It’s time to film your first big action scene. A swarm of insect-like aliens has descended upon the hero’s peaceful martial arts dojo, forcing the kung-fu monk to take up arms and defend his home. Unfortunately, each alien bug can periodic split into two bugs, allowing their numbers to overwhelm the poor monk. You will start the scene at an innocuous 50alien bugs, but by the end of the scene you want to have a huge swarm of 1,000 bugs on screen. The swarm will grow by having each bug split into two bugs. If the scene is 1minute and 20 seconds long, how frequently should each bug split into two bugs to achieve the desired swarm size by the scene’s end?

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

(2,20) means that the 2 is 2 hours and 20 is 20$. so there both 20$ for 2 hours only the same price. like it says. so, the dunk tank and the bounce house are the same price (20$) for the same amount of time (2 hours).

hope i

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west