121.3 grams of caffeine is remaining in her blood
<h3>How to determine the amount of caffeine?</h3>
The given parameters are:
- Initial, a = 175 mg
- Rate, r = 11.5%
- Time, t = 3 hours
The amount of caffeine is calculated as:
A(t) = a(1 - r)^t
This gives
A(t) = 175 * (1 - 11.5%)^t
At 3 hours, we have:
A(3) = 175 * (1 - 11.5%)^3
Evaluate
A(3) = 121.3
Hence, 121.3 grams of caffeine is remaining in her blood
Read more about exponential functions at:
brainly.com/question/2456547
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Answer:
See Below.
Step-by-step explanation:
We are given the function:

And we want to show that it has at least one zero between <em>x</em> = 1 and <em>x</em> = 2.
Because the function is a polynomial, it is everywhere continuous.
Evaluate the function at <em>x</em> = 1 and <em>x</em> = 2:

And:

Therefore, because the function changes signs from <em>x</em> = 1 to <em>x</em> = 2 and is continuous on the interval [1, 2], by the intermediate value theorem, there must exist at least one zero in the interval.
Answer:
a
b

c

Step-by-step explanation:
From the question we are told that
The population proportion is p = 0.53
The sample size is n = 10
Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution
i.e
and the probability distribution function for binomial distribution is
Here C stands for combination hence we are going to be making use of the combination function in our calculators
Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as
=>
=>
Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

=> ![P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]](https://tex.z-dn.net/?f=P%28%20X%20%5Cge%20%206%20%29%3D%20%20%5B%5E%7B10%7DC_6%20%2A%20%20%5B0.53%5D%5E6%20%2A%20%20%281-%200.53%29%5E%7B10-6%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_7%20%2A%20%20%5B0.53%5D%5E7%20%2A%20%20%281-%200.53%29%5E%7B10-7%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_8%20%2A%20%20%5B0.53%5D%5E8%20%2A%20%20%281-%200.53%29%5E%7B10-8%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_9%20%2A%20%20%5B0.53%5D%5E9%20%2A%20%20%281-%200.53%29%5E%7B10-9%7D%5D%20%2B%20%5B%5E%7B10%7DC_%7B10%7D%20%2A%20%20%5B0.53%5D%5E%7B10%7D%20%2A%20%20%281-%200.53%29%5E%7B10-10%7D%5D)
=> ![P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]](https://tex.z-dn.net/?f=P%28%20X%20%5Cge%20%206%20%29%3D%20%20%5B0.227%5D%20%2B%20%20%5B0.1464%5D%20%2B%20%20%5B0.0619%5D%20%2B%20%20%5B0.0155%5D%20%2B%20%5B0.00082%5D)
=> 
Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

=> ![P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]](https://tex.z-dn.net/?f=P%28%20X%20%3C%20%204%20%29%3D%20%20%5B%5E%7B10%7DC_3%20%2A%20%200.53%5E3%20%2A%20%20%281-%200.53%29%5E%7B10-3%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_2%20%2A%20%200.53%5E2%20%2A%20%20%281-%200.53%29%5E%7B10-2%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_1%20%2A%20%200.53%5E1%20%2A%20%20%281-%200.53%29%5E%7B10-1%7D%5D%20%2B%20%20%5B%5E%7B10%7DC_0%20%2A%20%200.53%5E0%20%2A%20%20%281-%200.53%29%5E%7B10-0%7D%5D%20)
=> 
=> 