What is a31 of the arithmetic sequence for which a5=3.2 and a9=12.0?

Answer:

a) Null hypothesis:

Alternative hypothesis:

b)

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c)

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d)

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

represents the average of the samples

denotes the standard deviation of the samples

indicates the number of samples

is the value we are examining

defines the significance level for the test.

t represents the specific statistic of interest

indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:

Alternative hypothesis:

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

(1)

Part b

In this scenario t=-2.3,

Initially, we need to calculate the degrees of freedom

Since this is a left-tailed test, the p-value is determined by:

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8,

We need to find the degrees of freedom

For the left-tailed test, the p-value is given by:

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6,

Firstly, we find the degrees of freedom

Since we are conducting a left-tailed test, the p-value is calculated as:

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

Answer: (3y - 5) • (2y - 3)

Step-by-step explanation: 6y2 - 10y - 9y - 15

2.1 Factoring 6y2-19y+15

The leading term is 6y2, with a coefficient of 6.

The middle term is -19y, having a coefficient of -19.

The last term is the constant, which is +15.

Answer:

The exact population of Vermont in 2008 could have been around 618,000

Step-by-step explanation:

* Here's how rounding to the nearest ten thousand works:

- Numbers ending with the last four digits between 0001 and 4999 are rounded down to the nearest lower multiple of ten thousand

- Example: 83,525 rounds down to 80,000.

- If the last four digits are 5000 or above, round up to the next higher ten thousand

- Example: 58,988 rounds up to 60,000

* Applying this rule to the problem given:

Since the rounded population is 620,000 to the nearest ten thousand, the actual population could be any value with the last four digits from 0001 to 4999 (like 618,000) or from 5000 upwards (like 624,000).

Therefore, 618,000 might represent the actual population of Vermont in 2008

To determine the rates at which the inlet and outlet pipes fill and empty the reservoir, we remember that work done equals rate multiplied by time. Let’s denote the inlet rate as i and for the outlet pipe as 0. Therefore,
i(24) = 1
o(28) = 1
In this context, the '1' represents the total number of reservoirs, since the problem states the time needed for each pipe to either fill or empty a singular reservoir. Solving for rates yields:
i = 1/24 reservoirs/hour
o = 1/28 reservoirs/hour

Over the first six hours, the inlet pipe fills (1/24)(6) = 1/4 reservoirs and during the same period, the outlet pipe empties (1/28)(6) = 3/14 reservoirs. To calculate the net volume of the reservoir filled, we subtract the emptying total from the filling total:
1/4 - 3/14 = 1/28 reservoirs (note that if emptying exceeds filling, a negative value results. In such cases, treat that negative value as zero, indicating that the outlet rate surpasses the inlet rate, leading to an empty reservoir).
Now we need to find out how long it will take to fill up one reservoir since we’ve already partially filled 1/28 of it, after closing the outlet pipe. In simpler terms, we need to determine the time required for the inlet pipe to finish filling the remaining 27/28 of the reservoir. Fortunately, we have already established the filling rate for the inlet pipe, leading to the equation:
(1/24)t = 27/28
Solving for t gives us 23.14 hours. Remember to add the initial 6 hours to this result since the question seeks the total time. Thus, the final total is 29.14 hours.

Please ask me any questions you may have!

Initial cost = 16000
depreciation rate = 35%
remaining value = 2000
depreciation over one year = 16000*35% = 5600
5600*2 = 11200 + 5600*6\12 = 2000
The car is aged 2.5 years, accounting for two full years and an additional half year of depreciation.