Answer: 0.31
Step-by-step explanation:
Let A denotes the event of Tampa Bay Buccaneers will score a touchdown on their opening drive and B denote the event that their defense will have 3 or more sacks in the game.
Given : P(A)=0.14 P(B) = 0.31 P(A or B)=0.14
Formula : P(A and B)= P(A) + P(B) - P(A or B)
Now, the probability that they will both score a touchdown on the opening drive and have 3 or more sacks in the game will be :-
P(A and B)= 0.14 + 0.31 - 0.14=0.31
Hence, the required probability : 0.31
2t+2d+2c=30
2(8.50)+2(3.50)+2(3.00)=$30
t representing tickets d representing drinks c representing candy
the answer really depends on how much of each item they buy
if they bought more than 2 of each just change the 2 to the number of items bought
To find the specification limit such that only 0.5% of the bulbs will not exceed this limit we proceed as follows;
From the z-table, a z-score of -2.57 cuts off 0.005 in the left tail; given the formula for z-score
(x-μ)/σ
we shall have:
(x-5000)/50=-2.57
solving for x we get:
x-5000=-128.5
x=-128.5+5000
x=4871.50
3x(x^2+4)
3x(x^2)+3x(4)
=3x^3+12x