Taylor charges $15 for the first hour of dog walking and $10 for each additionalhour or fraction of an hour. The rates that Lucy charges for x hours of dog walking

are modeled with the function shown.

f(x) =

{

20x, 0
10x. 2
5x, x >4

Who will charge more to walk a dog for 2.5 hours, Taylor or Lucy? How much more?

Who?:

How much more?:

Answers

Answer 1
Answer:

Answer:

The answer is below

Step-by-step explanation:

Let x represent the number of hours of dog walking. Taylor charges $15 for the first hour and $10 for each additional hour. Therefore:

Let g(x) represent how much Taylor charges. Hence:

g(x) = 15   for 0 ≤ x ≤ 1

g(x) = 15 + 10(x - 1) = 15 + 10x - 10 = 10x + 5   for x > 1

g(x) can be represented by the piecewise function:

g(x)=\left \{ {{15\ \ \ \ \ \ for\ 0\leq x\leq 1} \atop {10x+5\ \ for\ x>1}} \right.

Lucy charges f(x) is modeled by the piecewise function:

f(x)=\left \{ {{20x\ \ \ \ \  \ for\ 0\leq x\leq 2} \atop {10x\ \ \ for\ 0< x< 4}} \right.

Therefore the charge for walking a dog for 2.5 hours by either Taylor or Lucy is:

For Taylor; f(2.5) = 10(2.5) + 5 = $30

For Lucy; g(2.5) = 10(2.5) = $25

Therefore Taylor charges more by walking a dog for 2.5 hours. Taylors charge is $5 more than Lucy charge ($30 - $25).


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A researcher examines 27 sedimentary samples for bromide concentration. The mean bromide concentration for the sample data is 0.383 cc/cubic meter with a standard deviation of 0.0209. Determine the 90% confidence interval for the population mean bromide concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
You go to school in a college town. You know that there are 2000 students enrolled in the school, but you don’t know the population of the town (without students). You walk up and down the main streets of the town, stop people, and ask them if they are students or not. You ask 100 people, and 60 of them say they are students. Estimate the nonstudent population of the town.

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. Historical data indicates that 20% of all potential purchasers select a day visit, 50% choose a one-night visit, and 30% opt for a two-night visit. In addition, 10% of day visitors ultimately make a purchase, 30% of onenight visitors buy a unit, and 20% of those visiting for two nights decide to buy. Suppose a visitor is randomly selected and is found to have made a purchase. How likely is it that this person made a day visit? A one-night visit? A two-night visit?

Answers

Answer:

0.087 = 8.7% probability that this person made a day visit.

0.652 = 65.2% probability that this person made a one-night visit.

0.261 = 26.1% probability that this person made a two-night visit.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = (P(A \cap B))/(P(A))

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Made a purchase.

Probability of making a purchase:

10% of 20%(day visit)

30% of 50%(one night)

20% of 30%(two night).

So

p = 0.1*0.2 + 0.3*0.5 + 0.2*0.3 = 0.23

How likely is it that this person made a day visit?

Here event B is a day visit.

10% of 20% is the percentage of purchases and day visit. So

P(A \cap B) = 0.1*0.2 = 0.02

So

P(B|A) = (P(A \cap B))/(P(A)) = (0.02)/(0.23) = 0.087

0.087 = 8.7% probability that this person made a day visit.

A one-night visit?

Event B is a one night visit.

The percentage of both(one night visit and purchase) is 30% of 50%. So

P(A \cap B) = 0.3*0.5 = 0.15

So

P(B|A) = (P(A \cap B))/(P(A)) = (0.15)/(0.23) = 0.652

0.652 = 65.2% probability that this person made a one-night visit.

A two-night visit?

Event B is a two night visit.

The percentage of both(two night visit and purchase) is 20% of 30%. So

P(A \cap B) = 0.2*0.3 = 0.06

Then

P(B|A) = (P(A \cap B))/(P(A)) = (0.06)/(0.23) = 0.261

0.261 = 26.1% probability that this person made a two-night visit.

What sample size is needed to give a margin of error within in estimating a population mean with 95% confidence, assuming a previous sample had .

Answers

Answer:

n = ((1.96√(\pi(1-\pi)))/(M))^2

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion \pi(if no previous estimate use 0.5) and M is the desired margin of error.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the z-score that has a p-value of 1 - (\alpha)/(2).

The margin of error is of:

M = z\sqrt{(\pi(1-\pi))/(n)}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - (0.05)/(2) = 0.975, so Z = 1.96.

