All categories

1 year ago

The probability that a vehicle entering the Luray

1 answer:

4 0

Probability that a vehicle entering the Luray Caverns has Canadian license plates = 0.12
Probability that it is a camper = 0.28
Probability that it is a camper with Canadian license plates = 0.09

a. Probability that a camper entering the Luray Caverns has Canadian
license plates = 0.09

b. Probability that a vehicle with Canadian
license plates entering the Luray Caverns is a camper = 0.09/0.12
= 0.75

c. Probability that a vehicle entering the Luray Caverns
does not have Canadian plates or is not a camper = 0.88 + 0.72 - 0.81
= 0.80

You might be interested in

What is the 5th term of the geometric sequence a1=120, a2=36, a3=10.8, a6=0.2916

Serggg [28]

The fifth term of the geometric sequence would be 0.972

6 0

1 year ago

Read 2 more answers

Titus works at a hotel. Part of his job is to keep the complimentary pitcher of water at least half full and always with ice. Wh

sertanlavr [38]

The Answer is 160 minutes

8 0

2 years ago

The Institute of Education Sciences measures the high school dropout rate as the percentage of 16- through 24-year-olds who are

n200080 [17]

Answer:

$Z_{H_0}= -1.85$

Step-by-step explanation:

Hello!

The high school dropout rate, as a percentage of 16- through 24- year-olds who are not enrolled in school and have not earned a high school credential was is 2009 8.1%.

To thest the claim that this percentage has decreased, a polling company takes a random sample of 1000 people between the ages of 16 and 24 and finds out that 6.5% of them are highschool dropouts.

The study variable is

X: Number of individuals with age between 16 and 24 years old that are highschool dropouts.

The parameter of interest is the proportion fo highschool dropouts p

And the sample proportion is p'= 0.065

The hypotheses are:

H₀: p ≥ 0.081

H₁: p < 0.081

To study the population proportion, you have to approximate the distribution of the sampling proportion to normal applying the Central Limit Theorem, then the statistic to use is an approximate standard normal:

$Z_{H_0}= \frac{(p'-p)}{\sqrt{\frac{p*(1-p)}{n} } } = \frac{0.065-0.081}{\sqrt{\frac{0.081*0.919}{1000} } } = -1.85$

I hope this helps!

3 0

2 years ago

Darryl wants to add enough water to 8 gallons of a 10% bleach solution to make a 5% bleach solution. Which equation can he use t

Sveta_85 [38]

Answer:- 0.8/8+x=5/100

Step-by-step explanation:

6 0

1 year ago

Read 2 more answers

Ice cream usually comes in 1.5-quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some v

Troyanec [42]

Answer:

(a) The expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b) The expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

Step-by-step explanation:

The random variable X and Y are defined as follows:

X = amount of ice cream in the box

Y = amount of ice cream scooped out

The information provided is:

E (X) = 48

SD (X) = 1

V (X) = 1

E (Y) = 2

SD (Y) = 0.25

V (Y) = 0.0625

(a)

The total amount of ice-cream served at the party can be expressed as:

X + 3Y.

Compute the expected value of (X + 3Y) as follows:

$E(X+3Y)=E(X)+3E(Y)\\= 48+(3\times2)\\=48+6\\=54$

Compute the variance of (X + 3Y) as follows:

$V(X+3Y) = V (X)+3^{2}V(Y)+2\times 3Cov (X,Y)\\=1+(9\times0.0625)+0\\=1.5625$

Then the standard deviation of (X + 3Y) is:

$SD(X + 3Y) =\sqrt{V(X + 3Y)}\\\sqrt{1.5625}\\=1.25$

Thus, the expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b)

The amount of ice-cream left in the box after scooping out one scoop is represented as follows:

X - Y.

Compute the expected value of (X - Y) as follows:

$E(X-Y)=E(X)-E(Y)\\=48-2\\=46$

Compute the variance of (X - Y) as follows:

$V(X - Y) =V(X)+V(Y)-2Cov(X,Y)\\=1+0.0625-0\\=1.0625$

Then the standard deviation of (X - Y) is:

$SD(X-Y) =\sqrt{V(X -Y)}\\\sqrt{1.0625}\\=1.031$

Thus, the expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

(c)

The variance of the sum or difference of two variables is computed by adding the individual variances. This is because the variance of each variable is dependent on the others.

7 0

1 year ago