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2 years ago

1. If an android phone is out of date or outdated and it requires an update. Explain the steps how I can update this phone.

Computers and Technology

2 answers:

BartSMP [9]2 years ago

7 0

An android phone is outdated .Like all brands there is

No more security patches.
No more Operating system updates.

If it still requires updates then follow the instructions

Disable OEM by going through developer options.It will directly turnoff all security restrictions from first (Your phone is not receiving any security patches so these restrictions are not of any use now)
Now try to search through web for finding custom roms of most trusted companies .
Some examples:-Pixel,Linux, Microsoft etc
Install them on phone and activate.
You will experience those UIS on your phone+There will be no harm to device as they are equipped with security patches for whole life .

The trusted antivirus apps

Norton
Bitdefender
Avast

Alexandra [31]2 years ago

6 0

Answer:

Question 1:

» Go to settings

» Go to phone or device system

» Go to system update

» If you have update files, go to local update

» If you lack update files, go to online update

» The phone will reboot.

${ \tt{ \red{make \: sure \: battery \: is \: above \: 50\%}}}$

or else, connect the OTP charger.

Question 2:

• Avast mobile security

• AVG antivirus [ an avast inc. ]

• Phone master

• Norton mobile security and firewall

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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

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Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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