Yes, the line can be used to make reasonable predictions of the number of cheese pizzas that would be sold in the upcoming weeks. This is because the line is the line of best fit
<h3>Line of best fit </h3>
From the question, we are to determine if the line can be used to make reasonable predictions of the number of cheese pizzas that would be sold in the upcoming weeks
In the graph, we have a scatterplot.
The line drawn is the <u>line of best fit</u>
Hence,
Yes, the line can be used to make reasonable predictions of the number of cheese pizzas that would be sold in the upcoming weeks. This is because the line is the line of best fit.
Learn more on Line of best fit here: brainly.com/question/1564293
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Answer:
6.28
Step-by-step explanation:
2 x 3.14
Answer:
The average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802
Step-by-step explanation:
We are given that
Standard deviation,
ounces
We have to find the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag.
![P(x\geq 10)=0.99](https://tex.z-dn.net/?f=P%28x%5Cgeq%2010%29%3D0.99)
Assume the bag weight distribution is bell-shaped
Therefore,
![P(\frac{x-\mu}{\sigma}\geq 10)=0.99](https://tex.z-dn.net/?f=P%28%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cgeq%2010%29%3D0.99)
We know that
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
Using the value of z
Now,
![\frac{10-\mu}{0.2}=0.99](https://tex.z-dn.net/?f=%5Cfrac%7B10-%5Cmu%7D%7B0.2%7D%3D0.99)
![10-\mu=0.99\times 0.2](https://tex.z-dn.net/?f=10-%5Cmu%3D0.99%5Ctimes%200.2)
![\mu=10-0.99\times 0.2](https://tex.z-dn.net/?f=%5Cmu%3D10-0.99%5Ctimes%200.2)
![\mu=9.802](https://tex.z-dn.net/?f=%5Cmu%3D9.802)
Hence, the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802
I think the answer is 2 13/20
Or, in decimals: 1.3
Answer: g=5
Step-by-step explanation: