The question is provided on the image

The distribution of the amount of money in savings accounts for Florida State students has an average of 1,200 dollars and a sta

Anestetic [448]

Answer:

By the Central Limit Theorem, the sampling distribution of the sample mean amount of money in a savings account is approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Average of 1,200 dollars and a standard deviation of 900 dollars.

This means that $\mu = 1200, \sigma = 900$

Sample of 10.

This means that $n = 10, s = \frac{900}{\sqrt{10}} = 284.6$

The sampling distribution of the sample mean amount of money in a savings account is

By the Central Limit Theorem, approximately normal with mean of 1,200 dollars and standard deviation of 284.6 dollars.

7 0

3 years ago

ILL MARK BRAINLIST pls help

kaheart [24]

You cut out the question-

6 0

2 years ago

5x minus 4 equals x squared minus 4x plus 4. What is x

Sauron [17]

Two solutions were found :

x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i

x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((0 - 5x2) - 4x) - 4 = 0

Step 2 :

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

-5x2 - 4x - 4 = -1 • (5x2 + 4x + 4)

Trying to factor by splitting the middle term

3.2 Factoring 5x2 + 4x + 4

The first term is, 5x2 its coefficient is 5 .

The middle term is, +4x its coefficient is 4 .

The last term, "the constant", is +4

Step-1 : Multiply the coefficient of the first term by the constant 5 • 4 = 20

Step-2 : Find two factors of 20 whose sum equals the coefficient of the middle term, which is 4 .

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step 3 :

-5x2 - 4x - 4 = 0

Step 4 :

Parabola, Finding the Vertex :

4.1 Find the Vertex of y = -5x2-4x-4

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -0.4000

Plugging into the parabola formula -0.4000 for x we can calculate the y -coordinate :

y = -5.0 * -0.40 * -0.40 - 4.0 * -0.40 - 4.0

or y = -3.200

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = -5x2-4x-4

Axis of Symmetry (dashed) {x}={-0.40}

Vertex at {x,y} = {-0.40,-3.20}

Function has no real roots

Solve Quadratic Equation by Completing The Square

4.2 Solving -5x2-4x-4 = 0 by Completing The Square .

Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:

5x2+4x+4 = 0 Divide both sides of the equation by 5 to have 1 as the coefficient of the first term :

x2+(4/5)x+(4/5) = 0

Subtract 4/5 from both side of the equation :

x2+(4/5)x = -4/5

Add 4/25 to both sides of the equation :

On the right hand side we have :

-4/5 + 4/25 The common denominator of the two fractions is 25 Adding (-20/25)+(4/25) gives -16/25

So adding to both sides we finally get :

x2+(4/5)x+(4/25) = -16/25

Adding 4/25 has completed the left hand side into a perfect square :

x2+(4/5)x+(4/25) =

(x+(2/5)) • (x+(2/5)) =

(x+(2/5))2

Things which are equal to the same thing are also equal to one another. Since

x2+(4/5)x+(4/25) = -16/25 and

x2+(4/5)x+(4/25) = (x+(2/5))2

then, according to the law of transitivity,

(x+(2/5))2 = -16/25

Note that the square root of

(x+(2/5))2 is

(x+(2/5))2/2 =

(x+(2/5))1 =

x+(2/5)

Now, applying the Square Root Principle to Eq. #4.2.1 we get:

x+(2/5) = √ -16/25

Subtract 2/5 from both sides to obtain:

x = -2/5 + √ -16/25

Since a square root has two values, one positive and the other negative

x2 + (4/5)x + (4/5) = 0

has two solutions:

x = -2/5 + √ 16/25 • i

or

x = -2/5 - √ 16/25 • i

Note that √ 16/25 can be written as

√ 16 / √ 25 which is 4 / 5

Solve Quadratic Equation using the Quadratic Formula

4.3 Solving -5x2-4x-4 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B2-4AC

x = ————————

2A

In our case, A = -5

B = -4

C = -4

Accordingly, B2 - 4AC =

16 - 80 =

-64

Applying the quadratic formula :

4 ± √ -64

x = —————

-10

In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)

Both i and -i are the square roots of minus 1

Accordingly,√ -64 =

√ 64 • (-1) =

√ 64 • √ -1 =

± √ 64 • i

Can √ 64 be simplified ?

Yes! The prime factorization of 64 is

2•2•2•2•2•2

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 64 = √ 2•2•2•2•2•2 =2•2•2•√ 1 =

± 8 • √ 1 =

± 8

So now we are looking at:

x = ( 4 ± 8i ) / -10

Two imaginary solutions :

x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i

or:

x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i

Two solutions were found :

x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i

x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i

<em>hope i helped</em>

6 0

3 years ago

Read 2 more answers

Write the equation for the area of a triangle with a height that is 5 more than twice its base using one variable. Please help m

zysi [14]

Answer:

$A=b(b+\frac{5}{2} )$

Step-by-step explanation:

hey there!!!

we know already that the area of a triangle is given as

$A=\frac{1}{2}b *h$

where b= base

and h= heigth

Given that the height is 5 times more than twice it base

let h= 2b+5

Now we substitute h=2b+5 in the area of the triangle

$A=\frac{1}{2}b * (2b+5)$

$A= \frac{2b^2}{2} +\frac{5b}{2}$

$A= b^2+\frac{5b}{2}\\\\A=b(b+\frac{5}{2} )$

7 0

3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

2 years ago

The question is provided on the image ​

The question is provided on the image