The first answer is 3/34 sec and the second answer is 15/34 sec.
Set up a proportion for these problems. For the first question,
340/1 = 30/x
Cross multiply:
340*x = 1*30
340x = 30
Divide both sides by 340:
340x/340=30/340
x = 30/340 = 3/34
For the second question,
340/1 = 150/x
Cross multiply:
340*x = 150*1
340x=150
Divide both sides by 340:
340x/340 = 150/340
x = 150/340 = 15/34
Answer: 
Step-by-step explanation:
The given geometric series are:
a)
on simplifying in decimals, we get
b) 
on simplifying in decimals, we get
c)
on simplifying in decimals, we get
Thus, this geometric series represent 0.4444.
d)
on simplifying in decimals, we get
First of all, lets consider that you made a litte mistake and you meant this problem.........
<span>"The combined average weight of an okapi and a llama is 450 kilograms. The average weight of 3 llamas is 190 kilograms more than the average weight of one okapi. On average, how much does an okapi weigh, and how much does a llama weigh?"
This is a system of two equations.
Let it be X the average weight of a LLAMA
And Y the average weight of an OKAPI
X + Y = 450 kg 1)
3X = 190 kg +Y 2)
So, with 1) we have that Y = 450 - X
We subsitute in 2) and we have
3X = 190 + (450 -X).............We solve for X ....==> 4X = 640kg ==> X = 160kg
..We replace X in 1 and get => Y = 450kg -X = 450kg -160kg = 290kg
</span>160kg....... average weight of a LLAMA
290kg........average weight of an OKAPI
Answer:
<h2>a) 1.308*10¹² ways</h2><h2>b)
455 way</h2>
Step-by-step explanation:
If there are 15 balls labeled 1 through 15 in a standard football game, the order of arrangement of the 15 balls can be done in 15! ways.
15! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2
15! = 1.308*10¹² ways
b) If 3 of the 15 balls are to be chosen if order does not matter, this can be done in 15C3 number of ways. Since we are selecting some balls out of the total number of balls, we will use the concept of combination.
Using the combination formula nCr = n!/(n-r)!r!
15C3 = 15!/(15-3)!3!
15C3 = 15!/12!3!
15C3 = 15*14*13*12!/12!*6
15C3 = 15*14*13/6
15C3 = 455 ways
To write the coefficients of the 8 terms, either start with a combination of 7 things taken 0 at a time and continue to 7 things taken 7 at a time or use the 7th row of Pascal’s triangle.
For the first term, write x to the 7th power and 3 to the 0 power. Then decrease the power on x and increase the power on y until you reach x to the 0 and y to the 7.
Simplify by evaluating the coefficients and powers 3
Step-by-step explanation:
