Let triangle FGH be A right triangle with right angle G and an altitude as drawn.....please help ASAP!!!

Whats is the relationship between <a and <B?

SashulF [63]

Angle A and angle B are congruent, because they are on a line, and if the middle angle is in the middle, then the two outer angles would be congruent.

Hope this helps! :)

6 0

3 years ago

Which function has two x-intercepts, one at (0,0) and one at (4,0)?

Tcecarenko [31]

Answer:

$f(x)=x(x-4)$

Step-by-step explanation:

Given: Function has two x-intercepts, one at (0,0) and one at (4,0)

To choose: the correct option

Solution:

x-intercepts of are the points whose x-coordinate is 0.

Consider $f(x)=x(x-4)$

Put $x=0$

$f(0)=0(0-4)=0$

Put $x=4$

$f(4)=4(4-4)=0$

So, the function $f(x)=x(x-4)$ has two x-intercepts, one at (0,0) and one at (4,0).

6 0

3 years ago

lin is shopping for a couch with her dad and hears him ask the salesperson how much is your commission the salesperson says that

jonny [76]

A. $27.23

B. $ 467.77

4 0

3 years ago

Read 2 more answers

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

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