The amount of caffeine in the body of a healthy adult after drinking espresso decreases exponentially, with a half-life of appro
ximately 6 hours. That is, every 6 hours, the amount of caffeine in the body decreases by 50%. Charlotte wants to go to sleep at 11 pm and does not want more than 20 milligrams (mg) of caffeine in her system at that time. What is the latest time Charlotte can drink a 2-ounce shot of espresso that contains 80 mg of caffeine?
A. 5 am
B. 11 am
C. 11:30 am
D. 12 pm
A. 5 am
B. 11 am
C. 11:30 am
D. 12 pm
2 answers:
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m□ebd=4 x-8 and m□ebc=5 x+20
This is solvable only if e b is the initial side and b d and b c lies on opposite side of each other and lies on a line i.e c,b,d are Collinear.
∠ebd and ∠ebc will form a linear pair.The meaning of linear pair is that angles forming on one side of a straight line through a common vertex which are adjacent is 180°.
i.e
∠ ebd + ∠ebc = 180°
4 x- 8 + 5x + 20= 180°
adding like terms
⇒ 9 x +12 =180°
⇒ 9 x = 180° - 12
⇒ 9 x = 168°
⇒ x =( 168/9)°=(56/3)°
now m□ebc =5 x +20
= 5 × 56/3 + 20
= 280/3 + 20
=340/3
m□ebc=( 340/3)°
So, solution set is x =(56/3)° and m□ebc =(340/3)°
Answer:
a) 14.46%
b) 0.00%
c) 1.54%
Step-by-step explanation:
According to the survey, 10% of Americans are afraid to fly.
This means and
.
If 1100 Americans are sampled, then the sample size is .
The mean of the distribution is .
This means
The standard deviation is
We substitute the values to get:
a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.
We first apply the continuity correction factor to get:
We now convert to Z-scores to get:
From the standard normal distribution table P(z>1.06)=0.1446
As a percentage, the probability is 14.46%
b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.
We apply the CCF to get:
We convert to z-scores:
From the normal distribution, P(z>164.5)=0
c) First, 8% of 1100 is 88.
We want to find the probability that 88 or less Americans in the survey are afraid to fly.
We apply the CCF to get:
We convert to z-scores:
From the normal distribution, P(z>\:-2.16)=0.0154
As a percentage, we get 1.54%