1. 10x
2. 18g
3. 6m
4. 4p
5. -5y
6. 9w
7. c^2+3d
8. 3a^2+2b^2+6a
9. 3x^2+x
10. 16x-y+3
11. 68x^2-34x-1
12. 32
13. 32
14. -9
15. 3m+3n
16. 6x-6y
17. 15f+5
Answer:
Step-by-step explanation:
The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve
Answer:
Approximately 6 observations are more than 74
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 50, \sigma = 12](https://tex.z-dn.net/?f=%5Cmu%20%3D%2050%2C%20%5Csigma%20%3D%2012)
Approximately how many observations are more than 74?
First step is finding the percentage of observations which are higher than 74, which is 1 subtracted by the pvalue of Z when X = 74. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{74 - 50}{12}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B74%20-%2050%7D%7B12%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
Out of 250
0.0228*250 = 5.7
Approximately 6 observations are more than 74
Answer:4 4/5
Step-by-step explanation:
To make A frequency table, you will need to look at the lowest average number of movies and the highest average number of movies and create equally space intervals.
You will need to include 0.5 to 4.5.
I would use the following equal intervals (frequencies are beside the intervals).
0 - 0.9(1)
1 - 1.9(5)
2 - 2.9(5)
3 - 3.9(2)
4 - 4.9(1)
Check to make sure they are all there by counting the frequencies. The total should equal the number of pieces of data that you had to start with.