Help me out !!!!!!!!!!!!

The circle will be enlarged by a scale factor of 4.

sertanlavr [38]

Answer:

The area of the new circle is $144\pi \ in^2$

Step-by-step explanation:

Given:

radius of the circle = 3 in.

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared.

Let

z-----> the scale factor

x-----> the area of the enlarged circle

y-----> the area of the original circle

So;

$z^2= \frac{x}{y}$

In this problem we have

$z=4$ ----> the scale factor

The area of the original circle is equal to

$y = \pi r^2= \pi (3)^2= 9\pi \ in^2$

Now substituting value of y and solve for x we get;

$4^2=\frac{x}{9\pi}\\ \\x= 16\times9\pi= 144\pi \ in^2$

Hence Area of the new Circle is $144\pi \ in^2$

Also The area of the enlarged circle is 16 times the area of the original circle.

4 0

4 years ago

On a given blueprint, 1 inch = 12 feet. if the dimensions of a recreation room on the blueprint are 2 inches × 1.75 inches, what

DedPeter [7]

24 feet × 21 feet is the answer

4 0

3 years ago

Solve the system y = 3x − 2 y = 4x − 4 HELPPPP MEHH WITH DIS PLZZ

ELEN [110]

Both x equal 0 (i’m not sure if that is what you where asking for)

8 0

3 years ago

Read 2 more answers

Plz help me out! what is the complement of m^0?

Leni [432]

Anything raised do 0 is always 1
I hope that’s correct

3 0

3 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

4 years ago