Answer:
6.18% of the class has an exam score of A- or higher.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:

What percentage of the class has an exam score of A- or higher (defined as at least 90)?
This is 1 subtracted by the pvalue of Z when X = 90. So



has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% of the class has an exam score of A- or higher.
The number 0 is basically nothing, so 7 divided by nothing is not defined. That's why calculators or even the most powerful computer can't divide anything by 0
The answer is 5. Hope this helps
3×2=6+4=d the 3×2 is how much the cost is and 6+4 is 10 so she paid 10 dollars
Using cos addition formula:
use x for theta
cos(x+π/6)=cosx*cos(π/6)-sinx*sin(π/6)
sinx=1/4
cosx=√15/4
cos(π/6)=√3/2
sin(π/6)=1/2
cos(x+π/6)=(√15/4*√3/2)-(1/4*1/2)
cos(x+π/6)=(√45/8)-(1/8 )
cos(x+π/6)=(√45-1)/8)