The label on a ceiling lighting fixture warns you to use a lightbulb of 60 watts or less. The voltage to the lightbulb is 120 vo
1 answer:
You might be interested in
<h2>Answer:</h2>
Ques 1)
Ques 2)
Ques 1)
We know that if a graph is stretched by a factor of a then the transformation if given by:
f(x) → a f(x)
Also, we know that the translation of a function k units to the right or to the left is given by:
f(x) → f(x+k)
where if k>0 then the shift is k units to the left
and if k<0 then the shift is k units to the right.
Here the graph of f(x) is transformed into the graph of g(x) by a vertical stretch of 4 units and a translation of 4 units right.
This means that the function g(x) is given by:
Ques 2)
We know that the transformation of the type:
f(x) → f(x)+k
is a shift or translation of the function k units up or down depending on k.
If k>0 then the shift is k units up.
and if k<0 then the shift is k units down.
Here, The graph of the function f(x)=|3x| is translated 4 units up.
This means that the transformed function g(x) is given by:
Answer:
a) 2.84% probability that he is late for his first lecture.
b) 5.112 days
Step-by-step explanation:
When the distribution is normal, we use 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.
In this question, we have that:
a. Find the probability that he is late for his first lecture.
This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So
has a pvalue of 0.9716
1 - 0.9716 = 0.0284
2.84% probability that he is late for his first lecture.
b. Find the number of days per year he is likely to be late for his first lecture.
Each day, 2.84% probability that he is late for his first lecture.
Out of 180
0.0284*180 = 5.112 days