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:
1/2
Step-by-step explanation:
There are 2 numbers less than 3. (1 and 2) and one 6 on a dice. Therefore there is a 3 out of 6 chance of rolling one of these 3 numbers (1, 2, and 6). This can be represented as 3/6 which would simplify to 1/2 or 50%.
It can also be found this way...
Probability = Number of desired outcomes ÷ number of possible outcomes. Therefore 3 ÷ 6 = 0.5 which is equal to 50% or 1/2.
I hope you choose my answer so you do well on your assignment! I've gotten an A in math every year :)
Hi I am jazmin...that's what she said :)