Answer:
10 ball caps
Step-by-step explanation:
In this question, we are trying to know the number of caps the manger would buy that would equate the cost at both companies.
How do we get this?
Since we do not know the number of caps, let’s assign a variable. Let the number of caps that is required be x. Let’s now make some costings in terms of x. We proceed;
Company X charges $50 fee plus $7 per cap. Total amount company X will charge on x caps will be; $50 + $7(x) = $50 + $7x
Company Y will charge $30 plus $9 per cap. Total amount company Y will charge on x caps will be $30 + $9(x) = $30 + $9x
We are trying to look at the value of x that will make both costs equal. What we do is to equate both costs.
30 + 9x = 50 + 7x
We simply by taking like terms to the same sides
9x-7x = 50-30
2x = 20
x = 20/2 = 10
X = 10
So what this means is that manger has to buy 10 caps to have the same cost in both companies
We would have the following sample space:
(1, 1), (1, 2), (1, 3), (1, 4)
(2, 1), (2, 2), (2, 3), (2, 4)
(3, 1), (3, 2), (3, 3), (3, 4)
(4, 1), (4, 2), (4, 3), (4, 4)
Those give us these sums:
2, 3, 4, 5
3, 4, 5, 6
4, 5, 6, 7
5, 6, 7, 8
P(sum of 2) = 1/16 =0.0625
P(sum of 3) = 2/16 = 0.125
P(sum of 4) = 3/16 = 0.1875
P(sum of 5) = 4/16 = 0.25
P(sum of 6) = 3/16 = 0.1875
P(sum of 7) = 2/16 = 0.125
P(sum of 8) = 1/16 = 0.0625
Answer:
A system has infinitely many solutions when it is consistent and the number of variables is more than the number of nonzero rows in the rref of the matrix. For example if the rref is ��has solution set (4-3z, 5+2z, z) where z can be any real number.
4.626,5.63,4811,9320.........
