Help with question b please​

The Minimum sample size table is attached below

Answer:

[tex]X=173[/tex]

Step-by-step explanation:

From the question we are told that:

Confidence Interval [tex]CI=99\%[/tex]

Variance [tex]\sigma^2=30\%[/tex]

Generally going through the table the

Minimum sample size is

[tex]X=173[/tex]

Answer:

First Choice: As the number of hours spent on homework increases, the tests scores increase.

Step-by-step explanation:

The definition of a positive correlation  is a relationship between two given variables, in which both variables are moving in the same direction. This can mean when one variable increases and the other variable increases, too, or one variable decreases and the other decreases as well.

The first choice is a positive correlation because both variables are changing (increasing) in the same direction. As you spend more time on homework, you're likely to get a higher test score.

The second choice cannot be a positive correlation because only one variable is having some kind of change (increasing). The doctor visits amount remains the same, so we can call this a zero-correlation relationship because the number of apples eaten yearly doesn't affect the amount of doctor visits. An apple a day keeps the doctor a way is just a proverb, not to be taken literally.

The third choice cannot be a positive correlation because the two variables are going different directions. Even though the number of times going to bed early is increasing, the number of times waking up late decreases, which is not moving in the same direction as the other variable.

The fourth choice cannot be a positive correlation because, similarly to the third choice, the two variables are going different directions. One variable is increasing, which is the amount of practice time. Meanwhile, the other variable is decreasing (going in the opposite direction), which is the number of games lost in a season.

Answer:

hi I am a Nepal

[tex] {233333}^{2332} [/tex]

Answer:

I think you can go with:

The margin of error is equal to half the width of the entire confidence interval.

so  try .74 ±   =   [ .724 , .756] as the confidence interval

Step-by-step explanation:

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]

Answer:

0.85 in²

Step-by-step explanation:

really ? you need help with that ? you could not find the formula for the area of a circle on the internet and type it into your calculator ? I can't do anything else here.

a circle area is

A = pi×r²

r being the radius.

and pi being, well, pi (3.1415....)

r = 0.52 in

so,

A = pi×0.52² = pi×0.2704 = 0.849486654... in²

the area of one side of the coin is 0.85 in²

Answer:

The last option

V = (-1.5,3)

other options dont lie where V is exact

V is only Exact at (-1.5,3)

Hence the time that the ball will be height than 12 feet off the ground is 4secs

Given the expression for calculating the height in  feet as;

h(t) = -4t²+16t

If the ball is higher than 12feet, h(t) > 12

Substituting h = 12 into the expression

-4t²+16t > 12

-4t²+16t - 12 > 0

4t²- 16t + 12 > 0

t²- 4t + 3 > 0

Factorize

(t²- 3t)-(t + 3) > 0

t(t-3)-1(t-3) > 0

(t-1)(t-3)>0

t > 1 and 3secs

Hence the time that the ball will be height than 12 feet off the ground is 4secs

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Answer:

123456-6-&55674

Step-by-step explanation:

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dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

Answer:

The dimensions of the rectangle are length 25 feet and width 15.92 feet

Step-by-step explanation:

Let L be the length of the rectangle and w be the width.

The area of the rectangular part of the shape is Lw while the area of the two semi-circular ends which have a diameter which equals the width of the rectangle is 2 × πw²/8 = πw²/4. The area of each semi-circle is πw²/4 ÷ 2 = πw²/8

So, the area of the shape A = Lw + πw²/4.

The perimeter of the shape, P equals the length of the semi-circular sides plus twice its length. The length of a semi-circular side is πw/2. So, both sides is 2 × πw/2 = πw

P = πw + 2L

Since the perimeter, P = 100 feet, we have

πw + 2L = 100

From this L = (100 - πw)/2

Substituting L into A, we have

A = Lw + πw²/4.

A = [(100 - πw)/2]w + πw²/4.

A = 50w - πw²/2 + πw²/4.

A = 50w - πw²/2

Now differentiating A with respect to w and equating it to zero to find the value of w which maximizes A.

So

dA/dw = d[50w - πw²/2]/dw

dA/dw = 50 - πw

50 - πw = 0

πw = 50

w = 50/π = 15.92 feet

differentiating A twice to get d²A/dw² =  - π indicating that w = 50/π is a value at which A is maximum since d²A/dw² < 0.

