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3 years ago

If you had the z-score, mean and standard deviation, how could you find the x - value? Please help, 99 points!

Mathematics

1 answer:

VladimirAG [237]3 years ago

8 0

Answer:

z * sigma+u = x

Step-by-step explanation:

z = (x-u)/ sigma

z is the z score, u is the mean, sigma is the standard deviation

We need to solve for x

Multiply each side by sigma

z * sigma = (x-u)/ sigma * sigma

z * sigma = x-u

Add u to each side

z * sigma+u = x-u+u

z * sigma+u = x

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Suppose <br> g(x) = x^2f(x)<br> and it is known that f(3) = 5 and f '(3) = −2. Evaluate g'(3).

vlabodo [156]

g'(x)=(x^2)'(f(x))+(f'(x))(x^2)

g'(x)=(2x)(f(x))+(f'(x))(x^2)

for g'(3)

g'(3)=(2(3))(f(3))+(f'(3))(3^2)

g'(3)=(6)(5)+(-2)(9)

g'(3)=30-18

g'(3)=12

7 0

4 years ago

I really need help! I'm stuck.

Dovator [93]

y-axis = altitude in feet

x-axis = minutes

slope = y/x = feet/minute

According to above data, we know that 1800 feet is y-intercept and slope is -300 feet per minute (why slope is negative? because its approaching ground at a constant speed)

a. slope = -300

y-intercept = 1800

b. y = -300 x + 1800

c. Here we are given x = 2 minutes

y = -300(2) + 1800

y = -600 + 1800

y = 1200

The balloon will be 1200 feet high above the ground in 2 minutes.

d. Here we are given 300 feet which is y, and find x minutes.

y = -300 x + 1800

300 = -300x + 1800

300 - 1800 = - 300x

-1500 = -300x

1500 = 300 x

x = 1500 / 300

x = 5

The balloon will reach 300 feet above the ground in 5 minutes.

e. It asking when the balloon will reach the ground, if the balloon reaches the ground than the height will be 0, therefore x-intercept (y=0).

y = -300 x + 1800

0 = -300 x + 1800

-1800 = -300x

1800 / 300 = x

x = 6

The balloon will reach the ground in 6 minutes.

<em>//Hope this helps</em>

3 0

4 years ago

Using the Pythagorean Theorem, which of the triangles shown are right triangles?

Feliz [49]

A) 2squared +3squared =13

7 0

4 years ago

Read 2 more answers

Solve 2x2 + 8 = 0 by graphing the related function.

Tpy6a [65]

Answer:

The values of x are x= 2i and x= -2i

Step-by-step explanation:

We need to solve the equation:

2x^2 + 8 =0

Taking 2 common

2(x^2 +4) =0

Dividing both sides by 2

x^2+4 =0

Adding -4 on both sides

x^2 +4 -4 = 0-4

x^2 = -4

Taking square root on both sides we get

√x^2 = √-4

we know √4 = 2 and √-1 = i so answer is:

x = ± 2i

The values of x are x= 2i and x= -2i

The graph is shown in figure attached.

7 0

3 years ago

\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Vinvika [58]

$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0$

Compute each remaining limit:

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23$

$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

$\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0$

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0

3 years ago