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Evgen [1.6K]
3 years ago
5

Why did the United States attack Afghanistan in 2001?

Mathematics
2 answers:
vivado [14]3 years ago
7 0

Answer:

The answer is: Osama bin Laden, aided by the Taliban, hid in Afghanistan after the 9/11.

Step-by-step explanation:

I took the test :)

Oduvanchick [21]3 years ago
6 0

Answer:

These are for the whole test. the first answer is everything but,Saddam Hussein and Iraq assisted al Qaeda in the terrorist attacks.

The second answer is all but, Members of al Qaeda believe they will be punished by God if they take the lives of non-Muslims.

and the answer to the final question is  

Osama bin Laden, aided by the Taliban, hid in Afghanistan after the 9/11 attacks.

Mark me brainiest please and this is for K12

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The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviat
aleksley [76]

Answer:

4.5 hour and 2 hour

Step-by-step explanation:

Given: The mean number of hours per day spent watching television, according to a national survey, is 3.5 hours, with a standard deviation of two hours.

To Find: If each time was increased by one hour, what would be the new mean and standard deviation.

Solution:

let the total numbers entries of hours in survey be   =   \text{N}

each entry in survey be  =  \text{x}_{i}

mean of survey is     =     \mu  =3.5 \text{hours}

standard deviation is  =   \sigma = 2 \text{hours}

if each entry in survey is increased by one hour then,

each new entry in survey  is =  \text{x}_{i}+1

the new mean is\mu_{new}  = \frac{\text{sum of all hours}}{total number of entries}

                                                \frac{\text{x}_{1} +1+\text{x}_{2}+1 +....+\text{x}_{N}+1 }{N}

                                                \frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})+\text{N}\times1 }{N}

                                                \frac{(\text{x}_{1} +\text{x}_{2} +....+\text{x}_{N})}{N}+\frac{\text{N}\times1 }{N}

                                             \mu_{new} = \mu + 1=4.5 \text{hours}

now,

standard deviation is      \sigma_{new}= \sqrt{\sum_{1}^{N}\frac{(\text{x}_{inew}-\mu_{new})^{2}}{N}}

                                   \text{x}_{inew}= \text{x}_{i}+1

                                   \mu_{new}=\mu+1        

putting values,

                                                \sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}+1-\mu-1)^{2}}{N}}

                                                \sqrt{\sum_{1}^{N}\frac{(\text{x}_{i}-\mu)^{2}}{N}}= \sigma

                                 \sigma_{new}        =2

new mean and standard deviation are 4.5 and 2 \text{hours}

                                               

5 0
3 years ago
Suppose that the function g is defined, for all real numbers, as follows.
lesantik [10]
<h3>Answers:</h3>
  • g(-1) = -1
  • g(1) = -3/2
  • g(5) = -7/2

======================================

Explanation:

The piecewise function g(x) is defined based on what the input is.

  • If the input x is less than -2, then g(x) = 2
  • Or if -2 \le x < 1 then g(x) = -(x-1)^2+3
  • Or if x \ge 1 then g(x) = (-1/2)x-1

---------------

In the case of g(-1), we have x = -1 which makes -2 \le x < 1 true.

So we'll use the second piece

g(x) = -(x-1)^2+3

g(-1) = -(-1-1)^2+3

g(-1) = -(-2)^2+3

g(-1) = -4+3

g(-1) = -1

----------------

For g(1), we have x = 1 which makes x \ge 1 true

We'll use the third piece

g(x) = (-1/2)x-1

g(1) = (-1/2)*1 - 1

g(1) = -1/2 - 1

g(1) = -1/2 - 2/2

g(1) = -3/2

-----------------

For g(5), we have x = 5 which makes x \ge 1 true

g(x) = (-1/2)x-1

g(5) = (-1/2)*5-1

g(5) = -5/2 - 1

g(5) = -5/2 - 2/2

g(5) = -7/2

8 0
3 years ago
Solve for X<br> x - y = 6
natka813 [3]
If I am not mistaken the answer should be
X=5
Y=-1
5-(-1)=6.
7 0
3 years ago
How many solutions does the following system of equations have?
Allushta [10]
I think it would be two solutions 
4 0
4 years ago
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hodyreva [135]
They would get 1 cheesecake a piece and there would be 2 left and if they used that then they would have to cut that into 4 pieces for both cheesecakes and they would have 1 2/8 of cheesecake
4 0
3 years ago
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