Answer:
82% of scores were between 286 and 322
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation , the z-score of a measure X is given by:
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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 310 and a standard deviation of 12.
This means that
What percent of scores were between 286 and 322?
The proportion is the pvalue of Z when X = 322 subtracted by the pvalue of Z when X = 286. So
X = 322
has a pvalue of 0.8413
X = 286
has a pvalue of 0.0228
0.8413 - 0.0228 = 0.8185
0.8185*100% = 81.85%
Rounding to the nearest whole number
82% of scores were between 286 and 322
If you multiply 5 by 7 you get 35 :)
Answer:
Step-by-step explanation:
Simplify the expression:
Grouping like terms, 10x² + 5x² + 4 x - 4 x - 1 + 3 = (10 x² + 5x²) + (4 x - 4 x) + (-1 + 3):
10x² + 5x² = 15x²:
3 - 1 = 2:
4 x - 4 x = 0:
Subst for f and t in...
9t+3ft+12 = 9×21 + 3×3×21 + 12 = 390