Answer:
53
Step-by-step explanation:
Given: The sum of two digit number is 8
Reversing the digit will get us number 18 less than the original.
Lets take x as tenth digit of our number and y as unit digit of our number.
As given sum of digit is 8
∴ 
∴ 
- equation 1
We also know that reversing the digit will get us number 18 less than the original.
∴ 
Now, lets put the value of y from equation 1
⇒ 
⇒ 
⇒ 
∴ 
Next, substituting the value of x in equation 1
⇒ 
∴ 
∴ The original number is 53, sum of the digit is 8 and if we reverse the digit of the number, we get 35, which is 18 less than the original number.
Answer: 3
Step-by-step explanation:
Let the smaller number be y.
Since the difference of two numbers is 5,then the larger number will be: y+5
The sum of three times the larger number and twice the smaller number is 30. This can be written as:
3(y+5) + 2y = 30
3y + 15 + 2y = 30
5y + 15 = 30
5y = 30 - 15
5y = 15
y = 15/5
y = 3
The smaller number is 3
Based on the amount predicted to be spent, the hypotheses will be:
- Null Hypothesis = $1,007.24
- Alternate Hypothesis ≠ $1,007.24
The Null Hypothesis is the one that confirms the prediction so in this case it will be that the average shopper will indeed spend $1,007.24.
The Alternate Hypothesis theorizes that the event being predicted will not happen so in this case that would mean that the shopper would not spend $1,007.24.
In conclusion, the null hypothesis confirms and the alternate denies.
<em>Find out more on the null hypothesis at brainly.com/question/15980493.</em>
Answer:I think it’s 16 because 4x4=16
Step-by-step explanation: :D
Answer:
0(n)
Step-by-step explanation:
Result previous exercise:
procedure count(a1a2...an : string with n > 1)
i:=0
for k:=1 to n
if ak =1 then i:=i + I
return i
<em>Note</em><em>: If you use a different algorithm, then you could possible get different results. </em>
<u>SOLUTION </u>
There is only one part of the code that contains an operation (comparison), namely if a_k =1
This comparison is executed in every iteration of the for-loop
k can take on the values from 1 to n (for k:=I to n), thus k can take on n values.
Thus in total there are then n comparisons, while n is 0(n).
