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Lina20 [59]
3 years ago
13

Catherine took her camera to the repair shop. The technician at the shop told her that acid had leaked into her camera. What cou

ld be the possible reason for acid leaking into Catherine’s camera?
A. She used the camera even though the battery was almost empty.
B. She wiped the camera lens with a moist cloth.
C. She left the camera in harsh daylight.
D. She left fully discharged batteries in her camera.
Computers and Technology
1 answer:
tino4ka555 [31]3 years ago
3 0

Answer:

B

Explanation:

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The term best describing these lines is called pseudocode. Essentially a “text-based” detail (algorithmic) design tool.
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3 years ago
Use the following cell phone airport data speeds​ (Mbps) from a particular network. Find P10. 0.1 0.1 0.3 0.3 0.3 0.4 0.4 0.4 0.
sasho [114]

Answer:

P_{10} =0.1

Explanation:

Given

0.1, 0.1, 0.3, 0.3, 0.3, 0.4, 0.4, 0.4, 0.6, 0.7, 0.7, 0.7, 0.8, 0.8

Required

Determine P_{10}

P_{10} implies 10th percentile and this is calculated as thus

P_{10} = \frac{10(n+1)}{100}

Where n is the number of data; n = 14

P_{10} = \frac{10(n+1)}{100}

Substitute 14 for n

P_{10} = \frac{10(14+1)}{100}

P_{10} = \frac{10(15)}{100}

Open the bracket

P_{10} = \frac{10 * 15}{100}

P_{10} = \frac{150}{100}

P_{10} = 1.5th\ item

This means that the 1.5th item is P_{10}

And this falls between the 1st and 2nd item and is calculated as thus;

P_{10} = 1.5th\ item

Express 1.5 as 1 + 0.5

P_{10} = (1 +0.5)\ th\ item

P_{10} = 1^{st}\ item +0.5(2^{nd} - 1^{st}) item

From the given data; 1st\ item = 0.1 and 2nd\ item = 0.1

P_{10} = 1^{st}\ item +0.5(2^{nd} - 1^{st}) item becomes

P_{10} =0.1 +0.5(0.1 - 0.1)

P_{10} =0.1 +0.5(0)

P_{10} =0.1 +0

P_{10} =0.1

3 0
3 years ago
Write an application that determines whether the first two files are located in the same folder as the third one. The program sh
marissa [1.9K]

Answer:

The program in Python is as follows:

fname1 = input("Filepath 1: ").lower()

fname2 = input("Filepath 2: ").lower()

fname3 = input("Filepath 3: ").lower()

f1dir = ""; f2dir = ""; f3dir = ""

fnamesplit = fname1.split("/")

for i in range(len(fnamesplit)-1):

   f1dir+=fnamesplit[i]+"/"

fnamesplit = fname2.split("/")

for i in range(len(fnamesplit)-1):

   f2dir+=fnamesplit[i]+"/"

fnamesplit = fname3.split("/")

for i in range(len(fnamesplit)-1):

   f3dir+=fnamesplit[i]+"/"

if f1dir == f2dir == f3dir:

   print("Files are in the same folder")

else:

   print("Files are in the different folder")

Explanation:

The next three lines get the file path of the three files

<em>fname1 = input("Filepath 1: ").lower()</em>

<em>fname2 = input("Filepath 2: ").lower()</em>

<em>fname3 = input("Filepath 3: ").lower()</em>

This initializes the directory of the three files to an empty string

f1dir = ""; f2dir = ""; f3dir = ""

This splits file name 1 by "/"

fnamesplit = fname1.split("/")

This iteration gets the directory of file 1 (leaving out the file name)

<em>for i in range(len(fnamesplit)-1):</em>

<em>    f1dir+=fnamesplit[i]+"/"</em>

This splits file name 2 by "/"

fnamesplit = fname2.split("/")

This iteration gets the directory of file 2 (leaving out the file name)

<em>for i in range(len(fnamesplit)-1):</em>

<em>    f2dir+=fnamesplit[i]+"/"</em>

This splits file name 3 by "/"

fnamesplit = fname3.split("/")

This iteration gets the directory of file 3 (leaving out the file name)

<em>for i in range(len(fnamesplit)-1):</em>

<em>    f3dir+=fnamesplit[i]+"/"</em>

This checks if the file directories hold the same value

This is executed, if yes

<em>if f1dir == f2dir == f3dir:</em>

<em>    print("Files are in the same folder")</em>

This is executed, if otherwise

<em>else:</em>

<em>    print("Files are in the different folder")</em>

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Explanation:

Thanks!!!!!

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3 years ago
"En la opción
Katyanochek1 [597]
Answer: I think email

Explanation: not rlly sure tho..
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