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kykrilka [37]
2 years ago
14

Ellen is downloading a file. After 2 minutes, there are 38.4 megabytes left to download. After 8 minutes, there is only 9.6 mega

bytes left to download. What is the size, in megabytes, of the file.?
Mathematics
2 answers:
atroni [7]2 years ago
7 0
Find the average download rate.
(38.4 - 9.6) / (2-8) = -4.8 mb/min

that means in the first 2 min she had downloaded 2(4.8)  = 9.6 mb

38.4 mb left + 9.6 mb already downloaded = 48 mb total
 

Oksana_A [137]2 years ago
3 0

Answer:

The size of the file is of 48 megabytes.

Step-by-step explanation:

8-2 = 6 minutes

38.4 - 9.6 = 28.8 megabytes

As we can see, 28.8 megabytes were downloaded in 6 minutes. Assuming a constanct and non stopped download speed, we can affirm that 4.8 megabytes were download per minute (28.8 / 6 = 4.8). As after 2 minutes, there were 38.4 megabytes left, in order to know the size of the file we have to do the following calculation:

38.4 + (4.8 x 2) = Size of the file

38.4 + 9.6 = 48 megabytes

The size of the file is of 48 megabytes.

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Suppose we want to choose 5 objects, without replacement, from 13 distinct objects. (a) How many ways can this be done, if the o
Arisa [49]

Answer:

A. 1, 287 ways

B. 154,440 ways

Step-by-step explanation:

A. We want to choose 5 objects from a total 13, without considering the order in which they are chosen.

The correct way to do this is by using the combination formula since order is not considered;

Thus we have ; 13 C 5 read as 13 combination 5;

Mathematically, n C r is ; n!/(n-r)!r!

Thus, we have ;

13!/(13-8)!8! = 13!/5!8! = 1,287 ways

B. By considering order, we shall be using the permutation formula;

Mathematically n P r = n!/(n-r)!

Read as n permutation r;

Using the numbers involved, we have ; 13 P 5

= 13!/(13-5)! = 13!/8! = 154,440 ways

8 0
2 years ago
Help please.<br><br> Find x.
Degger [83]
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5 0
2 years ago
A random sample of 50 units is drawn from a production process every half hour. the true fraction of nonconforming products manu
Naya [18.7K]

solution:

The probability mass function for binomial distribution is,

 

Where,

X=0,1,2,3,…..; q=1-p

find the probability that (p∧ ≤ 0.06) , substitute the values of sample units (n) , and the probability of nonconformities (p) in the probability mass function of binomial distribution.

Consider x   to be the number of non-conformities. It follows a binomial distribution with n   being 50 and p  being 0.03. That is,

binomial (50,0.02)

Also, the estimate of the true probability is,

p∧  = x/50

The probability mass function for binomial distribution is,

 

Where,

X=0,1,2,3,…..; q=1-p

The calculation is obtained as

P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)

         = 50cx ₓ (0.03)x ₓ (1-0.03)50-x  

=    (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ      (0.03)3 ₓ (1- 0.03)50-3 )

=(   ₓ (0.03)0 ₓ (1-0.03)50-0 +  ₓ (0.03)1 ₓ (1-0.03)50-1 +   ₓ (0.03)2 ₓ (1-0.03)50-2   ₓ (0.03)3 ₓ (1-0.03)50-3 )



5 0
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Answer:

The answer for edgednuity is

The length of side AD is 8 units.

The length of side A'D' is 4 units.

The scale factor is 1/2

Step-by-step explanation:


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