<span><span><span>−<span>36n</span></span>−75</span>=<span>n−65</span></span><span>−36n−75−n</span><span>=</span><span>n−65−n</span>
<span><span><span>−<span>37n</span></span>−75</span>=<span>−65</span></span><span><span><span><span>−<span>37n</span></span>−75</span>+75</span>=<span><span>−65</span>+75</span></span><span><span>−<span>37n</span></span>=10</span><span><span><span>−<span>37n</span></span><span>−37</span></span>=<span>10<span>−37
</span></span></span><span>n= <span><span>−10/</span>37 <--------- answer</span></span>
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
41% of students ran in 12 minutes or less.
Using the histogram we see that there were 2+12+11+9=34 students that ran. Out of those, 2+12=14 ran in 12 minutes or less (the first two bars).
14/34 = 0.411 = 41%.
