If a certain coin is flipped, the probability that the coin will land heads up is \small \frac{1}{2}. if the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

The probability of the heads up on the first three flips and not on the last two flips will be 1/32.

Step-by-step explanation:

As, the probability of head up when one time coin is flipped is 1/2. And it is said in the question that you must get heads in the first three flips.

So, for the first three flips the probability is = (1/2)^3 = 1/8

And for the last two flips you want to get not heads up.

Then, the probability is = (1/2)^2 = 1/4

Hence, your overall probability will be = 1/8 × 1/4 = 1/32 .

1 is c

2 is a

3 is a

step-by-step explanation:

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.