Answer:
the probability is 2/9
Step-by-step explanation:
Assuming the coins are randomly selected, the probability of pulling a dime first is the number of dimes (4) divided by the total number of coins (10).
p(dime first) = 4/10 = 2/5
Then, having drawn a dime, there are 9 coins left, of which 5 are nickels. The probability of randomly choosing a nickel is 5/9.
The joint probability of these two events occurring sequentially is the product of their probabilities:
p(dime then nickel) = (2/5)×(5/9) = 2/9
_____
<em>Alternate solution</em>
You can go at this another way. You can list all the pairs of coins that can be drawn. There are 90 of them: 10 first coins and, for each of those, 9 coins that can be chosen second. Of these 90 possibilities, there are 4 dimes that can be chosen first, and 5 nickels that can be chosen second, for a total of 20 possible dime-nickel choices out of the 90 total possible outcomes.
p(dime/nickel) = 20/90 = 2/9
Answer:
20 chairs
Step-by-step explanation:
After 136 people are seated in the bleacher, there can be 514 people seated in chairs. We know that 514 = 25×20 +14, so there can be 20 rows of 25 chairs. We require an equal number of chairs in each row, so there cannot be some rows with 21 chairs, nor can there be a 26th row with 14 chairs.
There can be 20 chairs in each row.
The cost of a cd after 6 year is =$ 15.12
Step-by-step explanation:
The present cost of the cd is $11.95 and inflation rate 4%
P = $11.95 , r = 4% and n = 6 years
The cost of a cd after 6 year is =
=$ 
=$ 15.12
Answer: true
Step-by-step explanation:
Z-tests are statistical calculations that can be used to compare the population mean to a sample mean The z-score is used to tellsbhow far in standard deviations a data point is from the mean of the data set. z-test compares a sample to a defined population and is typically used for dealing with problems relating to large samples (n > 30). Z-tests can also be used to test a hypothesis. Z-test is most useful when the standard deviation is known.
Like z-tests, t-tests are used to test a hypothesis, but a t-test asks whether a difference between the means of two groups is not likely to have occurred because of random chance. Usually, t-tests are used when dealing with problems with a small sample size (n < 30).
Both tests (z-tests and t-tests) are used in data with normal distribution (a sample data or population data that is evenly distributed around the mean).
