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2 years ago

8

Cody started a fire in a field. The fire was expanding outward in all directions forming a larger and larger circle. The radius

of the circular fire was 10 feet and was increasing at 2 feet/minute when Cody called the fire department. How many minutes will it take from the time Cody called the fire department until the area of the circular fire is 45401 square feet? Type your answer to the nearest hundredth (2 decimal places).

1 answer:

Dimas [21]2 years ago

8 0

Answer:

<h2> 55.10min </h2>

Step-by-step explanation:

This problem is on the mensuration of flat shapes.

given that the initial radius before Cody alerted the fire department is

10 ft

Then we can calculate the area of the fire as

Area= πr^2

Area=3.142*(10)^2

Area=3.142*100

Area=314.2ft^2

Also, given that the area of the circular fire has grown to 45401 square feet

let us solve for the radius first, then we can calculate the time taken since the radius was increasing by 2 ft/min

let the radius of the large fire be R

45401=πR^2

R^2=45401/3.142

R^2=14449.714

square both sides we have

R=√14449.714

R=120.2ft

The difference between the radius of the initial fire and the large fire is

R-r=120.2-10

=110.2ft

Hence the time taken can be calculated as

time= 110.2/2

time=55.10min

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The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical

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Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538$

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

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

By the Central Limit Theorem

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

$Z = \frac{36 - 38.4}{0.8538}$

$Z = -2.81$

$Z = -2.81$ has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

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

$Z = \frac{41 - 38.4}{0.8538}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now $n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08$

So

a)

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

b)

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

$Z = \frac{36 - 38.4}{1.08}$

$Z = -2.22$

$Z = -2.22$ has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

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

$Z = \frac{41 - 38.4}{1.08}$

$Z = 2.41$

$Z = 2.41$ has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

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2 years ago

kifflom [539]

Answer:

The answers to the questions about problem 2 are:

The fraction of the length after Priya stops two times is 3/4.
The fraction of the length after Priya stops four times is 15/16.
Priya will never reach the end of the hallway.

Step-by-step explanation:

The explanation about each answer is below:

1. In the first stop, Priya has walked half of the total length, and there would still be half the hallway, however, when she stops the second time, she has walked the half of the half, I mean:

$\frac{1}{2}/2=\frac{1}{4}$

If you add the two values you obtain the total distance traveled by Priya:

$\frac{1}{2}+\frac{1}{4}= \frac{3}{4}$

And the distance traveled in two stops is 3/4 of the total length of the hallway.

2. With four stops the system is the same, we know with two stops Priya travels 3/4 of the hallway, now with the third stop, she will travel half of the remaining distance:

$\frac{1}{4}/2=\frac{1}{8}$

And in the fourth stop she will travel half of the remaining, I mean:

$\frac{1}{8}/2=\frac{1}{16}$

Now, we add all the values of the distances obtained:

$\frac{3}{4}+\frac{1}{8}+\frac{1}{16}= \frac{15}{16}$

So, the distance traveled by Priya in the fourth stop is 15/16 of the total length of the hallway.

3. How we know, the numbers are infinite, in the same forms the distances, by this reason, how the problem says that Priya walks just half of the distance, she will never reach the end because despite she has very near of the end, she will continue walking just half and ever smaller distances.

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3 years ago

The differnce between 3 and 11

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Answer:

-8

Step-by-step explanation:

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75% of the temperatures are below what value? How do you know? 75% of the temperatures are above what value? How do you know? Wh

Savatey [412]

Answer:

Complete Question:

Pick a city in Maryland and determine the average high temperatures for each month. Record this in a table.

a. Create a box-and-whisker plot for the data.

b. What is the median temperature?

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Step-by-step explanation:

TASK 3

The city picked is Baltimore, the year 2017.

The table is attached in picture #1.

a) The box-and-whisker plot is attached in picture #2.

b) The median is the central value of the distribution i.e:

42.4, 45.3, 46.8 | 54.7, 57.6, 66.0 | 69.1, 76.5, 80.6 | 85.8, 87.8, 90.0

where we marked with a tally the quartiles.

As we have 12 values, the average between the two central values will be the median :

M = (66.0 + 69.1) / 2 = 67.1 °F

c) 75% of the temperatures are below 83.2 °F.

Indeed, this value represents the third quartile. The position of the third quartile can be found by the formula:

3/4 · (n + 1) = 3/4 · (12 + 1) = 3/4 · 13 = 9.75

Therefore, since we did not get in integer position, the third quartile will be the average between the numbers in position 9 and 10:

q₀.₇₅ = (80.6 + 85.8) / 2 = 83.2 °F

d) 75% of the temperatures are above 50.8 °F.

Indeed, this value represents the first quartile. Similarly to point c), the position can be found by the formula:

1/4 · (n + 1) = 1/4 · 13 = 3.25

And therefore:

q₀.₂₅ = (46.8 + 54.7) / 2 = 50.8 °F.

e) Baltimore in 2017 had a median high temperature of 67.1 °F.

25% of the year the high temperatures were warmer than 83.2 °F, with no outlier above 90.0 °F which was the hottest temperature.

25% of the high temperatures were colder than 50.8 °F, with no outlier below 42.4 °F, which was the coldest high temperature.

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