Mia enlarged a plan for an outdoor stage. The original plan is shown below. She dilated the outdoor stage by a scale factor of 4
with the center of dilation at the origin. Which ordered pair will be the coordinates of one of the new vertices? A (2,1) B (8,16) C (32,4) D (32,16)
2 answers:
Answer:
D (32,16)
Step-by-step explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please see the attached photo.
In the photo, the four vertices are:
(-5,4) by a scale factor of 4 (-20, 16)
(8, 4) by a scale factor of 4 (32, 16)
(-6, -2) by a scale factor of 4 (-24, -8)
(8, -2) by a scale factor of 4 (31, -8)
So only D is correct.
Hope it will find you well.
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Answer:
The probability that the whole package is uppgraded in less then 12 minutes is 0,1271
Step-by-step explanation:
The mean distribution for the length of the installation (in seconds) of the programs will be denoted by X. Using the Central Limit Theorem, we can assume that X is normal (it will be pretty close). The mean of X is 15 and the variance is 15, hence, the standard deviation is √15 = 3.873.
We want to find the probability that the full installation process takes less than 12 minutes = 720 seconds. Then, in average, each program should take less than 720/68 = 10.5882 seconds to install. Hence, we want to find the probability of X being less than 10.5882. For that, we will take W, the standariation of X, given by the following formula
We will work with , the cummulative distribution function of the standard Normal variable W. The values of
can be found in the attached file.
Since the density function of a standard normal random variable is symmetrical, then
Therefore, the probability that the whole package is uppgraded in less then 12 minutes is 0,1271.
Answer:
Confidence limit = [52.8%, 75.2%]
Step-by-step explanation:
±
where the value will be taken from the z-table for 95% confidence interval
1-0.95= 0.05/2= 0.025
0.95+0.025= 0.0975
From the z-table the value of corresponding to 0.0975 is 1.96
±
±
±
% ±
%
so the confidence interval is
%
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