I cant see the question, id help if i could
The number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.
<h3>How to determine the number of ways</h3>
Given the word:
ESTABROK
Then n = 8
p = 6
The formula for permutation without restrictions
P = n! ( n - p + 1)!
P = 8! ( 8 - 6 + 1) !
P = 8! (8 - 7)!
P = 8! (1)!
P = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 × 1
P = 40, 320 ways
Thus, the number of ways in which the name 'ESTABROK' can be made with no restrictions is 40, 320 ways.
Learn more about permutation here:
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That's definitely an example of exponential decay, since the base (1/2) (also called the "common ratio") is greater than 0 but less than 1.
Answer:
There are 70 sheep in the farm.
Step-by-step explanation:
Consider the provided information.
Let c represents the cows, s represents the sheep and p represents the pigs.
The number of cows and the number of sheep are in the ratio 6:5.


The number of sheep and the number of pigs are in the ratio 2:1.


The total number is 189.

Substitute the value of c and p in above.




Hence, there are 70 sheep in the farm.
Answer:
f(x) = 0.43 *
*
*(x + 10)
Step-by-step explanation:
We have a 6th degree polynomial f(x)
r = 3 is a root of f with multiplicity 2
r = 1 is a root of f with multiplicity 3
f(-5) = -29721.6
f(-10) = 0
Then: f(x) = a*((x -3)^2 ) * ((x - 1)^3)*(x + 10)
f(-5) = a * (-8)^2 * (-6)^3 * (5) = -29,721.6
a* (64) * (-216)* 5 = -29,721.6
-a*69,120 = -29,721.6
a = -29,721.6/-69,120
a = 0.43
so
f(x) = 0.43 *
*
*(x + 10)