The police department of UCR is looking to collect more fines for speeders on campus. (They are turning to speeding because they cannot extract any more money out of people for parking illegally!) They record the speeds on West Campus Drive for a year and find that the average speed is 31.0 miles per hour with a standard deviation of 2.0. The data is approximated by a normal distribution. The top five percent (5%) of speeders will get tickets. What speed must you be going before getting a ticket

≥ 34.29 mph

Step-by-step explanation:

Given that:

Distribution is normal :

Mean(m) = 31 mph

Standard deviation = 2 mph

Speed of top 5%

Top 5% = bottom (100% - 5%) = 95% = 0.95

Zscore which corresponds to 0.95 = 1.645

Using the Z formula:

Zscore = (score - mean) / standard deviation

1.645 = ( score - 31) / 2

Croas multiply:

2 * 1.645 = score - 31

3.29 = score - 31

3.29 + 31 = score

34.29 = score

Hence, speed must be greater than or equal to 34.29 mph

< 6 cm to > 26 cm i believe

step-by-step explanation:

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints