  144 + 49

            = 189

x = √189

  =  13.75 to nearest hundredth

So the perimeter

  =    18 + 12 + 25 + 13.75

  =    68.75 cm.

Area of a trapezoid = 1/2 (a + b) * h.

Area =  1/2 *(18 + 25) * 12

        =  258 cm^2.

Answer:

2/5 or 0.4

Step-by-step explanation:

First we have to make equation into the gradient slope form :

y=mx+c    m is gradient and c is y-intercept

15y-6x=8

First we add 6x to both sides :

15y = 6x+8

Now we divide both sides by 15 :

y = 6x+8 / 15

Split the fraction on the right :

y = 6x/15 + 8/15

6x/15 = 6/15x

Our gradient will be 6/15 = 2/5 or 0.4

Hope this helped and have a good day

The option c is correct. The linear representation model is best for this situation out of the exponential, quadratic model.

According to the statement

we have given that:

Monthly Rate = $20, Number of customers = 5000

If there is a decrease of $1 in the monthly rate, the number of customers increase by 500.

And we have to find that The type of model that best fits the given situation.

So, according to the linear representation model

Monthly Rate = $20, Number of customers = 5000

Let us decrease the monthly rate by $1.

Monthly Rate = $20 - $1  = $19, Number of customers = 5000 + 500 = 5500

Let us decrease the monthly rate by $1 more.

Monthly Rate = $19 - $1  = $18, Number of customers = 5500 + 500 = 6000

Here, we can see that there is a linear change in the number of customers whenever there is decrease in the monthly rate.

We have 2 pair of values here,

x = 20, y = 5000

x = 19, y = 5500

Let us write the equation in slope intercept form:

y = mx + c

And Slope of a function is

m = y₂ - y₁ / x₂ - x₁

Then

m = 5500 - 5000 / 19 - 20

m = -500

So, the equation become

y = -500x +c

And find the value of by putting the values.

So, c = 15000

And the equation become

y = -500x + 15000.

According to this linear model is perfect for this situation.

So, The linear representation model is best for this situation.

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A one-tailed test's p-value is half that of a two-tailed test.

Whether it is positive or negative, when you find your test statistic, you also find a tail probability to the right or left of that test statistic. What is the likelihood that, if the null hypothesis is correct, you will obtain a value that is more extreme (far out in the tail) than the observed test statistic? If the test is two-tailed, the phrase "probability...more extreme" applies to both tails, beyond the positive and negative values of your test statistic. Your calculations will fall either on the right or left, but if the test is two-tailed, it will also apply to both tails.

Two possibilities sharing similarities is the null hypothesis. That the observed discrepancy is solely the result of chance is the null hypothesis. Calculating the probability that the null hypothesis is correct using statistical testing is achievable.

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The equations intersect at the point (3 ,3 ).

An equation can be represented by a linear function. The standard form for the linear equation is: ax+b , for example, y=2x+7. Where:

a= the slope

b=constant term that represents the y-intercept.

System of linear equations is the given term math for two or more equations with the same variables. The solution of these equations represents the point at which the lines intersect.

The question gives:

4y –x = 9 (1)

y = 2x − 3 (2)

You should replace the variable y of equation 1 to equation 2. Then, you will have:

4*(2x-3)-x=9

8x-12-x=9

8x-x=9+12

7x=21

x=3

If x=3, from equation 2 you can find the value of y. Thus, y=2*3-3=6-3=3

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Check the picture below.

Make sure your calculator is in Degree mode.

The view of war doves with respect to the Vietnam War was that: B. as a localized civil war that should not include the United States of America (US).

The Vietnam War is also referred to as Second Indochina War or the Vietnam conflict and it can be defined as a short, costly, protracted and divisive conflict between the communist government of North Vietnam and South Vietnam, which ended on the 30th of April, 1975.

Based on historical information and records, the Vietnam War officially started on the 1st of November, 1955 in the following locations:

In conclusion, we can infer and logically deduce that the view of war doves with respect to the Vietnam War was that it is a a localized civil war and as such should not include the United States of America.

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The value of M is:

M = ± (1 / 2) · √[(k² + 5)² /(4 · k²) - 5] ∓ (1 / 2) · √[(k² / 100) · (25 / k² + 5)² - 5] + √3, for k ≠ 0.

