The resonance frequency f in an electronic circuit containing inductance L and capacitance C in series is given by f=1/2π√LC. Determine the inductance L in an electric circuit if the resonance frequency is 6.2 and the capacitance is 0.0001. Round your answer to the nearest tenth.

10270.1
6.6
256.8
0

Answer:

Option C. 140

Step-by-step explanation:

Base of a pyramid which has four triangular sides is four sided.

If pyramid contains seven layers of oranges then the layer at the bottom will have 7 oranges.

Since base is a square then other side of the square base will have the same number of oranges = 7.

Therefore number of oranges at the first layer will be = 7×7

Similarly other layers above this layer will have the same sequence

6×6, 5×5, 4×4, 3×3, 2×2, 1×1

Therefore total number of oranges = 7×7 + 6×6 + 5×5 + 4×4 + 3×3 + 2×2 + 1×1 = 49 + 36 + 25 + 16 + 9 + 4 + 1 = 140

Option C. 140 is the answer.

A grocer stacks oranges in a four-sided pyramid that is 7 layers high. Many oranges are in the pile are C.140

The orange is the fruit of the citrus species Citrus and sinensis in the family Rutaceae, native to China. Orange is also called sweet orange, to distinguish it from the related Citrus and aurantium, referred to as bitter orange.

The proper name of four-sided pyramid is a "tetrahedron". The tetrahedron has the extra interesting property of having all four triangular sides congruent

1st Layer: The 1st layer contains only 1 orange so it has only 1 orange

2nd layer: The 2nd layer has total 4 oranges. Calculation:   [tex]2*2=4[/tex]

3rd layer: The 3rd layer has total 9 oranges. Calculation: [tex]3 * 3 = 9[/tex]

4th Layer: The 4th layer has total 16 oranges. Calculation: [tex]4 * 4 = 16[/tex].

5th layer: The 5th layer has total 25 oranges. Calculation:  [tex]5 * 5 = 25[/tex]

6th layer: The 6th layer has total 36 oranges. Calculation: [tex]6 * 6 = 36[/tex]

7th layer: The 7th layer has total 49 oranges. Calculation: [tex]7 * 7 = 49[/tex].

Sum of all layers:

[tex]1+4+9+16+25+36+49 = 140[/tex]

Grade:  9

Subject:  mathematics

Chapter:  pyramid

Keywords:  oranges, pyramid, four-sided, layers, pile

Answer:

The answer is A. The events are independent because P(red) • P(green) = P(red and green).

Answer:

A. The events are independent because P(red) • P(green) = P(red and green).

Step-by-step explanation:

EDGE 2021 AQR

Answer:

the answer is 675

Step-by-step explanation:

1. there is a tank with 1000 gallons, and we have 65 cans with a capacity of 5 gallons, so if we multiply the 65 cans with the capacity we would have as result

  65 x 5 = 325

2-  already having the value of gallons used (325) we can subtract this difference to the 1000 gallons that were in the tank at first.

 1000 - 325 = 675

Solution: The test statistic under the null hypothesis is:

[tex]t=\frac{\bar{x_{1}}-\bar{x_{2}}}{s_{p}\sqrt{(\frac{1}{n_{1}})+(\frac{1}{n_{2}})}}[/tex]

Where:

[tex]s_{p} = \sqrt{\frac{(n_{1}-1)s^{2}_{1}+(n_{2}-1)s^{2}_{2}}{n_{1}+n_{2}-2} }[/tex]

[tex]s_{p} = \sqrt{\frac{(16-1)14.8^{2}+(18-1)16^{2}}{16+18-2} }[/tex]

               [tex]=15.45[/tex]

[tex]\therefore t=\frac{85.4-74.9}{15.45\sqrt{(\frac{1}{16})+(\frac{1}{18})}}[/tex]

                   [tex]=\frac{10.5}{5.31}[/tex]

                   [tex]=1.98[/tex]

Now, to find the critical values, we need to use the t distribution table at 0.01 significance level for [tex]df=n_{1} + n_{2} -2 = 16+18-2=32[/tex] and is given below:

[tex]t_{critical}=-2.738,2.738[/tex]

Since the t statistic is less than the  t critical value, we therefore fail to reject the null hypothesis and conclude that the two population means are equal.

Answer:

Step-by-step explanation:

How do do a 3/4 as a fraction, decimal, and percent?

1) 3/4 is just a fraction

2) 3:4 = 0.75 (0.75 = 75%)

3) 75%

1. The dataset with the smallest IQR is the Mid-term exams. 2. The exam median is only 1–2 points higher than the class median. 3. The standard deviation is the best measurement of spread for games and movies.

1. The dataset with the smallest IQR is the Mid-term exams. To find the IQR, first, calculate the first quartile (Q1) and the third quartile (Q3). Then, find the difference between Q3 and Q1. By comparing the IQR values for the Mid-term and Final exams, it can be determined that the Mid-term exams have the smallest IQR.

