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2 years ago

I really need help. I need to go to sleep but I also need to finish this first plz help

Mathematics

1 answer:

WITCHER [35]2 years ago

8 0

Since it doesn't give us a specific variable to use I'll just use y. and my phone doesn't have the less than or equal to sign so ill just say how to draw it i guess.
41. y (less than sign with line) 4.
42. y × (-7) > 42
43. y ÷3 - 5 (greater than sign with line) 11

and im not 100% sure on 44 so... anyway you probably know what the greater than or equal to sign and the less than or equal to sign are so. yeah hope it helped.

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Find the measure of the nlissing angle.<br> 220<br> a

Goryan [66]

Answer:

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Step-by-step explanation:

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2 years ago

Uggh!! help pls last question

RideAnS [48]

Answer:

The answer is 46.

Step-by-step explanation:

You have to substitute the value of a and c into the expression :

$a + 9c$

Let a = 10,

Let c = 4,

$10 + 9(4)$

$= 10 + 36$

$= 46$

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2 years ago

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The sign of the product of –35 and –625 is (positive, negative, zero)

ziro4ka [17]

Step-by-step explanation:

-35 x - 625 = 21875 so positive

263 x 0 = 0 so zero

-21 x 451 = - 9471 so negative

-350 x 89 = - 31150 so negative

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2 years ago

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Which Trig. Function do you use to find X?

zvonat [6]

Answer:

Cosine

Step-by-step explanation:

Cosine = $\frac{adjacent}{hypothenuse}$

Cos 70 = $\frac{x}{8}$

0.633 = $\frac{x}{8}$

0.633 x 8 = $\frac{x}{8}$ x 8

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2 years ago

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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

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

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3 years ago