So, substituting w = 50/π into L, we have

L = (100 - πw)/2

L = 50 - π(50/π)/2

L = 50 - 50/2

L = 50 - 25

L = 25 feet

So, the dimensions of the rectangle are length 25 feet and width 15.92 feet

Answer:

6.986.

Step-by-step explanation:

6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

We do the multiplications first    ( according to PEMDAS):-

= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001

= 6 + 0.9 + 0.08 + 0006

= 6.9 + 0.086

= 6 986.

The value of the equation in the decimal form is A = 6.986

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

On simplifying the equation , we get

The value of 6 x 1 = 6

The value of 9 x 1/10 = 0.9

The value of 9 x 1/100 = 0.08

The value of 6 x 1/1000 = 0.006

So , substituting the values in the equation A , we get

A = 6 + 0.9 + 0.08 + 0.006

On simplifying the equation , we get

A = 6.986

Therefore , the value of A is 6.986

Hence , the value of the equation is 6.986

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Using the z-distribution, it is found that the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).

The confidence interval is:

[tex]\overline{x} \pm zs[/tex]

In which:

In this problem, we are given that z = 2.576, s = 0.0407. The sample mean is the difference of the proportions, hence:

[tex]\overline{x} = \frac{28}{162} - \frac{82}{272} = -0.129[/tex]

Then, the bounds of the interval are given by:

[tex]\overline{x} - zs = -0.129 - 2.576(0.0407) = -0.234[/tex]

[tex]\overline{x} + zs = -0.129 + 2.576(0.0407) = -0.024[/tex]

The 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).

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Answer:

a + b = ⟨-4, -9, 0⟩

a - b = ⟨6, 1, 4⟩

2a = ⟨2, -8, 4⟩

3a + 4b = ⟨-17, -32, -2⟩

|a| = √21

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Pre-Calculus

Vectors

Step-by-step explanation:

Adding and subtracting vectors are follow the similar pattern of normal order of operations:

a + b = ⟨1 - 5, -4 - 5, 2 - 2⟩ = ⟨-4, -9, 0⟩

a - b = ⟨1 + 5, -4 + 5, 2 + 2⟩ = ⟨6, 1, 4⟩

Scalar multiplication multiplies each component:

2a = ⟨2(1), 2(-4), 2(2)⟩ = ⟨2, -8, 4⟩

Remember to multiply in the scalar before doing basic operations:

3a + 4b = ⟨3(1), 3(-4), 3(2)⟩ + ⟨4(-5), 4(-5), 4(-2)⟩ = ⟨3, -12, 6⟩ + ⟨-20, -20, -8⟩ = ⟨-17, -32, -2⟩

Absolute values surrounding a vector signifies magnitude of a vector. Follow the formula:

|a| = √[1² + (-4)² + 2²] = √21

Answer:

the answer is d

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{a(b+c)}{bc} ,\frac{b(c+a)}{ca} ,\frac{c(a+b)}{ab} ~are~in~A.P.\\if~\frac{ab+ca}{bc} ,\frac{bc+ab}{ca} ,\frac{ca+bc}{ab} ~are~in~A.P.\\add~1~to~each~term\\if~\frac{ab+ca}{bc} +1,\frac{bc+ab}{ca} +1,\frac{ca+bc}{ab} +1~are~in~A.P.\\if~\frac{ab+ca+bc}{bc} ,\frac{bc+ab+ca}{ca} ,\frac{ca+bc+ab\\}{ab} ~are~in~A.P.\\\\divide~each~by~ab+bc+ca\\if~\frac{1}{bc} ,\frac{1}{ca} ,\frac{1}{ab} ~are ~in~A.P.\\if~\frac{a}{abc} ,\frac{b}{abc} ,\frac{c}{abc} ~are~in~A.P.\\if~a,b,c~are~in~A.P.\\which~is~true.[/tex]

Answer:

None

Step-by-step explanation:

The given angles aren't equal which is needed for the polygon to be similar

No, the following polygons are not similar.

Used the concept of a similar figure that states,

In terms of Maths, when two figures have the same shape but their sizes are different, then such figures are called similar figures.

Given that,

Two polygons EFGH and JKLM are shown in the image.

Now the corresponding sides of both figures are,

EF = 27

JK = 63

And, EH  = 27

JM = 63

Hence, the ratio of corresponding sides is,

EF/JK = 27/63

= 9/21

= 3/7

EH/JM = 27/63

= 3/7

So their corresponding sides are equal in ratio.

