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3 years ago

Urgent!!! Which of the following was most crucial in bringing about US abandonment of neutrality during WWI?

History

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

It was because of Germany's use of submarines against neutral passenger and merchant ships. The unrestricted submarine warfare was agitating for the US, the last thing that provoked the Americans was the Zimmerman telegram.

Hope this helps.

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Gumawa ng sanaysay gamit ang mga matatalinghagang salitang nabangit sa alamat ng daragang magayon

Anna [14]

I’m confused...lol :)

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3 years ago

The tribes of central and southern Africa are based on which Customs European Asian Arabic or Africa

ad-work [718]

Its Africa bro cause I looked it up in google lol

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3 years ago

Was Columbus typical of explorers during that time or did other explorers have different or additional motives?

Inga [223]

He was a typical explorer. He would explore for three main reasons mother countries explored, gold, god and glory. To find wealth in spices and precious metals, to spread his catholic religion, and to find glory in his adventures.

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2 years ago

The text describes the March on Washington for Jobs and Freedom. According to the text, the event was designed to do what?

Nitella [24]

Answer:

To stand up for rights for African-American citizens.

Explanation:

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2 years ago

An hourly wage is normally distributed with a mean of $6.75 and a standard deviation of $0.55. What is the probability that an e

monitta

Answer:

Option A. 81.5%

Explanation:

Population mean, $\mu = 6.75$

Standard deviation, $\sigma = 0.55$

The question is to find $p(5.65 < \bar{x} < 7.30)$

The z value is given by the formula:

$z = \frac{\bar{x} - \mu}{\sigma}$

z - value corresponding to the mean, $\bar{x} = 5.65$ :

$z_1 = \frac{5.65 - 6.75}{0.55}\\z_1 =-2.0$

z - value corresponding to the mean, $\bar{x} = 7.30$ :

$z_1 = \frac{7.30 - 6.75}{0.55}\\z_1 =1.0$

$p(5.65\leq \bar{x} \leq 7.30) = p(-2.0 \leq z \leq 1.0)\\p(-2.0 \leq z \leq 1.0) = p(z\leq 1) - p(z\leq -2)\\ p(-2.0 \leq z \leq 1.0) = 0.84134 - 0.02275\\p(-2.0 \leq z \leq 1.0) = 0.81859 \\p(-2.0 \leq z \leq 1.0) = 0.82$

Probability that an employee’s hourly wage is between $5.65 and $7.30 = 81.5%

4 0

3 years ago