Please help :(( I’m having a hard time with these

The University of Virginia has an acceptance rate of 24%. Another 20% get waitlisted, and the rest get denied. If you were going

Bas_tet [7]

sStep-by-step explanation:

hi

4 0

2 years ago

8 9 10 11 12 13 14 15 16 17 18 19 20You and two friends (Adam and Alana) wonly play a game if it is fair for all three of you. T

pav-90 [236]

Well I don't know !
Let's take a look and see:

The idea is that there could be more than one way
for a roll of the dice to land with the same number.

-- If the sum is from 1-4, you get the point.
There are 6 different ways for a roll of the dice to come up 1-4.

-- If the sum is from 5-8, Adam gets the point.
There are 20 different ways for a roll of the dice to come up 5-8.

-- If the sum is 9-12, Lana gets the point.
There are 10 different ways for a roll of the dice to come up 9-12.

-- The game is not fair to all three of you.
-- Lana has a distinct advantage over you.
-- Adam has a big advantage over Lana.
-- Adam has an even bigger advantage over you.
-- You are at a big disadvantage. (Notice that one of your
numbers ... 1 ... can never come up unless one of the dice
falls off of the table.)
_______________________________

Here's how to figure it:

Ways to roll a 2:
1 ... 1

Ways to roll a 3:
1 ... 2
2 ... 1

Ways to roll a 4:
1 ... 3
2 ... 2
3 ... 1

Ways to roll a 5:
1 ... 4
2 ... 3
3 ... 2
4 ... 1

Ways to roll a 6:
1 ... 5
2 ... 4
3 ... 3
4 ... 2
5 ... 1

Ways to roll a 7:
1 ... 6
2 ... 5
3 ... 4
4 ... 3
5 ... 2
6 ... 1

Ways to roll an 8:
2 ... 6
3 ... 5
4 ... 4
5 ... 3
6 ... 2

Ways to roll a 9:
3 ... 6
4 ... 5
5 ... 4
6 ... 3

Ways to roll a 10:
4 ... 6
5 ... 5
6 ... 4

Ways to roll 11:
5 ... 6
6 ... 5

Ways to roll 12:
6 ... 6

5 0

2 years ago

A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs

Makovka662 [10]

Answer:

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be 95% confident that the mean cost, μ, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this interval, but we can be 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be 95% confident that the mean cost, μ,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this interval, but we can be 95% confident that it does.

Hope this Helps!!!

4 0

2 years ago

Ppppppppeeeelelelelele help

kirill115 [55]

11/3*6/11=66/33=2
6 24/20-2 15/20=2 9/20

5 0

2 years ago

Solve for z<br> -p(d+z) = -2z+59−p(d+z)=−2z+59

alexira [117]

$-p(d+z)=-2z+59\qquad\text{use distributive property}\\\\-pd-pz=-2z+59\qquad\text{add pd to both sides}\\\\-pz=-2z+pd+59\qquad\text{add 2z to both sides}\\\\2z-pz=pd+59\\\\z(2-p)=pd+59\qquad\text{divide both sides by}\ (2-p)\neq0\\\\\boxed{z=\dfrac{pd+59}{2-p}}$

6 0

2 years ago