The two figures above are congruent. Find the measure of the angle that isn't labeled on either figure.

In the adjoining figure O is the centre of the circle the tangent to the circle of radius 6 cm from the exterior point P of the

slamgirl [31]

Answer:

Let P be the external point. O be the origin. join O and P get OP and nearest point on the circle from P be A.

Let Q be the point onthe circle in which, tangent make 90° with radius at Q.

PQ = 8 and OQ = 6

we get a right angled triangle PQO right angled at Q.

so, OP^2 = OQ^2 + PQ^2= 8^2 + 6^2 = 64 + 36 =1==

therefore OP =10cm

we need nearest point from P, which is PA

PA = OP - OA= 10 -6=4cm

7 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

3 years ago

a triangle pyramid with an equilateral base has a side length of 10 centimeters and a surface area of 214.5 square meters. find

sweet [91]

Area of the base = 1/2 * 10 * 5sqrt3 = 25 sqrt3

Total surface area = 25 sqrt3 + 3 * 1/2 * 10 * slant height = 214.5

25 sqrt3 + 15h = 214.5
15h = 214.5 - 25 sqrt3

h = (214.5 - 25sqrt3() / 15

= 11.41 cm to nearest hundredth

6 0

3 years ago

A scale factor that is less than 1, but greater than 0 is

swat32

Since the scale factor is smaller than one (the original), the resulting image is smaller than the original.

Hope this helped, and feel free to ask more questions if needed(:

7 0

2 years ago

Read 2 more answers

X/6 + 4 = 15 what does x equal? What is x?

Umnica [9.8K]

Hello there! x = 66.

x/6 + 4 = 15

This means that since 15 - 4 = 11, x/6 = 11.

Given the two numbers and a variable, we can multiply:

6 x 11 = 66

66/6 = 11

Therefore, x = 66

Hope this helps!

4 0

1 year ago

Read 2 more answers