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3 years ago

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Prove that the equation is an identity. tan A tan B = tan A + tan B cot A + cot B Working from the right-hand side, use the trig

onometric identities for tangent and cotangent. RHS = tan A + tan B cot A + cot B

1 answer:

Yuliya22 [10]3 years ago

7 0

Answer:

Stedadasadjdjsjxjnakcp-by-step explanation:bdbdnxhsjd xhdvsh

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What is the equation for the perpendicular that passes through the point (-1,-2) ?

tangare [24]

Answer:

Perpendicular with respects to what?

You can't find it with only a point, you also need a slope to know which line is which.

There are an infinite number of lines that go through any point. Each has a slope, without slopes, we cant solve such a question.

It's line asking, how old is Jane's dog?

Who knows man? There's like thousands of Janes, which one is the one?

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3 years ago

HELP ME PUT THESE IN ORDER TO LEAST TO GREATEST PLS ( i will give breanliest) HELP PLS

gregori [183]

Answer: $\sqrt{40}$ , $7\frac{1}{5}$ , $3\sqrt{6}$ , 15/2, 7.5

Step-by-step explanation: This is a bit tricky because 15/2 is equal to 7.5 !!

√40 = 6.32455532

7 1/5 = 7.2

3√6 = 7.348469228

15/2 = 7.5

7.5 = 7.5

6 0

2 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

Can someone help answering my questions in English and Math

vagabundo [1.1K]

Answer:

yo

Step-by-step explanation:

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2 years ago

What is the missing value in the equation shown □ x1/10=0.034

grin007 [14]

Answer:

0.34

Step-by-step explanation:

Do the inverse operation so;

0.034 / 1/10 = 0.34

Now go back and subsitute,

0.34 * 1/10 = 0.034

4 0

3 years ago