The required probability of success is given by 0.98906500
Assume the random variable X has a binomial distribution with the given probability of obtaining a success.
P(X<5), n=6, p=0.3
<h3>What is binomial distribution?</h3>
In binomial distribution for number trials we are investigating the probability of getting a success remain the same.
n = number of trails,
p = probability of success
x = the number of success
p(x<5) = P(x=0)+ p(x=1) + p(x=2)+p(x=3)+p(x=4)
= 
p(x<5) = 0.98906500
Thus the required probability of success is given by 0.98906500
Learn more about Binomial distribution here:
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We need to cancel out the discontinuity in x = 4
In order to do that, factorize the numerator:
5x⁴ - 20x³ = 5x³(x-4)
This way, your function will be:
f(x) = <span>5x³(x-4) / (x-4)
and the two parentheses cancel out, leaving
</span>f(x) = <span>5x³
</span>
which at x = 4 gives:
f(4) = <span>5·4³</span> = 5 · 64 = 320
Therefore K = 320.
55% is 55 percent of 100 any time a number is below 100 and is out of 100 it is that % of 100 for example 45 is 45% of 100
The answer is F. Hope you pass.