But their corresponding angles are not the same.

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B.The failures of the past should inspire people to accomplish more in the future.

1st rectangle:

width: 2.48cm

length: 4.96

2nd rectangle:

width: 4.96 (equals to the length of the 1st rectangle)

area: 9.92

length: 9.92/4.96 = 2

Answer:

Hello,

742/27 (ft)

Step-by-step explanation:

[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]

[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]

The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

What is the total distance?

Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.

Here given that,

A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.

So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].

Then again it hits the ground and covers the distance  [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is

[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]

Then it falls the same distance and goes back to the height

[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]

So, the total distance travelled is

[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]

We take the sum is twice because it goes back to the particular height and falls to the same distance.

[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]

Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

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9514 1404 393

Answer:

   $40,615.20

Step-by-step explanation:

The amortization formula will tell you Michelle's monthly payment.

  A = P(r/12)/(1 -(1 +r/12)^(-12t)) . . . . loan value P at interest rate r for t years

  A = $124,500(0.059/12)/(1 -(1 +0.059/12)^(-12·10)) ≈ $1375.96

__

The total of Michelle's 120 monthly payments is ...

  12 × $1375.96 = $165,115.20

This amount pays both principal and interest, so the amount of interest she pays is ...

  $165,115.20 -124,500 = $40,615.20

Michelle will pay $40,615.20 in interest over the course of the loan.

__

A calculator or spreadsheet can figure this quickly.

Expanding each square on the left side, you have

(x cos(A) + y sin(A))² = x² cos²(A) + 2xy cos(A) sin(A) + y² sin²(A)

(x sin(A) - y cos(A))² = x² sin²(A) - 2xy sin(A) cos(A) + y² cos²(A)

so that adding them together eliminates the identical middle terms and reduces to the sum to

x² cos²(A) + y² sin²(A) + x² sin²(A) + y² cos²(A)

Collecting terms to factorize gives us

(y² + x²) sin²(A) + (x² + y²) cos²(A)

(x² + y²) (sin²(A) + cos²(A))

and sin²(A) + cos²(A) = 1 for any A, so we end up with

x² + y²

as required.

Answer:

35 cm

Step-by-step explanation:

is the correct answer

Answer:

A.x<-8

Step-by-step explanation:

=1/2x<−4

=2*(1/2x)< (2)*(-4)

= x<-8

Answer:

B

Step-by-step explanation:

Answer: 10m, 33m, and 29m

Step-by-step explanation:

n + 3n+3 + 3n-1 = 72m

7n+2=72m

7n = 72-2

n = 70/7

n = 10

Answer:

Step-by-step explanation:

The combinations are:

Total number of combinations:

Correct choice is B

there are 8different combinations are modeled by the diagram.

Answer:

Solution given:

orange:2

grape:2

apple:2

grapefruit:2

no of term:4

now

total no. of combination ia 4*2=8

Answer:

Step-by-step explanation:

Our friend asking what the actual function is has a point. I completed this under the assumption that what we have is:

[tex]f(x)=\frac{e^{3x}}{5x-2}[/tex] and used the quotient rule to find the derivative, as follows:

[tex]f'(x)=\frac{e^{3x}(5)-[(5x-2)(3e^{3x})]}{(5x-2)^2}[/tex] and simplifying a bit:

[tex]f'(x)=\frac{5e^{3x}-[15xe^{3x}-6e^{3x}]}{(5x-2)^2}[/tex]and a bit more to:

[tex]f'(x)=\frac{5e^{3x}-15xe^{3x}+6e^{3x}}{(5x-2)^2}[/tex] and combining like terms:

[tex]f'(x)=\frac{11e^{3x}-15xe^{3x}}{(5x-2)^2}[/tex] and factor out the GFC in the numerator to get:

[tex]f'(x)=\frac{e^{3x}(11-15x)}{(5x-2)^2}[/tex]  That's the derivative simplified. If we want f'(0), we sub in 0's for the x's in there and get the value of the derivative at x = 0:

[tex]f'(0)=\frac{e^0(11-15(0))}{(5(0)-2)^2}[/tex] which simplifies to

[tex]f'(0)=\frac{11}{4}[/tex] which translates to

The slope of the function is 11/4 at the point (0, -1/2)

Answer:

y=5x-1 I think because the snd option doesn't make sense but you should try y =5x-1