What is 50 divided by 48 in all kinds of forms

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 sampling distribution of the sampling proportions.

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

Sara is mixing together a fruit punch for a party. She's made 4 gallons of punch with a mixture of 50% juice. Her mother tells h

Nata [24]

$\bf \begin{array}{lccclll}
&amount(gallons)&juice&\textit{juice amount}\\
&--------&-----&-----\\
\textit{50\% punch}&4&0.50&(4)(0.50)\\
\textit{pure juice}&x&1.00&(x)(1.00)\\
-----&-----&-----&-----\\
mixture&4+x&0.60&(4+x)(0.60)
\end{array}$

notice, that, pure juice is 100% juice, dohhh, thus 100/100 = 1.00
50% is 50/100 or 0.50 in decimal format

so..... whatever those two quantities amount to, that is, the 50% and pure juice, or (4)(0.50) + (x)(1.00)
they will equal the mixture desired 60% juice, or 0.60, namely (4+x)(0.60)

thus (4)(0.50) + (x)(1.00) = (4+x)(0.60)

solve for "x"

4 0

3 years ago

Given: 5n - 42 - 12n: Prove: n = -6 Statements : Reasons :

OlgaM077 [116]

5*-6-42-12*12*-6=0
the answer is: 0

6 0

2 years ago

The factors of 55 are________________________________.I know that 55 is a (prime/composite) number because...

statuscvo [17]

Answer:

11 & 5 are the factors of 55

Step-by-step explanation:

Factors of 55

11 × 5 = 55

Thus, 11 & 5 are the factors of 55

-TheUnknownScientist 72

7 0

2 years ago

DATA ANALYSIS AND STATISTICSProbabilities involving two rolls of a dieAn ordinary (falr) die is a cube with the numbers 1 throug

Zepler [3.9K]

Solution

The table below is the required sample space of the to fair die

From the above table

The sample space contain 36 outcomes

Event A: The sum is greater than 9

we will look at the table and count all the elements that are greater than 9

There are 6 elements (they are 10, 10, 10, 11, 11, 12 from the table)

The probability for event A will be

$\begin{gathered} p(A)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(A)=\frac{6}{36} \\ p(A)=\frac{1}{6} \end{gathered}$

P(A) = 1/6

Event B: The sum is an even number.

We will look at the table and count the number of elements that are even

There are 18 elements (notice that there are 3 even number on each of the 6 rows of the table)

The probability for event B will be

$\begin{gathered} p(B)=\frac{\text{required outcome}}{\text{total outcome}} \\ p(B)=\frac{18}{36} \\ p(B)=\frac{1}{2} \end{gathered}$

p(B) = 1/2

5 0

1 year ago