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.
value of x is
and value of y is 
Step-by-step explanation:
We need to find Which expression could be substituted for y in the second equation to find the value of x
The given equations are:

For putting value of y in second equation, we will find from equation 1
From equation 1:

Putting value of y i.e y=10+2x in equation (2) to find value of x

So, now finding value of y by putting value of x= -29/4 in eq(1)

So, value of x is
and value of y is 
Keywords: System of equations
Learn more about System of equations at:
#learnwithBrainly
What is the question if it 98-49 than it would be 49
Answer:
$8511.11
Step-by-step explanation:
Each year, the amount Walter owes is multiplied by 1.06, so at the end of 6 years, Walter owes 1.06^6 times the amount he borrowed.
he will pay $6,000×1.06^6 ≈ $8511.11
_____
At the end of the first year, Walter owes the original loan amount plus 6% interest. That total is ...
$6000 + 0.06×6000 = $6000×1.06
At the end of the following year, he owes 1.06 times that amount, or ...
6000×1.06²
The amount owed is multiplied by 1.06 each year until Walter pays off the loan.