40.6 in numerical form

We all hate to bring change to a store. By using random numbers, we can eliminate the need for change and give the store and the

Sunny_sXe [5.5K]

Answer:

Round off to closest value.

Step-by-step explanation:

In order to remove the requirement for change, every time the client buys a product for $0.20 the random number that lies between 0 and 1 would be produced also the same would be rounded to the closest value i.e. 0 or 1. In the case when 0 arise so the customer would not have to pay any amount but if 1 arise so the customer have to pay $1. So this type of method would remove the requirement for a minor change

6 0

2 years ago

At a coffee shop, 70% of the customers buy a coffee drink, 30% buy food, and 20% buy both a coffee drink and food. Are buying fo

Shkiper50 [21]

Yes.

70% of the customers buy a coffee drink and 30% buy food.

The sum of it is 100% already.

So 20% of it just another addition of information since it cannot exceed 100% of percentage (100% + 20% = 120%).

It also contains a subset for both percentage.

8 0

3 years ago

Read 2 more answers

|y| x-3 <br> what is the vertex

lutik1710 [3]

Answer:

(3,0)

Step-by-step explanation:

put x-3=0 and solve for it whic hwould be x=3

8 0

3 years ago

If one pound is the same weight as 16 ounces, how many ounces equal 3.5 pounds? (only input whole numbers.)

astra-53 [7]

Answer: well 56 is the answer

Step-by-step explanation:

5 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