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3 years ago

A normal distribution has a mean of 15 and a standard deviation of 2. Find the value that corresponds to the 75th

Mathematics

1 answer:

Arada [10]3 years ago

7 0

I am sorry, I am so confused
I wish I could help

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Urgent please help quick!!!

Radda [10]

Answer:

$x=-1\\y=-1$

Step-by-step explanation:

$5x+5y=-10\\3x-7y=4$

Let's solve the first equation for either x or y. I'll do it for x.

$5x+5y=-10$

Begin by subtracting 5y.

$5x=-5y-10$

Now divide by 5.

$x=\frac{-5y}{5}-\frac{10}{5}$

Simplify:

$x=-y-2$

Now substitute x in the second equation for this value.

$3x-7y=4\\3(-y-2)-7y=4$

Distribute;

$-3y-6-7y=4$

Add 6

$-3y-7y=4+6$

Combine like terms;

$-10y=10$

Divide by -10.

$y=\frac{10}{-10}\\ y=-1$

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Take this value of y and replace it in the first equation to find the value of x.

$5x+5y=-10\\5x+5(-1)=-10\\5x-5=-10\\5x=-10+5\\5x=-5\\x=\frac{-5}{5}\\ x=-1$

5 0

3 years ago

The population of a city is 578,992. The land area of the city is 67.2 mi².

Inessa05 [86]

Sorry man I don't know if im write but try 28.78

5 0

3 years ago

Read 2 more answers

Ethan opened a savings account that compounds interest yearly. The value of Ethan's account is given by the equation y = 2000(1.

ivanzaharov [21]

Answer:

Step-by-step explanation:

hope this helps but I'm not sure if it's right sorry if wrong

4 0

3 years ago

Read 2 more answers

What is the length of -5,-4 and -3,3 to the nearest hundredth?

Sergeu [11.5K]

Answer:

-6

Step-by-step explanation:

cause there was an...negative 6

6 0

4 years ago

Read 2 more answers

What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

3 years ago