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Sindrei [870]
2 years ago
7

A pet store donated 50 pounds of food for adult dogs, puppies, and cats to an animal shelter. 19 and 3/4 pounds was adult dog fo

od and 18 and 7/8 pounds was puppy food. How many pounds of cat food did the pet store donate?
Mathematics
2 answers:
Sati [7]2 years ago
8 0
So first we will add together the fractions. 3/4 is equal to 6/8, so we will add 6/8 to 7/8 and get 13/8, or 1 5/8.  Then we can add together the whole numbers, 19 and 18, to get 37. Then add 37 + 1 5/8 and get 38 5/8. Lastly, we subtract that amount from 50 to get a finale number of 11 3/8. The pet store donated 11 3/8 pounds of cat food! Hope this helps!
madam [21]2 years ago
3 0
50 - (19 3/4 + 18 7/8) = the cat food
50 - (19 6/8 + 18 7/8) 
50 - (37 13/8) =
50 - (38 5/8) = 
11 3/8 lbs of cat food <==
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1 year ago
Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in
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a) There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c) The expected number of Chrome users is 4.074.

d) The variance for the number of Chrome users is 3.2441.

The standard deviation for the number of Chrome users is 1.8011.

Step-by-step explanation:

For each Internet browser user, there are only two possible outcomes. Either they use Chrome, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

Google Chrome has a 20.37% share of the browser market. This means that p = 0.2037

20 Internet users are sampled, so n = 20.

a.Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is P(X = 8).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243

There is a 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b.Compute the probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

Either there are less than 3 Chrome users, or there are three or more. The sum of the probabilities of these events is decimal 1. So:

P(X < 3) + P(X \geq 3) = 1

P(X \geq 3) = 1 - P(X < 3)

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105

P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538

P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.1950

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1950 = 0.8050

There is an 80.50% probability that at least 3 of the 20 Internet browsers users use Chrome as their Internet browser.

c.For the sample of 20 Internet browser users, compute the expected number of Chrome users

We have that, for a binomial experiment:

E(X) = np

So

E(X) = 20*0.2037 = 4.074

The expected number of Chrome users is 4.074.

d.For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

We have that, for a binomial experiment, the variance is

Var(X) = np(1-p)

So

Var(X) = 20*0.2037*(0.7963) = 3.2441

The variance for the number of Chrome users is 3.2441.

The standard deviation is the square root of the variance. So

\sqrt{Var(X)} = \sqrt{3.2441} = 1.8011

The standard deviation for the number of Chrome users is 1.8011.

6 0
2 years ago
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