0.3, 30% is the correct answer
Answer:
Step-by-step explanation:
Th average rate of change is the slope of the secant line that goes through those 2 values of x. Of course, each value of x also has a value of y. The coordinates for these combinations of x's and y's are:
(-1, 5) and (4, 0). We can use the slope formula to find the average rate of change of this function without having to know what the function's equation is:

So the average rate of change, aka slope, between those 2 points is -1
Question:
In a neighbourhood pet show, each of the animals entered is equally likely to win. if there are 7 dogs, 6 cats, 3 birds, and 2 gerbils entered, what is the probability that a bird will win the top prize?
Answer:
Probability that a bird will win the top prize is 0.167
Step-by-step explanation:
Given:
The number of dogs = 7
The number of cats = 6
The number of birds = 3
The number of gerbils = 2
To Find:
Probability that a bird will win the top prize = ?
Solution:
Let us first find the total number of pets .
The Total number of pets = 7 + 6 + 3 + 2 = 18
Now the probability of a bird will win the top prize is
=> 
=>
=> 
=>0.167
9514 1404 393
Answer:
a = 3, b = 12, c = 13
Step-by-step explanation:
The applicable rules of exponents are ...
(a^b)(a^c) = a^(b+c)
(a^b)/(a^c) = a^(b-c)
(a^b)^c = a^(bc)
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You seem to have ...

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<em>Additional comment</em>
I find it easy to remember the rules of exponents by remembering that <em>an exponent signifies repeated multiplication</em>. It tells you how many times the base is a factor in the product.

Multiplication increases the number of times the base is a factor.

Similarly, division cancels factors from numerator and denominator, so decreases the number of times the base is a factor.
