The experimental probability of the arrow stopping on Section 2 will be 
. is the ratio of the favorable event to the total number of events.
<h3>What is probability?</h3>
Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1.
No of arrows given in the different section ;
Section 1=18
Section 2=30
Section 3=32
The probability that the arrow stopping on Section 2 is found as;
Hence, the experimental probability of the arrow stopping on Section 2 will be .
More about the probability link is given below.
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F(x) = -x^2 + 6x + 6
where f(x) represents the height, in feet, of the water from the fountain at different times x, in seconds
rate of change of a distance is speed
at x=5, distance = 11
at x=2, distance = 14
so average rate of change = (14-11)/(5-2) = 1
so ans is D
Answer:
81.66 repeating
Step-by-step explanation:
86+72+65+82+91+94=490
490÷6=81.66 repeating
EDIT: this answer is false
Answer:
The sum is 493.4
Step-by-step explanation:
In order to find the value of the sum, you have to apply the geometric series formula, which is:
where i is the starting point, n is the number of terms, a is the first term and r is the common ratio.
The finite geometric series converges to the expression in the right side of the equation. Therefore, you don't need to calculate all the terms. You can use the expression directly.
In this case:
a=40
b= 1.005
n=12 (because the first term is 40 and the last term is 40(1.005)^11 )
Replacing in the formula:
Solving it:
The sum is 493.4
This should be quite easy to explain. So the first thing you want to do is multiply them. This isn't to hard if you know your times tables. Don't worry about the decimal till the end.
15.37
× 5
---------
76.85
15.37 times 5 is 76.85
Now on to the decimal. All you need to do is move it two times to the left. You see how it's in between the 5 and 3? The same goes for your answer. So it is 76.85
I hope this helped!! ^^ I can make this a little more easier to explain if you want me to just in case, ok?
