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3 years ago

The graph of a function is shown:

Mathematics

2 answers:

Viefleur [7K]3 years ago

8 0

AB. It sounds complicated but is really simple to learn. For a function to increase, it must have an upward slope. So the only part of the graph that is going up is AB

Paladinen [302]3 years ago

6 0

AB because its the only part of the graph that increases

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a. 314

Step-by-step explanation:

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3 years ago

Cos (90-theta) • cosec90-theta) =tano. How?

denis-greek [22]

Answer:

______________________________________________________

TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-

For any right angled triangle with one angle α ,

$\cos (90 - \alpha ) = \sin \alpha$ or $\sin(90 - \alpha ) = \cos\alpha$

$cosec \: (90 - \alpha ) = \sec\alpha$ or $\sec(90 - \alpha ) = cosec\:\alpha$

SOME GENERAL TRIGNOMETRIC FORMULAS :-

$\sin \alpha = \frac{1}{cosec \: \alpha }$ or $cosec \: \alpha = \frac{1}{\sin \alpha }$

$\cos \alpha = \frac{1}{\sec \alpha }$ or $\sec \alpha = \frac{1}{\cos \alpha }$

______________________________________________________

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

$\cos (90 - \alpha ) \times cosec(90 - \alpha )$

$=> sin\alpha \times \sec\alpha$

$=> \sin\alpha \times \frac{1}{\cos\alpha }$

$=> \frac{\sin\alpha }{\cos\alpha }$

$=> \tan\alpha$ = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

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3 years ago

Suppose three marksmen shoot at a target. The ith marksman fires ni times, hitting the target each time with probability Pi, ind

astraxan [27]

Answer:

Distribution is NOT binomial.

Step-by-step explanation:

In order to be a binomial distribution, the probability of success for each individual trial must be the same. Since each marksman hits the target with probability Pi, the probability of success (hitting the target) is not necessarily equal for all trials. Therefore, the distribution is not binomial.

In this case, the distribution would only be binomial if Pi was the same for every "ith" marksman.

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3 years ago