Could anybody help with this question? My teacher didn’t give a review so I’m not sure what to do. Any help is appreciated!

Your friend Alex is planning a fundraising game night to raise money for a local children's hospital. She invites a few of your

katrin [286]

Step-by-step explanation:

(a) You win $5 if you roll a six, $1 if you roll an odd number, and $0 if you roll a 2 or a 4, and you pay $1.50 for every roll. The expected value is the sum of each outcome multiplied by its probability.

E = (5.00)(1/6) + (1.00)(3/6) + (0)(2/6) + (-1.50)(1)

E = -0.167

You are expected to lose on average about $0.17 per roll, which means Alex has the advantage.

(b) The probability of rolling a 2 or 4 on a fair die is 2/6 or 1/3. The probability of this happening five times is:

P = (1/3)⁵

P = 1/243

P ≈ 0.41%

There is approximately a 0.4% probability that a fair die will roll a 2 or 4 five times.

(c) The confidence interval for a proportion is:

CI = p ± ME

ME = CV × SE

The margin of error is the critical value times the standard error.

The critical value for 95% confidence is z = 1.960.

The standard error for a proportion is:

SE = √(pq/n)

Given p = 1/6, q = 5/6, and n = 100:

SE = √((1/6) (5/6) / 100)

SE = 0.037

So the confidence interval is:

CI = 1/6 ± (1.960) (0.037)

CI = 0.167 ± 0.073

0.094 < p < 0.240

Since the observed proportion of 0.08 is outside of this interval, we can conclude with 95% confidence that the die is not fair.

(d) Under the current game rules and die probabilities, the expected value is:

E = (5.00)(0.08) + (1.00)(0.33) + (0)(0.59) + (-1.50)(1)

E = -0.77

To make the game fairer, but to still give Alex the advantage so she can make money for her fundraiser, we need to change the rules of the game so that the expected value is less negative.

One simple way to do this is to pay players $2.00 if they roll a 2.

Now the expected value is:

E = (5.00)(0.08) + (1.00)(0.33) + (2.00)(0.29) + (0)(0.30) + (-1.50)(1)

E = -0.19

Now instead of expecting to lose on average $0.77 per roll, players can expect to lose on average $0.19 per roll. This means they have a better chance of winning money, but Alex still has the advantage.

8 0

2 years ago

A homemade loaf of bread turns out to be a perfect cube. Five slices of bread, each 0.6 inch thick, are cut from one end of the

erastovalidia [21]

A cube has all sides of equal length.

So let the side be x

that will be length = width = height = x

Now, when we cut the loaf into 5 equal slices, the length and height of the slices are the same as the original loaf. But the width will become less with each slicing. Here the width corresponds to the thickness of the slices.

For each slice the dimensions are -

Length = x

height = x

width = 0.6

Volume of the original loaf is x³.

Volume of 5 slices is 5(x * x * 0.6).

Volume that is left after the cutting the slices is 235 inch³.

So, equation of the volume is=

x³ - 5(0.6x²) = 235

Solve for x using this equation.

x³ - 3x² = 235

x³ - 3x² - 235 = 0

Using a graph calculator to find the roots, we get

x = 7.35

Hence the dimensions of the original loaf is 7.35 inches for each side.

Please find attached the graph.

7 0

2 years ago

In a school with 117 seventh-grade students and about 29 students per class, the number of seventh-grade classes is about

notka56 [123]

Answer - 4 classes

117 ÷ 29 = 4.03...

That is the answer. hope this helps.

5 0

2 years ago

Read 2 more answers

I need help with this question

4vir4ik [10]

Answer:

y = 3

Step-by-step explanation:

y/9 = 2/6

y/9 = 1/3 (dividing by 2in both numerator and denominator.)

CROSS MULTIPLYING WE GET

3y = 9*1

3y = 9

y = 3

4 0

2 years ago

You are dealt one card from a satndard 52-card deck. Find the probability of being dealt an ace or a 7

ruslelena [56]

Answer:

2/13 or 15.38%

Step-by-step explanation:

Number of Aces in a deck = 4

Number of 7s in a deck = 4

Number of cards in a deck = 52

Probability of A or 7 = (4 Aces + 4 Sevens) / 52 cards 8 cards / 52 cards

This simplifies to 2/13 or 15.38%

3 0

1 year ago