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3 years ago

PLEASE HELP GIVING BRAINLIST

Mathematics

1 answer:

melisa1 [442]3 years ago

3 0

I’m pretty sure it’s the number of wrenches in the toolbox because that gonna affect the amount of space he has left

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Futhe Mathematics <img src="https://tex.z-dn.net/?f=%28Cos%20%7B%7D%5E%7B4%7Dt%20-Sin%20%7B%7D%5E%7B4%7Dt%20%29%20%20%5Cdiv%2

Nastasia [14]

Answer:

cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

$\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t }$

We can write the numerator as ,

$\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t }$

Recall the identity ,

$\longrightarrow (a-b)(a+b)=a^2-b^2$

Using this we have ,

$\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}$

Again , as we know that ,

$\longrightarrow sin^2\phi + cos^2\phi = 1$

Therefore we can rewrite it as ,

$\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}$

Again using the first identity mentioned above ,

$\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}$

Or else we can also write it using ,

$\longrightarrow cos2\phi = cos^2\phi - sin^2\phi$

Therefore ,

$\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}$

And we are done !

$\rule{200}{4}$

Additional info :-

Derivation of cos²x - sin²x = cos2x :-

We can rewrite cos 2x as ,

$\longrightarrow cos(x + x )$

As we know that ,

$\longrightarrow cos(y + z )= cosy.cosz - siny.sinz$

So that ,

$\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x)$

On simplifying,

$\longrightarrow cos(x+x) = cos^2x - sin^2x$

Hence,

$\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }}$

$\rule{200}{4}$

7 0

3 years ago

A research team at Cornell University conducted a study showing that approximately 10% of all businessmen who wear ties wear the

LenKa [72]

Answer:

a) The probability that at least 5 ties are too tight is P=0.0432.

b) The probability that at most 12 ties are too tight is P=1.

Step-by-step explanation:

In this problem, we could represent the proabilities of this events with the Binomial distirbution, with parameter p=0.1 and sample size n=20.

a) We can express the probability that at least 5 ties are too tight as:

$P(x\geq5)=1-\sum\limits^4_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\geq5)=1-(0.1216+0.2702+0.2852+0.1901+0.0898)\\\\P(x\geq5)=1-0.9568=0.0432$

The probability that at least 5 ties are too tight is P=0.0432.

a) We can express the probability that at most 12 ties are too tight as:

$P(x\leq 12)=\sum\limits^{12}_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\leq 12)=0.1216+0.2702+0.2852+0.1901+0.0898+0.0319+0.0089+0.0020+0.0004+0.0001+0.0000+0.0000+0.0000\\\\P(x\leq 12)=1$