Answer:
![N(T(t)) = 1408t^2 - 385.6t - 105.52](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%201408t%5E2%20-%20385.6t%20-%20105.52)
Time for bacteria count reaching 8019: t = 2.543 hours
Step-by-step explanation:
To find the composite function N(T(t)), we just need to use the value of T(t) for each T in the function N(T). So we have that:
![N(T(t)) = 22 * (8t + 1.7)^2 - 123 * (8t + 1.7) + 40](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%2022%20%2A%20%288t%20%2B%201.7%29%5E2%20%20-%20123%20%2A%20%288t%20%2B%201.7%29%20%2B%2040)
![N(T(t)) = 22 * (64t^2 + 27.2t + 2.89) - 984t - 209.1 + 40](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%2022%20%2A%20%2864t%5E2%20%2B%2027.2t%20%2B%202.89%29%20-%20984t%20-%20209.1%20%2B%2040)
![N(T(t)) = 1408t^2 + 598.4t + 63.58 - 984t - 169.1](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%201408t%5E2%20%2B%20598.4t%20%2B%2063.58%20-%20%20984t%20-%20169.1)
![N(T(t)) = 1408t^2 - 385.6t - 105.52](https://tex.z-dn.net/?f=N%28T%28t%29%29%20%3D%201408t%5E2%20-%20385.6t%20-%20105.52)
Now, to find the time when the bacteria count reaches 8019, we just need to use N(T(t)) = 8019 and then find the value of t:
![8019 = 1408t^2 - 385.6t - 105.52](https://tex.z-dn.net/?f=8019%20%3D%201408t%5E2%20-%20385.6t%20-%20105.52)
![1408t^2 - 385.6t - 8124.52 = 0](https://tex.z-dn.net/?f=1408t%5E2%20-%20385.6t%20-%208124.52%20%3D%200)
Solving this quadratic equation, we have that t = 2.543 hours, so that is the time needed to the bacteria count reaching 8019.
Answer:
The equation of line with given points and perpendicular to y-axis is
y = - 7
Step-by-step explanation:
Given as :
The given points as ( - 10 , - 7)
The equation of line is Y = mX + c
So The line will satisfy given points
Or, - 7 = m ( -10 ) + c
Now This line is perpendicular to y- axis
∴ The slop of line perpendicular to y axis is 0
So, - 7= 0 + c
or, c = - 7
∴ Equation of line is y = 0 + c
Or, y = - 7
Hence The equation of line with given points and perpendicular to y-axis is y = - 7 Answer
Answer: 9.6
Step-by-step explanation: A rational number can be written as a ratio of two integers (9.6 can write as 96/10)
This is a problem of Standard Normal distribution.
We have mean= 12 grams
Standard Deviation = 2.5 grams
First we convert 8.5 to z score. 8.5 converted to z score for given mean and standard deviation will be:
![z= \frac{8.5-12}{2.5} =-1.4](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7B8.5-12%7D%7B2.5%7D%20%3D-1.4)
So, from standard normal table we need to find the probability of z score to be less than -1.4. The probability comes out to be 0.0808
Thus, the <span>
probability that the strawberry weighs less than 8.5 grams is 0.0808</span>