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kogti [31]
3 years ago
5

Timmykat is on the prowl. The kitten crusader has caught 5 bad guys already today and hopes to be able to catch 2.5 criminals ev

ery hour. What are possible amounts of hours that Timmykat needs to be on the prowl to catch at least 30 bad guys?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
5 0
30 divided by 2.5 
so 12 hours until he catches 30 bad guys 
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The length of the rectangle is 7/(x-4) feet, while its width is 5/x feet. Find its perimeter. Please help asap!!!!! :(
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The perimeter would be (24x - 40)/(x^2 - 4x).

In order to find this, first double the length and width as you would to find any perimeter.

7/(x - 4) * 2 = 14/(x - 4)

5/x * 2 = 10/x

Now to add those together, we need to give them common denominators. In order to do that with the first one, we need to multiply by x/x

14/(x - 14) * x/x = 14x/(x^2 - 14x)

Then we can do the same with the second part by multiplying by (x - 4)/(x - 4)

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2 years ago
Slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 67.7 grams an
Alexxx [7]

Answer:

<u>a. s.e. = 0.338</u>

<u>b. The  probability of finding a random slice of pizza with a mass of less than 67.2 grams is 0.5871 or 58.71%</u>

<u>c. The probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams is 0.0694 or 6.94%</u>

d. <u>The sample mean of 62.75 would represent the 15th percentile</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question properly:

μ of the mass of slices of pizza for a certain brand = 67.7 grams

σ of the mass of slices of pizza for a certain brand = 2.28 grams

2. For samples of size 20 pizza slices, what is the standard deviation for the sampling distribution of the sample mean?

Let's recall that the standard deviation of the sampling distribution of the mean is called the standard error of the mean and its formula is:

μσs.e.= √σ/n, where n is the sample size.

s.e. = √2.28/20

<u>s.e. = 0.338</u>

3. What is the probability of finding a random slice of pizza with a mass of less than 67.2 grams?

Let's find the z-score for X = 67.2, this way:

z-score = (X - μ)/σ

z-score = (67.2 - 67.7)/2.28

z-score = 0.5/2.28

z-score = 0.22 (rounding to the next hundredth)

Now, using the z-table, let's find p, the probability:

p (z = 0.22) = 0.5871

<u>The  probability of finding a random slice of pizza with a mass of less than 67.2 grams is 58.71%</u>

4.  What is the probability of finding a 20 random slices of pizza with a mean mass of less than 67.2 grams?

Let's use the central limit theorem to find the z-score, this way:

z-score = X - μ/s.e

z-score = 67.2 - 67.7/0.338

z-score = -0.5/0.338

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Now, using the z-table, let's find p, the probability:

<u>p (z = -1.48) = 0.0694</u>

5. What sample mean (for a sample of size 20) would represent the bottom 15% (the 15th percentile)?

For p = 0.150, let's find the z-score:

z-score = - 2.17

z-score = X - μ/s.e

- 2.17 = (X - 67.7)/2.28

- 4.95 = X - 67.7

X = 67.7 - 4.95

X = 62.75 (rounding to the next hundredth)

<u>The sample mean of 62.75 would represent the 15th percentile</u>

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