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2 years ago

Which are examples of unit rates? Check all that apply.

Mathematics

2 answers:

mylen [45]2 years ago

5 0

I believe the answer is $100 for 5 tickets. I hope this helps!

spayn [35]2 years ago

5 0

Answer:

The correct answer is A and C.

Step-by-step explanation:

I know there is other people saying something different but i just did it and i got it correct. Have a nice day <3

You might be interested in

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

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 samples 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

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

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

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

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

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

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago

Which statements describe an object in motion that has no external force acting on it? Check all that apply.

Step2247 [10]

Answer: it moves at a constant speed and stays in the same direction

Step-by-step explanation:

It does this because there is no friction or nothing pushing on it therefore there is nothing to slow it down or speed it up

5 0

2 years ago

Read 2 more answers

2. Tess picked 12 flowers on Monday morning In the afternoon she picked some more flowers. If Tess picked 27 flowers in all, how

sammy [17]

27 in total
12 in the morning
27-12 = 15
She picked 15 in the afternoon

5 0

2 years ago

A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto

ruslelena [56]

X(u, v) = (2(v - c) / (d - c) + 1)cos(pi * (u - a) / (2b - 2a))
y(u, v) = (2(v - c) / (d - c) + 1)sin(pi * (u - a) / (2b - 2a))

As v ranges from c to d, 2(v - c) / (d - c) + 1 will range from 1 to 3, which is the perfect range for the radius. As u ranges from a to b, pi * (u - a) / (2b - 2a) will range from 0 to pi/2, which is the perfect range for the angle. So, this maps the rectangle to R.

6 0

2 years ago

A campus deli serves 300 customers over its busy lunch period from 11:30 a.m. to 1:30 p.m. A quick count of the number of custom

Irina-Kira [14]

Complete question is;

A campus deli serves 300 customers over its busy lunch period from 11:30 am to 1:30 pm. A quick count of the number of customers waiting in line and being served by the sandwich makers shows that an average of 10 customers are in process at any point in time. What is the average amount of time that a customer spends in process?

Answer:

4 minutes

Step-by-step explanation:

For this question, we will apply Little's law which is is a theorem that determines the average number of items in a stationary queuing system, based on the average waiting time of an item within a system and the average number of items arriving at the system per unit of time.

The formula for the law is:

Inventory = flow rate × flow time.

We are given;

Inventory = 300 customers

flow time is from 11: 30am to 1:30pm which is 2 hours = 120 minutes

Flow rate = 300/120 = 2.5 persons/minute

Now, Making flow rate the subject of the formula earlier given, we have;

flow rate = inventory/ flow time

Flow time is the time each person spends in the process

Thus, plugging in the relevant values, we get ;

We are told that an average of 10 customers are in process at any point in time.

Thus;

Average flow time = average inventory/flow rate = 10/2.5 = 4 minutes

3 0

2 years ago