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otez555 [7]
3 years ago
9

The stalemate in the western front began after Germany failed to quickly defeat

Mathematics
2 answers:
Andrei [34K]3 years ago
5 0

Answer:

B) France

Step-by-step explanation:

The stalemate in the western front began after Germany failed to quickly defeat 'France'

Vlad1618 [11]3 years ago
4 0

Germany failed to defeat France quickly on the western front in World War I

Germany began to offer a ceasefire and was asked to disarm. The losing Germans were then forced to accept heavy conditions, including paying repairs totaling $ 37 billion (nearly $ 492 billion in current dollars).

<h2>Further Explanation </h2>

The war ended with a truce, an agreement in which both parties agreed to stop the battle, rather than surrender. In November 1918, both the Allied Bloc and the Central Bloc, which had been attacking each other for four years, had run out of supply. The German attack that year had been crushed with many casualties, and in the late summer and autumn, British, French and US forces forced them steadily to retreat.

The Germans agreed to withdraw their troops out of France, Belgium, and Luxembourg within 15 days, or risk becoming Allied captives. They had to surrender their weapons, including 5,000 artillery pieces, 25,000 machine guns, and 1,700 airplanes, along with 5,000 train locomotives, 5,000 trucks, and 150,000 carts.

Learn More

World War I  brainly.com/question/13688955

Battle of the Western Front  brainly.com/question/13688955

Details

Grade: Middle School

Subject: Mathematics

Keyword: World War I, western front, Germany

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4 years ago
A random sample of 21 observations is used to estimate the population mean. The sample mean and the sample standard deviation ar
stepladder [879]

Answer:

a. CI=[128.79,146.41]

b. CI=[122.81,152.39]

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Step-by-step explanation:

a. -Given the sample mean is 137.6 and the standard deviation is 20.60.

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\bar X\pm \ z\frac{s}{\sqrt{n}},

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we then calculate our confidence interval as:

\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.05/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm1.960\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]

Hence, the 95% confidence interval is between 128.79 and 146.41

b. -Given the sample mean is 137.6 and the standard deviation is 20.60.

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\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.01/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm3.291\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]

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c. -Increasing the confidence has an increasing effect on the margin of error.

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Answer:

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