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3 years ago

I only need the Evens ;-;

Mathematics

1 answer:

Simora [160]3 years ago

7 0

Supplementary is when two angles add up to 180
Complementary is when two angles add up to 90
Note: I'm gonna be giving the evens and odds xP

20. Angle 2 is 37

21. Supplementary

22. Angle 1 is 87

23. Angle 2 is 141

24. Complementary

25. Angle 1 is 4<span>5</span>

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On a coordinate plane, a line goes through (0, negative 1) and (3, 1). A point is at (negative 3, 0).

nekit [7.7K]

Answer:

it the 2nd because it has the 2 in it

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

TODIGIllici<br> 8. Solve: - 5y+30=15

nasty-shy [4]

Answer:

y=3

Step-by-step explanation:

Subtract 30 from both sides

-5y+30=15

-5y=-15

Divide by -5 on both sides

y=3

Hope this helps! :)

4 0

2 years ago

Write the ratio in three different formats:a to b, a:b, and a/b.

Aleks04 [339]

D.
24 is the distance Carlos biked
30 is the total distance
24:30 divided by 6
4:5

7 0

3 years ago

Pls anyone can help?

Luba_88 [7]

I think it’s 9/10 mile

3 0

2 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