Answer:
10 quarters = $2.50
10 nickels = $0.50
that leaves $0.20 for other coins (dimes / pennies)
Step-by-step explanation:
First, suppose she has only quarters and nickels and no other coins. Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this. So she must have at least one other type of coin.
Assume she has only quarters, nickels, and dimes. Then if D is the number of dimes, 5C + 25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32. The smallest D can be is 2, leaving 3C = 30 and thus C = 10. So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.
This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.
Answer:
<em>There is no significant difference in the amount of rain produced when seeding the clouds.</em>
Step-by-step explanation:
Assuming that the amount of rain delivered by thunderheads follows a distribution close to a normal one, we can formulate a hypothesis z-test:
<u>Null Hypothesis
</u>
: Average of the amount of rain delivered by thunderheads without seeding the clouds = 300 acrefeet.
<u>Alternative Hypothesis
</u>
: Average of the amount of rain delivered by thunderheads by seeding the clouds > 300 acrefeet.
This is a right-tailed test.
Our z-statistic is
We now compare this value with the z-critical for a 0.05 significance level. This is a value such that the area under the Normal curve to the left of is less than or equal to 0.05
We can find this value with tables, calculators or spreadsheets.
<em>In Excel or OpenOffice Calc use the function
</em>
<em>NORMSINV(0.95)
</em>
an we obtain a value of
= 1.645
Since 1.2845 is not greater than 1.645 we cannot reject the null, so the conclusion that can be drawn when the significance level is 0.05 is that there is no significant difference in the amount of rain produced when seeding the clouds.
Answer:
Step-by-step explanation:
total sides = 6
p (rolling a 1) =
p (rolling a 2) =
Note:
or - add
and - multiply
∴
∴
hence, p (rolling a 1 or 2) =