I'm hoping it might be around the .65 percentile?
Answer:
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
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:

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:

The percentage of cockroaches weighing between 77 grams and 83 grams
This is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 83. So
X = 83



has a pvalue of 0.7734
X = 77



has a pvalue of 0.2266
0.7734 - 0.2266 = 0.5468
Rounded to the nearest whole number, 55%
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
The slope of the line is: m=(5-0)/(-3-2)=-1
y=-x+b
find out b by using a point: 0=-2+b, b=2
so the equation of the line is y=-x+2
it is dotted, so they should be an equal sign.
the shaded part is to the right, above the line, so it is a >
the inequality is y>-x+2
Answer:
The answer is in the above image
Answer:
Millimeters per hour.
Step-by-step explanation:
Since its a water sprinkler, you would know that it doesn't cover miles of land, nor is it a food, so it isn't grams. Meters would seem like the right answer, but if you calculate meters to millimeters, you would see that meters are not covered in seconds. So the correct answer is millimeters per hour.