Needed sample size:

The needed sample size is n. We have that:

M = z\sqrt{(\pi(1-\pi))/(n)}

M = 1.96\sqrt{(\pi(1-\pi))/(n)}

√(n)M = 1.96√(\pi(1-\pi))

√(n) = (1.96√(\pi(1-\pi)))/(M)

(√(n))^2 = ((1.96√(\pi(1-\pi)))/(M))^2

n = ((1.96√(\pi(1-\pi)))/(M))^2

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion \pi(if no previous estimate use 0.5) and M is the desired margin of error.

At the U.S. Open Tennis Championship a statistician keeps track of every serve that a player hits during the tournament. The statistician reported that the mean serve speed was 100 miles per hour (mph) and the standard deviation of the serve speeds was 15 mph. Assume that the statistician also gave us the information that the distribution of serve speeds was mound- shaped and symmetric. What percentage of the player's serves were between 115 mph and 145 mph

Answers

Answer:

15.74% of the player's serves were between 115 mph and 145 mph

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

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

Z = (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.

In this question, we have that:

\mu = 100, \sigma = 15

What percentage of the player's serves were between 115 mph and 145 mph

This is the pvalue of Z when X = 145 subtracted by the pvalue of Z when X = 115.

X = 145

Z = (X - \mu)/(\sigma)

Z = (145 - 100)/(15)

Z = 3

Z = 3 has a pvalue of 0.9987

X = 115

Z = (X - \mu)/(\sigma)

Z = (115 - 100)/(15)

Z = 1

Z = 1 has a pvalue of 0.8413

0.9987 - 0.8413 = 0.1574

15.74% of the player's serves were between 115 mph and 145 mph

Final answer:

A total of 27% of the player's serves at the U.S. Open Tennis Championship were between 115mph and 145mph. This was found using the Empirical Rule which applies to a normal distribution of serve speeds.

Explanation:

This problem is a classic example of the use of the Empirical Rule in statistics. The Empirical Rule, also known as the 68-95-99.7 rule, applies to a normal distribution, which is a bell-shaped curve (mound-shaped and symmetric) as mentioned in the problem. This rule states that approximately 68% of the data falls within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

Given that the mean serve speed is 100 mph and the standard deviation is 15 mph, serves of 115 mph are one standard deviation above the mean and serves of 145 mph are three standard deviations above the mean. Therefore, we are looking for the percentage of serves between these two values. According to the Empirical Rule, this would be 95% (coverage for up to 2 standard deviations) minus 68% (coverage for up to 1 standard deviation), which equals 27%. So, 27% of the player's serves were between 115 mph and 145 mph.

Learn more about Empirical Rule here:

brainly.com/question/35669892

#SPJ3

Write an explicit equation for the arithmetic sequence
3,7,11,15,19

Answers

Answer:

Arithmetic Sequence:  

d=4

Step-by-step explanation:

This is an arithmetic sequence since there is a common difference between each term. In this case, adding  4 o the previous term in the sequence gives the next term. In other words, aₙ=a₁+d(n-1)

Find an equation in standard form for the ellipse with the vertical major axis of length 10 and minor axis of length 8

Answers

Answer:   The required equation of the ellipse in standard form is (y^2)/(25)+(x^2)/(16)=1.

Step-by-step explanation:  We are given to find the equation of an ellipse in standard form with the vertical major axis of length 10 units and minor axis of length 8 units.

Since the major axis is vertical, so it will lie on the Y-axis. Let the standard form of the ellipse be given by

(y^2)/(a^2)+(x^2)/(b^2)=1,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

where the length of major axis is 2a units and length of minor axis is 2b units.

According to the given information, we have

2a=10\n\n\Rightarrow a=(10)/(2)\n\n\Rightarrow a=5

and

2b=8\n\n\Rightarrow b=(8)/(2)\n\n\Rightarrow b=4

Substituting the values of a and b in equation (i), we get

(y^2)/(5^2)+(x^2)/(4^2)=1\n\n\n\Rightarrow (y^2)/(25)+(x^2)/(16)=1.

Thus, the required equation of the ellipse in standard form is (y^2)/(25)+(x^2)/(16)=1.

(x/h)^2+(y/v)^2=1   where h is the horizontal radius and v is the vertical radius

In this question it seem that they are saying the length of the axis and not radius so I would cut them in half so that they are radii...then:

(x/4)^2+(y/5)^2=1

x^2/16+y^2/25=1

A basket of fruit contains 4 bananas, 3 apples, and 5 oranges. You intend to draw a piece of fruit from the basket, keep it, and then draw a 2nd piece of fruit from the basket. What is the probability of selecting two oranges in a row from the basket if you are blindfolded?

Answers

Answer: 5/33

Step-by-step explanation: There are 12 fruits and 5 are oranges.

If you draw an orange, then there are 11 fruits and 4 oranges. (5/12)x(4/11)=20/132 or 5/33 in simplest form.