1279???
Because the gcf does not leave any remainders behind. The gcf could be 1280, but that would not leave a remainder behind if you divided 1280 by 1280. So, I think it is 1279, but i am probably wrong
solution:
The probability mass function for binomial distribution is,
Where,
X=0,1,2,3,…..; q=1-p
find the probability that (p∧ ≤ 0.06) , substitute the values of sample units (n) , and the probability of nonconformities (p) in the probability mass function of binomial distribution.
Consider x to be the number of non-conformities. It follows a binomial distribution with n being 50 and p being 0.03. That is,
binomial (50,0.02)
Also, the estimate of the true probability is,
p∧ = x/50
The probability mass function for binomial distribution is,
Where,
X=0,1,2,3,…..; q=1-p
The calculation is obtained as
P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)
= 50cx ₓ (0.03)x ₓ (1-0.03)50-x
= (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ (0.03)3 ₓ (1- 0.03)50-3 )
=( ₓ (0.03)0 ₓ (1-0.03)50-0 + ₓ (0.03)1 ₓ (1-0.03)50-1 + ₓ (0.03)2 ₓ (1-0.03)50-2 ₓ (0.03)3 ₓ (1-0.03)50-3 )
1/10 is the same as saying 10 times less
Refer to the diagram below
The digit as the first decimal place is worth 1/10
The digit as the second decimal place is worth 1/100 which is 1/10 worth the digit as the first decimal place
Example of two decimal numbers:
2.56 and 2.68
The digit 6 in 2.56 is 6/100
The digit 6 in 2.68 is 6/10
The digit 6 in 2.56 is 1/10 as much as the digit 6 in 2.68
Answer:
The value of q that maximize the profit is q=200 units
Step-by-step explanation:
we know that
The profit is equal to the revenue minus the cost
we have
---> the revenue
---> the cost
The profit P(q) is equal to

substitute the given values



This is a vertical parabola open downward (because the leading coefficient is negative)
The vertex represent a maximum
The x-coordinate of the vertex represent the value of q that maximize the profit
The y-coordinate of the vertex represent the maximum profit
using a graphing tool
Graph the quadratic equation
The vertex is the point (200,-120)
see the attached figure
therefore
The value of q that maximize the profit is q=200 units
Square's side range is (√0,√100) =(0,10); this more than 0 and less than 10.