Answer:
46
Step-by-step explanation:
Let 
represent a number with 
in the ten's position and 
is the one's position.
This means actually has value of 
.
We are given the sum of those digits of is 10; this means 
.
It says if 18 is added to the number , then the result is 
.
So has value and
has value 
.
We are given then:
Subtract 
on both sides:
Simplify:
Subtract on both sides:
Divide both sides by 9:
Rearrange by commutative property:
So the system of equations we want to solve is:
-------------------------Add equations together (this will eliminate the variable and allow you to go ahead and solve for :
Divide both sides by 2:
Simplify:
If and , then 
. since 4+6=10.
So the original number is (46).
18 more than 46 is 18+46=(64) which is what we wanted.
We also have the sum of 4 and 6 is 10 as well.
To solve this first we have to find out how many seconds are in 15 minutes.
So, 15 x 60 since there is 60 seconds in a minute, and of course there are 15 minutes.
15 x 60 = 900 seconds.
Now we need to find out about how many seconds does it take George to blink once. So we divide 900 seconds by 85 blinks.
900 / 85 = About 11 seconds for one blink.
Now we use the rate of 1 blink = 11 seconds and multiply. Since they said 51 blinks you would multiply 51 blinks by 11 seconds.
51 x 11 = About 561 seconds.
Answer: About 561 seconds passed in 51 blinks.
Answer:
x = √146 = 12.1 (nearest tenth)
Step-by-step explanation:
Apply pythagorean theorem, c² = a² + b², where,
c = x
a = √82 km
b = 8 km
Plug in the values
x² = (√82)² + 8²
x² = 82 + 64
x² = 146
x = √146 = 12.1 (nearest tenth)
Answer:
a) 1/2
b) 1/n
c) 1/4
Step-by-step explanation:
a) For each permutation, either 1 precedes 2 or 2 precedes 1. For each permutation in which 1 precedes 2, we can swap 1 and 2 to obtain a permutation in which 2 preceds 1. Thus, half of the total permutations will involve in 1 preceding 2, hence, the probability for a permutation having 1 before 2 is 1/2.
c) If 2 is at the start of the permutation, then it is impossible for 1 to be before 2. If that is not the case, then 1 has a probability of 1/n-1 to be exactly in the position before 2. We can divide in 2 cases using the theorem of total probability,
P( 1 immediately preceds 2) = P (1 immediately precedes 2 | 2 is at position 1) * P(2 is at position 1) + P(1 immediately precedes 2 | 2 is not at position 1) * P(2 is not at position 1) = 0 * 1/n + (1/n-1)*(n-1/n) = 1/n.
d) We can divide the total of permutations in 4 different groups with equal cardinality:
- Those in which n precedes 1 and n-1 precedes 2
- those in which n precedes 1 and 2 precedes n-1
- those in which 1 precedes n and n-1 precedes 2
- those in which 1 precedes n and 2 precedes n-1
All this groups have equal cardinality because we can obtain any element from one group from another by making a permutations between 1 and n and/or 2 and n-1.
This means that the total amount of favourable cases (elements of the first group) are a quarter of the total, hence, the probability of the event is 1/4.
Answer: it’s less 5<10
Step-by-step explanation:
