Find The last two digit of 36^2015

To determine the time it takes to ride around the neighborhood, equations are set up from the given scenarios and solved algebraically, resulting in 10 minutes for one neighborhood lap.

To solve the problem, we need to set up equations based on the information provided. Let's let N represent the time it takes to ride around the neighborhood once and B represent the time it takes to ride around the block twice. The first scenario gives us:

B + N = 15 minutes

In the second scenario, the only difference is the number of neighborhood laps, so:

B + 5N = 55 minutes

Subtracting the first equation from the second gives us:

4N = 40 minutes

Dividing both sides by 4, we find:

N = 10 minutes

Therefore, it takes 10 minutes to ride around the neighborhood once.

The Future value is $ 15162.80 and Interest is $6258.26

The worth of a current asset at some point in the future based on an estimated rate of growth is known as future value (FV). For investors and financial planners, the future value is crucial because they use it to predict how much an investment made now will be worth in the future.

Investors can make wise investment choices based on their projected demands by knowing the future worth. However, external economic forces that depreciate an asset's value, such inflation, might have a negative impact on the asset's future worth.

Given:

P = ​$8904.56

R= 6%

T= 9

n= 2

Now, Future Value = P [tex](1 + r/n)^ {(nt)}\\[/tex]

FV= 8904.56 [tex](1 + 0.06/2)^ {(2)(9)}\\[/tex]

FV = 8904. 56 [tex]( 1+ 0.03)^{18[/tex]

FV= 8904.56[tex](1.03)^{18}[/tex]

FV = $15162.80

And Interest = FV - P

Interest = 15162. 80 - 8906.4

Interest = $6258.26

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To calculate the stock required for continuous operation for 2000 hours with components having a mean lifetime of 100 hours and a standard deviation of 30 hours, use the Z-score for a 0.95 probability and solve for the number of components. This involves applying the normal distribution.

The given problem requires calculating the number of electrical components that must be in stock to ensure continuous operation of an electrical system for 2000 hours. We are to assume that the components have a mean lifetime of 100 hours and a standard deviation of 30 hours. The aim is to keep the probability of the system running without interruption at 0.95 or higher.

Firstly, determine the number of lifetimes within the 2000-hour period:

2000 hours / 100 hours = 20 lifetimes.

Next, apply the concept of the normal distribution to find the z-score that corresponds to the 0.95 cumulative probability. Looking at z-tables or using statistical software, we find that the z-score for 0.95 is approximately 1.645.

Now, calculate the stock required using this z-score. The formula for the z-score in this context is:

Z = (X - mean) / (standard deviation / sqrt(n))

where X is the total hours (2000), mean is 100, standard deviation is 30, and n is the number of components.

Rearrange the formula to solve for n:

1.645 = (2000 - (100n)) / (30 / sqrt(n))

Solve for n to get the required number of components in stock.

This is a complex statistics problem that requires knowledge of z-scores and the normal distribution to solve. Given that this requires more complex calculations and potentially the use of calculators or software, a more detailed step-by-step calculation may go beyond the scope of this platform.

Final answer:

To ensure the probability of continuous operation for the next 2000 hours is at least 0.95, at least one component must be in stock.

Explanation:

To calculate the number of components that must be in stock to ensure the probability of continuous operation for the next 2000 hours is at least 0.95, we need to use the concept of the normal distribution. Given that the mean lifetime of the component is 100 hours and the standard deviation is 30 hours, we can calculate the Z-score for 2000 hours.

Z-score = (X - mean) / standard deviation = (2000 - 100) / 30 = 63.33

Using a Z-table or calculator, we can find that the probability of the component lasting for 2000 hours or more is approximately 1. With a probability of 1, we can say that the system is guaranteed to be in continual operation for the next 2000 hours. Therefore, at least one component must be in stock.

Answer: $178.70

Step-by-step explanation: $125.36 + $153.53 = $278.89

$278.89 - $100.19 = $178.70

Answer:

Total purchase price of of the home is $160000

Step-by-step explanation:

We know that

Percent of down payment June Elloy makes = 22 %

Down payment is = $35200

Let x be the total purchase price.

