2 years ago

In the ______, President Lincoln declared all slaves who lived in states rebelling

1 answer:

7 0

In the Emancipation Proclamation, President Lincoln declared all slaves who lived in states rebelling
<span> against the government were free.</span>

You might be interested in

The people heading west called themselves emigrants, but we call them ____________________.

max2010maxim [7]

We now call them pioneers

5 0

3 years ago

Please help! history!!!!!

Alik [6]

A - The Articles of Confederation

8 0

2 years ago

You have researched how to become an American citizen. The first web site has great information. What do you need to do next? A.

Liula [17]

I believe that the best answer is C

3 0

2 years ago

Read 2 more answers

Who is credited for using his wealth to build up the steel industry?

Nostrana [21]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

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

2 years ago