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188 ASU students, faculty and staff were randomly surveyed and asked if they have an ASU parking decal, ride their bike to campu

bulgar [2K]

Part A.
The venn diagram is attached. In the venn diagram, circle A represent the number of responds that have an ASU parking deal. Circle B represent the number of respondents that ride their bike to campus and circle C represent the number of respondents that that ride the Light Rail.

Part B.
From the venn diagram, the number of respondents that ride the Light Rail and ride their bike to campus is given by 39 – 12 = 27

Part C.
From the venn diagram, the number of respondents that only have an ASU parking decal is given by <span>77 – 12 – 8 – 6 = 51

Part D.
The number of respondents that ride the Light Rail or have an ASU decal is given by the sum of the number of respondents that ride the Light Rail only and those that have an ASU parking decal and those that have both ASU parking decal and ride their bike to school.
This is given by 51 + 8 + 28 = 87.

Part E.
The venn diagram showing the area </span><span>that represents people who have a parking decal or who ride their bike to campus but who do not ride the Light Rail</span> sgaded is attached.

8 0

2 years ago

Critical Thinking: Empirical/Quantitative Skills

aliya0001 [1]

Answer:

1. 0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. 0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. 0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume that there is a 0.905 probability that a passenger with a ticket will show up for the flight.

This means that $p = 0.905$

Also assume that the airline sells 200 tickets

This means that $n = 200$

Question 1:

Exactly, so we can use the P(X = x) formula, to find P(X = 180).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 180) = C_{200,180}.(0.905)^{180}.(0.095)^{20} = 0.0910$

0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. When 200 tickets are sold, calculate the probability that at most 180 passengers show up for the flight.

Now we have to use the approximation.

Mean and standard deviation:

$\mu = E(X) = np = 200*0.905 = 181$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.905*0.095} = 4.15$

Using continuity correction, this is $P(X \leq 180 + 0.5) = P(X \leq 180.5)$ , which is the p-value of Z when X = 180.5. Thus

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{180.5 - 181}{4.15}$

$Z = -0.12$

$Z = -0.12$ has a p-value of 0.4522.

0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. When 200 tickets are sold, calculate the probability that more than 180 passengers show up for the flight.

Complementary event with at most 180 passengers showing up, which means that the sum of these probabilities is 1. So

$p + 0.4522 = 1$

$p = 1 - 0.4522 = 0.5478$

0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

4 0

2 years ago

A sock drawer contains eight navy blue socks and five black socks with no other socks. If you reach in the drawer and take two s

Rzqust [24]

Answer:

a. the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b. the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c. the probability of either 2 navy socks is picked or one black & one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

Step-by-step explanation:

A sock drawer contains 8 navy blue socks and 5 black socks with no other socks.

If you reach in the drawer and take two socks without looking and without replacement, what is the probability that:

Solution:

total socks = N = 8 + 5 + 0 = 13

a) you will pick a navy sock and a black sock?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick the black sock, total number of socks left is 12.

Let B be the probability of picking a black sock again.

P (B) = 5/12.

Then, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

b) the colors of the two socks will match?

Let A be the probability of picking a navy socks first.

Then P (A) = 8/13

without replacing the navy sock, will pick another navy sock, total number of socks left is 12.

Let B be the probability of another navy sock again.

P (B) = 7/12.

Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

Let D be the probability of picking a black socks first.

Then P (D) = 5/13

without replacing the black sock, will pick another black sock, total number of socks left is 12.

Let E be the probability of another black sock again.

P (E) = 4/12.

Then, the probability of picking 2 black sock = P (D & E)

= (5/13 ) * (4/12) = 5/39 = 0.128

Now, the probability of picking two navy or two black is

= 56/156 + 20/156 = 76/156 = 0.487

c) at least one navy sock will be selected?

this means, is either you pick one navy sock and one black or two navy socks.

so, if you will pick a navy sock and a black sock, the probability of picking a navy sock and a black sock = P (A & B)

= (8/13 ) * (5/12) = 40/156 = 0.256

also, if you will pick 2 navy sock, Then, the probability of picking 2 navy sock = P (A & B)

= (8/13 ) * (7/12) = 56/156 = 0.359

now either 2 navy socks is picked or one black one navy socks.

= 40/156 + 56/156 = 96/156 = 0.615

4 0

2 years ago

I NEED HELP PLEASE!!!

meriva

Answer:

a

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Evaluate (solve) the power: 2^9

german

Answer:

The corrext answer is, 512.

3 0

3 years ago

Read 2 more answers