In this question we have a product of two nonzero real numbers equal to 5, whose mathematical form is:

p · q = 5, where p = 2 · x + √(4 · x² + 5) = k and q = 2 · y + √(4 · y² + 5) = 5 / k

Then, we proceed to find the solutions for x and y in terms of k:

2 · x + √(4 · x² + 5) = k

2 · x = k - √(4 · x² + 5)

4 · x² = k² - 2 · k · √(4 · x² + 5) + 4 · x² + 5

2 · k · √(4 · x² + 5) = k² + 5

4 · k² · (4 · x² + 5) = (k² + 5)²

4 · x² + 5 = (k² + 5)² /(4 · k²)

4 · x² = (k² + 5)² /(4 · k²) - 5

x = ± (1 / 2) · √[(k² + 5)² /(4 · k²) - 5], for k ≠ 0.

2 · y = 5 / k - √(4 · y² + 5)

4 · y² = 25 / k² - (10 / k) · √(4 · y² + 5) + 4 · y² + 5

(10 / k) · √(4 · y² + 5) = 25 / k² + 5

(100 / k²) · (4 · y² + 5) = (25 / k² + 5)²

4 · y² + 5 = (k² / 100) · (25 / k² + 5)²

4 · y² = (k² / 100) · (25 / k² + 5)² - 5

y = ∓ (1 / 2) · √[(k² / 100) · (25 / k² + 5)² - 5], for k ≠ 0.

Therefore, the value of M is:

M = ± (1 / 2) · √[(k² + 5)² /(4 · k²) - 5] ∓ (1 / 2) · √[(k² / 100) · (25 / k² + 5)² - 5] + √3, for k ≠ 0.

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The maximum value of c=4x+2y subject to the constraints is 14

The objective function is given as:

c = 4x+2y

The constraints are given as:

x≥0 y≥0

2x+2y≤10

3x+y≤9

Rewrite 2x+2y≤10 and 3x+y≤9 as equations

2x+2y = 10

3x+y = 9

Divide 2x+2y = 10  through by 2

x+y = 5

Subtract x+y = 5 from 3x+y = 9

3x - x + y - y = 9 - 5

Evaluate the difference

2x = 4

Divide by 2

x = 2

Substitute x = 2 in x+y = 5

2+y = 5

Solve for y

y = 3

So, we have

(x, y) = (2, 3)

Substitute (x, y) = (2, 3) in c = 4x+2y

c = 4 * 2 + 2 * 3

Evaluate

c = 14

Hence, the maximum value of c=4x+2y subject to the constraints is 14

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The measure of Arc AU of the given Circle Geometry is; 50°

The circle theorem we will apply to solve for the arc measure states that “Measure of angles subtended to any point on the circumference of the circle from the same arc is equal to half of the angle subtended at the center by the same arc.”

We can express the above theorem as;

Angle at the center = 2 × Angle at the circumference

From the attached image below  and from the given statement in the question, we are given that; ∠QUA = 111°

Therefore, applying the circle theorem earlier quoted, we can say that; m∠QDA = 2 × ∠QUA

m∠QDA = 2 × 111° = 222°

Which gives;

∠QOA =  = 360° - 222° = 138°

∠QOA = ∠QOU + ∠UOA (by angle addition property)

Thus;

∠QOA = 138° = 88° + ∠UOA

∠UOA = 138° - 88° = 50°

Thus, the measure of Arc AU is;

Arc AU = ∠UOA = 50°

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Answer: b

Step-by-step explanation:

Answer:

[tex]\displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x=2(e-1)[/tex]

Step-by-step explanation:

Given definite integral:

[tex]\displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x[/tex]

Integration by substitution:

Substitute u for one of the functions to give a function that's easier to integrate.

[tex]\textsf{Let }u=\sqrt{x}[/tex]

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

[tex]\implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{2\sqrt{x}}[/tex]

[tex]\implies \text{d}x=2 \sqrt{x}\:\text{d}u}[/tex]

[tex]\implies \text{d}x=2u\:\text{d}u[/tex]

Use the substitution to change the limits of the definite integral from x-values to u-values:

[tex]\textsf{When }x=0 \implies u=\sqrt{0}=0[/tex]

[tex]\textsf{When }x=1 \implies u=\sqrt{1}=1[/tex]

Therefore, the limits are unchanged.

[tex]\boxed{\begin{minipage}{3 cm}\underline{Integrating $e^x$}\\\\$\displaystyle \int e^x\:\text{d}x=e^x+\text{C}$\end{minipage}}[/tex]

Substitute everything into the original integral and solve:

[tex]\begin{aligned}\implies \displaystyle \int^1_0 \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\:\:\text{d}x & =\displaystyle \int^1_0 \dfrac{e^u}{u} \cdot 2u\:\:\text{d}u \\\\ & = \displaystyle \int^1_0 2e^u\:\:\text{d}u \\\\ & = 2 \int^1_0 e^u\:\:\text{d}u\\\\ & = 2 \left[ e^u\right]^1_0\\\\ & = 2 \left[e^1-e^0\right]\\\\& = 2(e-1)\end{aligned}[/tex]

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The reflection of the word DAD across the y-axis is not the same

The statement is given as:

MOM, when reflected over the y-axis = MOM

The above is true because the letters M and O have reflectional symmetry across the y-axis

However, the letter D do not have a reflectional symmetry across the y-axis

So, the letters would not remain the same when reflected

Hence, the reflection of the word DAD across the y-axis is not the same

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The value added to the equation [tex]$x^2-6x=1[/tex] exists [tex]$x^2-6x+9=10[/tex].

