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vazorg [7]
3 years ago
10

Please help i am give brainliest

Engineering
2 answers:
Rus_ich [418]3 years ago
8 0
The answer will be b
Korolek [52]3 years ago
6 0

Answer:

A C power is the answer

hope this helps

You might be interested in
The typical Canadian worker is able to produce 100 board feet (a unit of measure) of lumber or 1000 light bulbs per year. The wo
faltersainse [42]

Answer:

Explanation:

In the Canadian country, it is possible that the country is able to produce either 100 units of lumber or 1000 lightbulbs. Then, 100 lumber is equal to 1000 bulbs in the home, and hence, one lumber is equivalent to 10 bulbs. Therefore, the opportunity cost of lumber in the home is ten bulbs.

Thus, the country's production possibilities frontier PPF is a linear combination of 100 lumber and 1000 bulbs. The graphical diagram of the production possibilities frontier PPF can be seen in the image below.

6 0
3 years ago
6.3.3 Marks on an exam in a statistics course are assumed to be normally distributed
bekas [8.4K]

Answer:

- The calculated p-value (0.392452) is higher than the significance level at which the test was performed, hence, the null hypothesis is true and μ = 60

- 95% Confidence interval for the population mean score = (47.4, 84.1)

Explanation:

The sample of 4 students had scores of 52, 63, 64, 84.

First of, we need to compute the sample mean, we do not need the sample standard deviation as the population variance is given as 5

Mean = (Σx)/N

x = each variable

N = number of variables = 4

Mean = (52 + 63 + 64 + 84)/4

Mean = 65.75

Sample Standard deviation = σ

= √[Σ(x - xbar)²/N]

xbar = mean = 65.75

Σ(x - xbar)² = 532.75

σ = √[532.75/4] = 11.54

in hypothesis testing, the first thing is usually to state the null and alternative hypothesis.

From the question, the null hypothesis has already been stated as

H₀: μ = 60

The alternative hypothesis would then be that the population mean score isn't equal to 60

Hₐ: μ ≠ 60

Since the population distribution is normal and the sample standard deviation is to be used, we use the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 65.75

μ₀ = Standard to be compared against = 60

σₓ = standard error = (σ/√n) = (11.54/√4) = 5.77

t = (65.75 - 60)/5.77 = 0.9965 = 1.00

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 4 - 1 = 3

Significance level = 0.05 (95% confidence level)

The hypothesis test uses a two-tailed condition because we're testing in two directions.

p-value (for t = 1.00, at 0.05 significance level, df = 3, with a two tailed condition) = 0.392452

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.392452

0.392452 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there is enough evidence to conclude that the populatiom mean score is equal to 60.

b) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 65.75

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 4 - 1 = 3

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 3) = 3.18 (from the t-tables)

Standard error of the mean = 5.77

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 65.75 ± (3.18 × 5.77)

CI = 65.75 ± 18.3486

95% CI = (47.4014, 84.0986)

95% Confidence interval = (47.4, 84.1)

Hope this Helps!!!

7 0
3 years ago
The electricity generated by wind turbines annually in kilowatt-hours per year is given in a file. The amount of electricity is
Anika [276]

Answer:

Steps:

1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year

2.  load the data file for example, in matlab, use ('fileame.txt') to load the file

3. create variables from each column of your data

  for example, in matlab,  

     x=t{1}

     y=t{2}

4. plot the wind velocity and electricity generated.

   plot(x, y)

5. Label the individual axis and name the graph title.

    title('Graph of wind velocity vs approximate electricity generated for the year')

     xlabel('wind velocity')

     ylabel('approximate electricity generated for the year')

5 0
3 years ago
Read 2 more answers
The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg
kiruha [24]

Answer:

percentage change in volume  = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × 10^{11}  N/m²

bulk stress = 10^{7}  N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = \frac{bulk\ stress}{bulk\ strain}   ...............1

put here value and we get

3.5 × 10^{11} = \frac{10^7}{bulk\ strain}  

solve it we get

bulk strain = 2.8571 × 10^{-5}

and

bulk strain = \frac{change\ volume}{original\ volume}  

so that percentage change in volume is = 2.8571 × 10^{-5}  × 100

percentage change in volume  = 0.00285 %

6 0
3 years ago
A common process for increasing the moisture content of air is to bubble it through a column of water. The air bubbles are assum
likoan [24]

Answer:

Explanation:

Assumptions is that

1. The flow is an unsteady one

2. Bubbles diameter is constant

3. The bubble velocity is slow

4. There is no homogenous reaction

5. It has a one dimensional flux model along the radial direction

5 0
3 years ago
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