First find the average number of hours worked by every employee: 11000/370 = 29.7 ~ 30 hrs/week. 

Next, recognize that a single standard deviation encompasses 34% of the population on 1 end of the curve, or 68% of the population both above and below the mean. Since a standard deviation of 6 hours means that 68% of the population works between 24 to 36 hours per week, finding the amount that works over 36 hours is 100-68 = 32; 32/2 = 16% of the entire workforce on 1 end of the curve.  

.16(370) = 59.2 or 59 employees work over 36 hrs/week.

First let's evaluate Quantity A. Max has 5 red marbles and 3 green marbles, so the probability of picking a green marble is 3/8.  

Next let's evaluate Quantity B. The bag of combined marbles now has 5 red, 7 green, and 4 white, so the probability of picking a green marble is 7/16.  

To see which fraction is larger, we can make the two fractions have the same denominator. 3/8 is equivalent to 6/16, so now we can compare 6/16 to 7/16. Clearly Quantity B is bigger. 

Since one blue balloon has already been popped, there are now 9 blue balloons left, and 24 balloons left overall. Therefore the probability that the next balloon Jill hits is also blue is 9/24 = 3/8.

The probability of choosing a spade is 13/52.  Now we have 51 cards to choose from, so the probability of then choosing a heart is 13/51.  Now we have 50 cards left, with 12 spades, 12 hearts, 13 diamonds, and 13 clubs, so the probability of choosing a diamond is 13/50.  Lastly we choose a club out of the remaining 49 cards with probability 13/49.

Therefore the answer is (13/52) * (13/51) * (13/50) * (13/49).

Here we simply add up the proportions of the photos with numbers greater than or equal to 3.

P(3 or 4 or 5 or 6)

= P(3) + P(4) + P(5) + P(6)

= 1/8 + 1/8 + 1/8 + 1/4

= 5/8

First let's find their individual probabilities of winning. Since B is twice as likely to win as C, P(B) = 2 * P(C). Since A is twice as likely to win as B, P(A) = 2 * P(B) = 2 * 2 * P(C) = 4 * P(C). We also know that the probabilities must sum to 1, so P(A) + P(B) + P(C) = 1, meaning P(C) + 2 * P(C) + 4 * P(C) = 1. Then P(C) = 1/7, so P(B) = 2 * P(C) = 2/7. Our answer is therefore P(B or C) = P(B) + P(C) = 2/7 + 1/7 = 3/7.

P(boy) = 10/30 = 1/3

P(blue eyes) = 1/2

P(boy and blue eyes) = 5/30 = 1/6, because we are told that half (or 5) of the 10 boys have blue eyes

P(boy or blue eyes) = P(boy) + P(blue eyes) – P(boy and blue eyes)

= 1/3 + 1/2 – 1/6

= 2/3

As the options suggest, this problem requires us to know how probabilities work. The probability of a single event happening is equal to the number of ways it can happen divided by the total number of outcomes. Because this problem asks about 31 students, we have to raise our probability to the 31st power because it's the probability of 31 different events happening.

Now we can figure out the probability of one January birthday (31 days in January/365 days in a year) and the probability of all Saturday birthdays (52 Saturdays in a year/365 days in a year). So thus the probabilities that ALL of the students will have either type of birthday is:

(31/365)³¹ and

(52/365)³¹

Since the numbers are raised to the same power, we can simply look at the base to determine which is larger and which is smaller.  Since the probability of having a birthday on Saturday (roughly 1/7) is greater than having a birthday in January (roughly 1/12), then Quantity B is greater.

There are 200 students in total, and 120 of them are male, so 80 must be female. We also know that there are 50 upper division students, and 40 of them are male, so 10 must be female. If 10 of 80 females are upper division, the other 70 females have to be lower division students, so the probability of choosing a lower division student, given the student is female, is 70/80 = 7/8.

We are given that the chosen dress has a defect, so we are looking at the 9 defective dresses. 3 of the defective dresses have rips, so the probability that a dress has a rip, given that is has a defect, is 3/9 = 1/3.