There is a type of ammo in f/o/r/t/n/i/t/e that can be magically used for every single gun. You open an ammo box, and out pops 141 rounds of that ammo. You have already collected 6 guns. How many rounds go into each gun?
The answer is 23.5
THERE IS A REMAINDER??
Nothing happens! Literally! It never happened! There is no ammo that can be used for all the guns. GG and hope to see you on the game sometime, cause...i am not allowed to have it.
Answer:
28
Step-by-step explanation:
8c - y
8( 3) - (- 4)
24 + 4
28
11,850 x .9 = 11,605
11605 x . 9 = 9,598.5 year 2
9,598.5 x .9 = 8,638.65
8,635 x .9 = 7,774.785 year 4
7,774.785 x .9 = 6,997.31
6,6997.31 x .9 = 6297.58 year 6
6297.58 x .9 = 5,667.82
5,667.82 x .9 = 5,101.04 year 8
The answer is D
Answer:
The Answer is 76.
Step-by-step explanation:
Given the normal distribution " 10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable'', we can see that exemplary employees are top 10% rated employees.
We have the formula for normal distribution:
z=(X-M)÷σ
where z is the <em>minimum z-score </em>for top 10% employee, X is the <em>minimum </em>score for top 10% employee, M is the <em>mean</em> of the score distribution, σ is the <em>standard deviation</em> of the score distribution.
The z-score we are looking for is the value "a" that separates the highest 10% from the lowest 90% i.e. P(z≤a)=0.90
If we look at z-table, corresponding value for a is 1.28155
We can now put the values in the formula:
1.28155=
So X=(1.28155×20)+50=75.631
Therefore minimum score for exemplary employee is 76.