Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when 
. So
Rouding up
We need a sample size of at least 719
The answers are the following:
<span><span><span>P(A)=0.75</span><span>
</span></span><span><span>P(B|A)=0.9
</span></span><span><span>P(B|<span>A′</span>)=0.8
</span></span><span><span>P(C|A∩B)=0.8
</span></span><span><span>P(C|A∩<span>B′</span>)=0.6
</span></span><span><span>P(C|<span>A′</span>∩B)=0.7
</span></span><span><span>P(C|<span>A′</span>∩<span>B′</span>)=0.3</span></span></span>
805 tens
805x10=8050
805 tens is 8050
Answer:
slope of M'N' = 1
Explanation:
First, we will need to get the coordinates of points M' and N':
We are given that the dilation factor (k) is 0.8
Therefore:
For point M':
x coordinate of M' = k * x coordinate of M
x coordinate of M' = 0.8 * 2 = 1.6
y coordinate of M' = k * y coordinate of M
y coordinate of M' = 0.8 * 4 = 3.2
Therefore, coordinates of M' are (1.6 , 3.2)
For point N':
x coordinate of N' = k * x coordinate of N
x coordinate of N' = 0.8 * 3 = 2.4
y coordinate of N' = k * y coordinate of N
y coordinate of N' = 0.8 * 5 = 4
Therefore, coordinates of M' are (2.4 , 4)
Then, we can get the slope of M'N':
slope = (y2-y1) / (x2-x1)
For M'N':
slope = (3.2-4) / (1.6-2.4)
slope = 1
Hope this helps :)
