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Lina20 [59]
3 years ago
15

The first Africans to be brought to British North America landed in which colony in the early 1600s?

Mathematics
2 answers:
madam [21]3 years ago
5 0
What  are the answer choices??????
Shtirlitz [24]3 years ago
3 0
Jamestown, Virginia, if I'm not mistaken.
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Help it gives me no numbers!!!
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Answer:

area= 15 u

Step-by-step explanation:

A=bh

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Match each description when z = 9 + 3i. 1. Real part of z 2. Imaginary part of z 3. Complex conjugate of z 4. 3i - z 5. z - 9 6.
sergey [27]
1)\quad \Re(9+3i)=9\qquad\text{F}\\\\
2)\quad \Im(9+3i)=3\qquad\text{E}\\\\
3)\quad \overline{(9+3i)}=9-3i\qquad\text{D}\\\\
4)\quad 3i-(9+3i)=3i-9-3i=-9\qquad\text{A}\\\\
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3 years ago
Four friends purchase a pineapple for $2.89 and 18.4 pounds of peaches. The peaches cost $1.75 per pound. The friends share the
umka21 [38]

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Step-by-step explanation:

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2 years ago
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Determining probability of events. please help! ​
Aneli [31]
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6 0
3 years ago
The Pew Research Center reported that 73% of Americans who own a cell phone also use text messaging. In a recent local survey, 1
AnnZ [28]

Answer:

z=\frac{0.775 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=1.433  

Now we can find the p value. Since we have a bilateral test the p value would be:  

p_v =2*P(z>1.433)=0.152  

Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

Do Not reject H0

Step-by-step explanation:

Information provided

n=200 represent the sample size slected

X=155 represent the cell phone owners used text messaging

\hat p=\frac{155}{200}=0.775 estimated proportion of cell phone owners used text messaging

p_o=0.73 is the value to verify

\alpha=0.1 represent the significance level

We need to conduct a z test for a proportion

z would represent the statistic

p_v represent the p value

System of hypothesis

We want to verify if the true proportion of cell phone owners used text messaging is different from 0.73 so then the system of hypothesis are:

Null hypothesis:p=0.73  

Alternative hypothesis:p \neq 0.73  

The statistic to check this hypothesis is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the data given we got:

z=\frac{0.775 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=1.433  

Now we can find the p value. Since we have a bilateral test the p value would be:  

p_v =2*P(z>1.433)=0.152  

Since the p value is higher than the significance level of 0.1 we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

Do Not reject H0

6 0
3 years ago
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