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2 years ago

I’ll give brainliest

Mathematics

1 answer:

skelet666 [1.2K]2 years ago

5 0

Answer:

I don't know nothing about this lol...

Step-by-step explanation:

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Proove that <br>1/1-sin theta + 1/1+cos theta = 2×sec square theta

Rina8888 [55]

Step-by-step explanation:

We need to prove that,

$\dfrac{1}{1-\sin\theta}+\dfrac{1}{1+\sin\theta}=2\times \sec^2\theta$

Taking LHS,

$\dfrac{1}{1-\sin\theta}+\dfrac{1}{1+\sin\theta}\\\\=\dfrac{(1+\sin\theta)+(1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}\\\\=\dfrac{2}{1-\sin^2\theta}\\\\\text{As}\ 1-\sin^2\theta=\cos^2\theta\\\\=\dfrac{2}{\cos^2\theta}\\\\\text{As}\ \dfrac{1}{\cos^2\theta}=\sec^2\theta\\\\=2\sec^2\theta\\\\=RHS$

Hence, LHS = RHS.

4 0

2 years ago

PLEASE HELP (TEST) The graphs below show measurements from cubes with

Vladimir [108]

Answer:

do you still need the answer?

Step-by-step explanation:

3 0

2 years ago

a pitcher contains 17.5 cups of iced tea. you drink 0.75 cup of tea each morning and 1.75 cups of tea each evening. when will yo

Annette [7]

Answer:

7 days

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In

Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the <em>sampling distribution of the sampling proportions</em>.

The mean and standard deviation of this sampling distribution is:

$\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}$

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

$P(0.45$

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

$z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)$

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.

For the second game, we win if we get a proportion over 80%.

The probability of winning is:

$P(p>0.8)=P(z>z^*)$

The z value is calculated as before:

$z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0$

As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).

If our chances of winnings depend on P(z>z*), they become lower as z* increases.

Then, we can conclude that our chances of winning decrease with the increase of the number of trials.

We prefer the option of 30 trials for this game.

8 0

2 years ago

9, 15, 1, 19, 4,6

mylen [45]

Answer:

is 7.5 hope i help

Step-by-step explanation:

u put them in order from least to greatest

1 , 4, 6, 9, 15, 19

then take out one number from each side after that u are left with 6 and 9 and whats in the middle of does two numbers ? well u just estimate and it might be 7.5

6 0

2 years ago