Therefore,

22 % of x = 35200

( 22 / 100 ) times x = 35200

0.22 times x =35200

x = 35200 / 0.22

  = 160000

Thus, total purchase price of of the home is $160000

The unknown number in the problem is 150. This was determined by creating an equation based on the information given, simplifying it, and solving for the unknown variable 'n'.

This is a math problem that involves working with percentages and algebraic equations. Let the unknown number we're trying to find be 'n'. The problem says that 50% of the number, which is 0.5*n, is added to the number (n), resulting in 225. This situation can be represented with the equation: n + 0.5*n = 225.

Combining like terms on the left side of the equation, we get 1.5*n = 225. To isolate 'n', divide both sides of the equation by 1.5. So, n = 225 / 1.5, which gives us the number 150. Therefore, the unknown number is 150.

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After rounding off the given number 0.125 to the nearest tenth it becomes 0.1.

We know that,

When you round a number to a certain digit, you have to check the value of the digit before the place you want to round off to.

1. If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up.

2. If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down.

The given number is:

0.125

To round off this digit nearest tenth,

See the number at the hundredth place, which is 2.

And 2 < 5  

Therefore,

0.125 is rounded down to 0.1

So after rounding 0.125 number becomes 0.1.

Hence,

The rounded-off number nearest tenth is 0.1.

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Final answer:

The area of the base is increasing at a rate of 200π m²/min after 5 minutes.

Explanation:

To find how fast the area of the base is increasing, we first need to calculate the rate at which the height and radius of the cone are changing. Given that the height of the pile is always twice the length of the base diameter, we can write the equation h = 2r, where h is the height and r is the radius of the cone. Differentiating both sides with respect to time, we get dh/dt = 2dr/dt. Since the height is changing at a constant rate of 20 m³/min, dh/dt = 20 m³/min. Using this information, we can find the rate at which the radius is changing, dr/dt. Since the base diameter is equal to 2r, the base radius changes at a rate of dr/dt = 10 m/min. Now, we can find the rate of change of the area A of the base using the formula A = πr². Differentiating both sides with respect to time, we get dA/dt = 2πr(dr/dt). Plugging in the values, we have dA/dt = 2π(10)(10) = 200π m²/min. Therefore, the area of the base is increasing at a rate of 200π m²/min after 5 minutes.

The probability of rolling a sum greater than 9 (Event A) on two fair dice is 1/6, and the probability of rolling a sum divisible by 5 or 6 (Event B) is 1/3.

To compute the probability of Event A, where the sum is greater than 9, we first identify the possible combinations that yield a sum greater than 9: (4,6), (5,5), (5,6), (6,4), (6,5), and (6,6). Since there are 6 possible outcomes and 36 total possible outcomes when rolling two dice (6 for the first die × 6 for the second die), the probability is 6/36 or 1/6.

For Event B, we look for sums divisible by 5 or 6. These sums are 5, 6, 10, 12, 15, 18, and 20. The combinations yielding these sums are (1,4), (2,3), (2,6), (3,2), (3,6), (4,1), (4,6), (5,1), (5,5), (6,2), (6,3), (6,6). By counting these results, we have 12 outcomes that meet the Event B criteria out of 36, so the probability is 12/36 or 1/3.

The probability of Event A, the sum being greater than 9, is 1/6, and the probability of Event B, the sum being divisible by 5 or 6, is 1/3.

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Applying it's concept, it is found that the rate of change for the price of a pie from 2006 to 2010 is of $2 a year.

It is given by the change in the output divided by the change in the input.

In this problem, in 4 years, from 2006 to 2010, the price increased $8, from $9.50 to $17.50. Hence, the rate of change is given by:

r = 8/4 = 2.

Thus, the rate of change for the price of a pie from 2006 to 2010 is of $2 a year.

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Final answer:

If a chi-square test produces a chi-square statistic near zero, it implies that there is minimal deviation between the observed and expected values, suggesting independence or a good fit. However, interpretation should consider the specific test and hypotheses.

Explanation:

If a chi-square test produces a chi-square statistic near zero, it implies that there is minimal deviation between the observed and expected values. This suggests that the variables being tested are independent of each other or that the data fits the assumed distribution very well.

For example, in a chi-square test of independence, if the chi-square statistic is close to zero, it indicates that there is no significant relationship between the two factors being tested.

However, it is important to interpret the results in the context of the specific test being conducted, considering the null and alternative hypotheses as well as the significance level.