Answer:
D. 81
Step-by-step explanation:
I think the attached photo supports for your question to be answered
Here is my answer:
Number of days: 30
Mean of the data set = (68 + 70*2 +74*2 +76*4 +78 +80*5 + 82*3 +84*5 + 88 +88*3 + 92*3) /30 = 81
The probability of an event is expressed as

Given:

The probability of drwing two blue balls one after the other is expressed as

For the first draw:

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).
Thus,

The probability of drawing two blue balls one after the other is evaluted as

The probablity that none of the balls drawn is blue is evaluted as

Hence, the probablity that none of the balls drawn is blue is evaluted as
Answer:
Quadrant 4
Step-by-step explanation:
x > 0 means towards right of the origin
y < 0 means below the origin
So Quadrant 4
Answer:

Step-by-step explanation:
Let the function of quantity in the lung of air be A(t)
So 
so, A(t) = Amax sin t + b
A(t) = 2.8t⇒ max
A(t) = 0.6t ⇒ min
max value of A(t) occur when sin(t) = 1
and min value of A(t) = 0
So b = 0.6
and A(max) = 2.2

at t = 2 sec volume of a is 0.6
So function reduce to

and t = 5 max value of volume is represent
so,

when t = 5

so the equation becomes
