In Gainesville, one in four people have blonde hair. What is the chance that if four people are chosen at random from Gainesville, at least one person has blonde hair?
1- the chance of not happening * the choices allowed (in this case the chance of not blond is 3/4 since 1/4 is blond so which means 3/4 is not blond) so take that number x it four time since we are picking four peopel. and take 1 and subtract that number.
What is the volume of a cube that has a surface area of 54?
know this formula
Area of cub = 6s^2
volume of cube= s^3
3^3 = 27
A stock falls in price by 30%. The next day it rises by 17%. The third day, it falls to 6% of the second day value. What is the approximate overall percent change in the stock price?
100 stocks x .70 = 70 stocks (day 1)
70 stocks x 1.17= 81.9 stocks (day 2)
81.9 stocks x .06= 4.914 stocks (day 3)
What is the approximate overall percent change in the stock price?
You had 100 stocks and now you have 4.914 stocks on day 3. 100-4.9= 95.1 or 95.1% decrease or you kept 4.91% of the original value.
The mean and standard deviation of a set of data is given by m and s, respectively. If 10 additional data points are added exactly at the mean, which of the following is true?
A. m stays the same, s decreases
B. m stays the same, s increases
C. m decreases, s stays the same
D. m increases, s stays the same
E. both m and s decreaseTerm
answer: A m stays the same, s decreases
A mean- measure of center and describes the “middle value” of a data set. Therefore, if you add ANY number of data points directly at the center, the mean value will not change.
Standard deviation,- a measure of spread. This describes how far the data set expands around the “middle value.” By adding 10 additional data points at the mean, you are “compressing” the data set around the mean. This decreases the standard deviation.
The length of symptoms of the flu is normally distributed with an average of 7 days and a standard deviation of 2 days. What is the probability that someone with the flu will have symptoms for less than 5 days?
To answer this problem, you must know the 68-95-99.7 rule of normal distributions. This is explained in the picture below.
In a normally distributed data set, 68% of the data lies between 1 standard deviation above and below the mean (μ ± σ), 95% of the data lies between 2 standard deviations above and below the mean (μ ± 2σ), and 99.7% of the data lies between 3 standard deviations above and below the mean (μ ± 3σ).
This problem gives you both the mean of experiencing flu symptoms, μ = 7 days, and the standard deviation, σ = 2 days, and asks you to determine the probability that someone with the flu will have symptoms for less than 5 days.
5 days is one standard deviation below the mean, μ-σ. You are trying to determine the percent of the population that is beneath this threshold. From the 68-95-99.7 rule and the picture below, you know that 68% of the population lies between 1 standard deviation above and below the mean; therefore, 32% of the population is both above and below 1 standard deviation from the mean. However, we only want the percent of the population below 1 standard deviation below the mean, which is half of 32%, or 16%, so the answer is [A].
If you have 2 distinct black playing cards and 2 distinct red playing cards, how many ways can you arrange the four cards given that the red cards can never be next to each other?
The first step is to determine the total number of all possible positions. Given that there are 4 distinct cards, the total number is 4! = 4 * 3 * 2 * 1 = 24.
Now we will determine how many combinations have the red cards NOT next to each other. At first, we will assume the red and black cards are the same, and then double the number of combinations to account for the fact that the red cards are different, and then double that value again to account for the fact that the black cards are different.
Possible combinations if the cards are identical:
RBRB, RBBR, BRBR
There are 3 possible combinations if the red and black cards are identical. If the red cards are different, we can double the number of combinations, so there are 6 total combinations. Since both the red and black cards are different, we will double it again and find that there are 12 possible configurations (I will enumerate them here, but on the test it is faster to not to do this!).