<h2>
Answer:</h2>
We know that,
<u>Area of semi circle = πr²/2</u>
Area = 22 × 2 × 2/7 × 2
Area = 44/7ft².
They've given me two categories of things: assessed values of properties, and the amounts of taxes paid. My ratios will then use these two categories. I will set up my ratios with the assessed valuation on top (because that's what I read first in the exercise), and I will use "v" to stand for the value that I need to find.
value
tax
:
70,000
1,100
=
v
1,400
tax
value
:
1,100
70,000
=
1,400
v
I'll use the shortcut method for solving, multiplying the 70,000 and the 1,400 in one direction, and then dividing by the 1,100 going in the other direction:
70,000 / 1,100 = v / 1,400; multiplying gives (70,000)(1,400); dividing gives v = [(70,000)(1,400)]/1,100
v
=
(
70,000
)
(
1,400
)
1,100
v=
1,100
(70,000)(1,400)
v
=
98,000,000
1,100
v=
1,100
98,000,000
v
=
89,090.9090909...
v=89,090.9090909...
Since the solution is a dollars-and-cents value, I must round the final answer to two decimal places; the "exact" form (whether repeating decimal or fraction) wouldn't make sense in this context. So my answer is: $ 89.090.91
The picture is very blurry I can’t see
Answer:
If we compare the p value and the significance level assumed we see that so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 140, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a one left tailed test the p value would be:
Conclusion
If we compare the p value and the significance level assumed we see that so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.
Answer:
Noo la chimba socio eso es culo de dificil xd.
Step-by-step explanation: