Which sequence of transformations proves that shape I is similar to shape II?
2 answers:
Answer:
B. a reflection across X axis and then a dilation by a scale factor of 1.5
Step-by-step explanation:
As clear from the graph all the coordinates of image 2 are 1.5 times of image 1 so shape 2 is dilated by the scale factor 1.5.
It is evident from the graph that lines AO and A"O"are parallel to each other so shape 2 is the reflection of shape 1.
Now we calculate the magnitude of line AB
m1 = (y2-y1)/(x2-x1)
=(-3+6)/(-4.5=9)
=3/4.5
=1/1.5
=2/3
Next we calculate magnitude m2 of A"B"
m2= (2-4) /(-3+6)
=(-2/3)
Then we know Tan(180-∅) = -tan∅
similarly if m1=(-m2)
then the one line having magnitude m2 is the rotated image through X axis by 180° of the line having magnitude m1.
So the answer is B.
You might be interested in
Answer:
54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches
Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.
Lesser than 207.9.
pvalue of Z when X = 207.9. So
By the Central Limit Theorem
has a pvalue of 0.2743
2*0.2743 = 0.5486
54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches