Please help clarence works at least 5 hours but not more than 7 hours. he earns 11.60 per hour. the function f(t)11.6t represent
1 answer:
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Answer:
Step-by-step explanation:
step 1
we have the points
(-1,0), (-2,0), and (0,2)
Plot the points
using a graphing tool
see the attached figure
The graph of a quadratic function must be a vertical parabola open upward
The vertex is a minimum
The quadratic function in general form is equal to
Substitute the value of x and the value of y of each given ordered pair in the general equation and solve for a,b and c
(0,2)
For x=0, y=2
substitute
(-1,0)
For x=-1, y=0
substitute
---->
----> equation A
(-2,0)
For x=-2, y=0
substitute
----> equation B
we have the system
----> equation A
----> equation B
substitute equation A in equation B
solve for b
Find the value of a
therefore
The quadratic function in general form is equal to
see the attached figure N 2 to better understand the problem
Answer:
approximately Normal, mean 8.1, standard deviation 0.063.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Let X the random variable weights of 8-ounce wedges of cheddar cheese produced at a dairy. We know from the problem that the distribution for the random variable X is given by:
We take a sample of n=10 . That represent the sample size.
What can we say about the shape of the distribution of the sample mean?
From the central limit theorem we know that the distribution for the sample mean is also normal and is given by:
approximately Normal, mean 8.1, standard deviation 0.063.