pls help! I need the answer fast!

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:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

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

recall see

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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You have to find the slope .

How?

Take 2points

[tex]\\ \rm\Rrightarrow \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Step-by-step explanation:

What the Slope Means A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y also decreases. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.

Answer:

35 cm

Step-by-step explanation:

is the correct answer

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:

hi I am a Nepal

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

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.

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:

1. coefficient of 3rd term = 1

2. constant term= 0

The coefficient of the third term is 1 while the constant term is 0 for the given expression.

An expression is a combination of some mathematical symbol such that an arithmetic operator and variable such that all are constrained and create an equation.

For example 3x +5y

As per the given polynomial,

(1/2)a⁴ + 3a³ + a

Here a is a variable.

(1)

The third term is a and its coefficient is 1 as (1)a.

(2)

All terms have variable "a" thus none of the terms is constant so the constant term is 0.

Hence "For the following statement, the constant term has a coefficient of 0 and the third term has a coefficient of 1".

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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.

Step-by-step explanation:

yah language mujhe samajh mein nahin a rahi hai kya karu aapki is bataiye

Answer:

x = -1

Step-by-step explanation:

1/3(6x-3)+2(x-1) = -7

Distribute

2x-1 +2x-2 = -7

Combine like terms

4x -3 = -7

Add 3 to each side

4x-3+3 = -7+3

4x = -4

Divide by 4

4x/4 = -4/4

x = -1

[tex] \frak{ \frac{1}{3}(6x - 3) + 2(x - 1) = - 7}[/tex]

[tex]\twoheadrightarrow \frak{2x - 1 + 2x - 2 = - 7}[/tex]

[tex]\twoheadrightarrow \frak{4x - 3 = - 7}[/tex]

[tex] \twoheadrightarrow \frak{4x = - 7 + 3}[/tex]

[tex]\twoheadrightarrow \frak{4x = - 4}[/tex]

[tex]\star \: \underline{ \boxed{ \frak \green{{x = - 1}}}}[/tex]

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:

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:

B

Step-by-step explanation:

A function is a relation in which each input, x, has only one output, y.

There are two ways to determine if a relation is a function:

1. If each x-input has only one, unique y-output, then it's a function. If some x-inputs share the same y-outputs, it's not a function.

2. Vertical Line Test on Graphs:

To determine whether y is a function of x, when given a graph of relation, use the  following criterion: if every vertical line you can draw goes though only 1 point, the relation can be a function. If you can draw a vertical line that goes though more than 1 point, the relation cannot be a function.

Since we're given a graph relation, let's test both of the answers out.

If I were to draw a vertical line in a specific place on the first graph, I'd be hitting more than one point in the coordinate plane.

If I were to draw a vertical line in a specific place on the second graph, I'd only be hitting one point in the coordinate plane.

Therefore, choice B is a function.

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:

The hottest month for the northern hemisphere is August.

The hottest month for the southern hemisphere is January and February (these top two might be the opposite)

It's globally warmer during the months of June July and August

During april and november, the southern hemisphere and northern hemisphere are the same, or very close.

During July and August the southern and northern hemispheres have the largest difference in temperature

Recall that

[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]

Differentiating the power series series for y(x) gives the series for y'(x) :

[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

Now, replace everything in the DE with the corresponding power series:

[tex]y'-2xy = 6\sin(3x) \implies[/tex]

[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.

Split up both series on the left into even- and odd-indexed series:

[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]

[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]

Next, we want to condense the even and odd series:

• Even:

[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]

• Odd:

[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].

The even series vanishes, so that

[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]

for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find

[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]

[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]

and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].

This leaves us with the odd series,

[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]

[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]

We have

[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]

[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]

[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]

[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]

So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then

[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]

[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]

[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]

and so the first four terms of series solution to the DE would be

[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/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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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:

B

Step-by-step explanation:

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:

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

Answer:

the answer is d

Step-by-step explanation:

Answer:

A.x<-8

Step-by-step explanation:

=1/2x<−4

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

= x<-8

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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6 minutes / 30 minutes

Divide the top and bottom by 6.

1 minute / 5 minutes

Fraction in lowest terms: 1/5

Hope this helps!

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:

[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]