2. The exam median is only 1–2 points higher than the class median. The box plot's median represents the middle value of the dataset. By comparing the medians of the exam and class data, it can be determined that the exam median is only 1–2 points higher than the class median.

3. The standard deviation is the best measurement of spread for games and movies. While the IQR can measure spread, the standard deviation is a more precise measurement. Comparing the spread of the data, the standard deviation is the best measurement for both games and movies.

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Various statistical measures such as IQR, Standard Deviation, Median and Mean were used to interpret the given data in numerous scenarios. Outliers were identified and impacts on datasets were evaluated.

To answer these questions, we need to understand few key statistical terms: 'Mean' is the simple average of data, 'Median' is the middle score of data,  'IQR' (Inter Quartile Range) is the difference between the upper quartile (Q3) and the lower quartile (Q1), which helps in understanding the spread and '.'Standard Deviation' measures the absolute variability of a dataset.

Question 1: IQR for the mid-term exams is Q3 (92) - Q1 (82) = 10. IQR for the final exams is Q3 (85) - Q1 (78) = 7. So, the final exams have the smallest IQR.

Question 2: The boxes showing median indicates that the exam median is only 1–2 points higher than the class median.

Question 3: Since the spread of the data at basketball games and local movie theaters demonstrates varied distributions, both the IQR and the standard deviation should be used to evaluate the data spread.

Question 4: Box plots do not provide direct information on the mean. So, there is no way of telling what the means are.

Question 5: For the video hours, we see that values 18, 21, and 26 lie far from the main part of the data, thus they can be considered as possible outliers.

Question 6: The male data set shows a high data value, which causes the data set to be asymmetrical. This could affect statistical measures like the mean and standard deviation.

Question 7: The spread of the college data, having a large standard deviation and IQR, is best described by the standard deviation. The high school data, with smaller numbers, is best described by the IQR.

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Answer:

D.3x(x-2)

Step-by-step explanation:

i already did this

a) What proportions of passes were completed?

b) How does this compare with what “should have happened” theoretically?"

First, take 100 numbers, i.e. from 00 to 99, and set that the first 54% are completed passes, therefore from 00 to 53, and the remaining 46% are not completed passes, therefore from 54 to 99.

Now, divide the random digits until you compose 20 numbers: 98, 72, 61, 09, 83, 56, 23, 94, 20, 42, 76, 52, 06, 82, 76, 58, 23, 94, 87, 29.  

Then, divide the numbers into the two set categories: 

completed: 09, 23, 20, 42, 52, 06, 23, 94, 87 = 9 outcomes
not completed: 98, 72, 61, 83, 56, 94, 76, 82, 76, 58, 29 = 11 outcomes

A) Therefore, from the random digits you get 9/20 = 45% of completed passes and 11/20 = 55% of not completed passes.

B) Using the random digits, the simulation gave an outcome exactly opposite to what we expected.

Answer:

y-axis... just look at the graph on desmos.com

The nearest tenth, the climber's distance to the horizon is approximately 156.6 miles.

The climber's distance to the horizon can be calculated using the Pythagorean theorem, where the Earth's radius and the climber's height above sea level form a right-angled triangle.

[tex]\[ d^2 + R^2 = (R + h)^2 \][/tex]

 Expanding the right side of the equation gives us:

[tex]\[ d^2 + R^2 = R^2 + 2Rh + h^2 \][/tex]

 Subtracting [tex]\( R^2 \)[/tex] from both sides, we get:

[tex]\[ d^2 = 2Rh + h^2 \][/tex]

 Since [tex]\( h^2 \)[/tex] is very small compared to [tex]\( 2Rh \)[/tex] (because [tex]\( h \)[/tex] is much smaller than [tex]\( R \))[/tex], we can neglect [tex]\( R^2 \)[/tex] in our calculation for a more practical approximation. This leaves us with:

[tex]\[ d^2 \approx 2Rh \][/tex]

 Taking the square root of both sides to solve for [tex]\( d \)[/tex], we have:

[tex]\[ d \approx \sqrt{2Rh} \][/tex]

Now, plugging in the values for [tex]\( R \) and \( h \):[/tex]

[tex]\[ d \approx \sqrt{2 \times 3959 \times 3.1} \][/tex]

[tex]\[ d \approx \sqrt{2 \times 3959 \times 3.1} \][/tex]

[tex]\[ d \approx \sqrt{24516.8} \][/tex]

[tex]\[ d \approx 156.6 \text{ miles} \][/tex]

Rounding to the nearest tenth, the climber's distance to the horizon is approximately 156.6 miles.

By calculating the probability of choosing a counter at random, we find that we are least likely to choose a white counter as it has the least probability.

Probability is simply how likely something is to happen.

Probability of choosing a red counter  = 6/25 = 0.24

Probability of choosing a white counter  = 4/25 = 0.16

Probability of choosing a blue counter  = 7/25 = 0.28

Probability of choosing a yellow counter  = 8/25 = 0.32

Learn more about Probability here

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