A perfect square exists as a number that can be described as the product of an integer by itself or as the second exponent of an integer.

The perfect square trinomial exists

[tex](a[/tex] ± [tex]b)^2[/tex] = [tex]a ^2[/tex] ± 2ab + [tex]b ^2[/tex]

[tex]$x^2-6x=1[/tex]

[tex]x^2-2*3x=1[/tex]

then [tex]$2ab = 2 * 3*x = 2 * x *3[/tex]

The value of a = x and b = 3

[tex]$b^2=3^2=9[/tex]

[tex]$x^2-6x+9=1+9[/tex]

[tex]$x^2-6x+9=10[/tex]

The value added to the equation [tex]$x^2-6x=1[/tex] exists [tex]$x^2-6x+9=10[/tex].

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The correct student's t- distribution requires knowing the degrees of freedom. The degrees of freedom are there for a sample of size n is B) n-1.

Distribution is described as the technique of having items to consumers. An instance of distribution is rice being shipped from Asia to the USA. The frequency of occurrence or quantity of lifestyles.

Distribution manner to spread the product for the duration of the market such that a big quantity of humans should purchase it. Distribution involves doing the following matters. An amazing delivery machine to take the goods into one-of-a-kind geographical regions.

Distribution is one of the four elements of the advertising and marketing mix. Distribution is the method of creating a service or product available for the consumer or enterprise user who needs it. this may be executed without delay through the manufacturer or service company or the usage of indirect channels with vendors or intermediaries.

Disclaimer: The question is incomplete. Please read below to find the missing content.

Question: To select the correct Student's t-distribution requires knowing the degrees of freedom. How many degrees of freedom are there for a sample of size n?

A) n

B) n-1

C) n+1

D) [X - μ / (s / n)]

The degrees of freedom depends on the number of parameters you are estimating.

8=

i-1

In student t distribution we estimate the population mean through sample mean and population standard deviation through sample standard deviation

So here both parameters depend on the sample mean i.e. we just calculate from a sample of size n so

Degrees of freedom for students t distribution is B) n-1

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Answer: a = 24 units²

Step-by-step explanation:

      Let us graph the figure given. This shape appears to be a trapezoid. Now, we can solve for the area. This shape also has a height of 4, which I forgot to show in the picture.

       In the formula, h is the height, a is a base, and b is the other base.

            [tex]\displaystyle a=\frac{a+b}{2} h[/tex]

            [tex]\displaystyle a=\frac{4+8}{2} 4[/tex]

            [tex]\displaystyle a=\frac{12}{2} 4[/tex]

            [tex]\displaystyle a=(6)4[/tex]

            [tex]\displaystyle a=24\; \text{units}^{2}[/tex]

The correct order is 3,1,4,2.

Marshaling of assets:

Therefore, the correct order is 3,1,4,2.

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The correct question is given below:
In accordance with the marshaling of assets provision of the uniform partnership act, the correct ranking of the following liabilities of a partnership in order of payment is:

(1) $20,000 loan from a partner.

(2) $30,000 of profits from the last year of operations.

(3) $3,000 payable to a supplier.

(4) $100,000 in capital balances of the partners.

first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.

let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so

[tex]cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill[/tex]

[tex]cos(67.5^o)\implies cos\left( \frac{135^o}{2} \right)\implies \pm \sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}} \\\\\\ \sqrt{\cfrac{ ~~ \frac{2-\sqrt{2}}{2} ~~ }{2}}\implies \sqrt{\cfrac{2-\sqrt{2}}{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{2}[/tex]

The coordinates of the missing points G. and H are G(a, y) and H(a,0) respectively

A right angled triangle is a triangle that has 3 sides and angle with one of its angles to be 90 degrees.

From the given triangle, the measure of the x-coordinate point of G and H will be equal.

Since the line GH intersects the x-axis and the point where the line intersects the x-axis is where y = 0, hence the coordinate H is (a, 0) and the measure of G is (a, +y)

The y-value of G can be any positive value on the y-axis

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