All categories

3 years ago

I need help thank you!!

Mathematics

2 answers:

Vedmedyk [2.9K]3 years ago

3 0

Answer:

Step-by-step Explanation

$\huge{ \red{ \csc \: \frac{\pi}{2}} = \purple{ 1}} \\$

Sphinxa [80]3 years ago

3 0

Answer is 1 like the guy above me said^

You might be interested in

If one endpoint is (-6,4) and the midpoint is (4,8) what would be the other endpoint ???? if you can explain it would be great

krek1111 [17]

I believe the next one would be 13.2 because if you add 8.4 to -6.4 it gets 4.8 so you just do the same thing

8 0

3 years ago

Please help! I'll rate 5 stars, if possible give brainliest, and give a thanks.

muminat

Answer:

2x+20 + x+40 = 180

Step-by-step explanation:

The angles are supplementary so they add to 180

2x+20 + x+40 = 180

3 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

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

7 0

3 years ago

I don’t understand this either

trasher [3.6K]

It is a part of slope form ( intercept

7 0

4 years ago

Read 2 more answers

.,..................................

Masteriza [31]

What do you want us to answer??

7 0

4 years ago