2,5,8,... Find the 41st term.

Mathematics

1 answer:

Snezhnost [94]2 years ago

8 0

Answer:

Step-by-step explanation:

hello :

this is the arithmetic -sequence because : 8-5=5-5=3 (comon dif)

An =A1+(n-1)r calculate......... (A1=2 and r=3 and n=41) continu...

You might be interested in

Symbol for taken away

daser333 [38]

Hi there!

In math terms, when we think of the words "Taken away", we immediately think of subtraction/subtracting. So, the symbol that you're looking for looks sort of like a dash.

- is what it looks like.

Hope this helps!

Message me if you need anything else! I'd be happy to help you! :D

6 0

3 years ago

Read 2 more answers

A tree struck by lightning broke at a point 12 ft above the ground as shown. What was the height of the tree to the nearest tent

forsale [732]

51, because 39+12=51. You just add both parts of the tree. Please Rate Brainliest.

6 0

3 years ago

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:

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